(i) $\tan 37^{\circ}=0.7536$ (ii) $\tan 42^{\circ} 18^{\prime}=0.9099$

### The mid-point of the segment AB, as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B.

Solution: As, \[point\text{ }A\text{ }lies\text{ }on\]\[x-axis\], we can assume the co-ordinates of \[point\text{ }A\text{ }to\text{ }be\text{ }\left( x,\text{ }0 \right).\] As, \[point\text{...

### Use tables to find cosine of: (v) 9° 23’ + 15° 54’

$(\mathrm{v}) \cos \left(9^{\circ} 23^{\circ}+15^{\circ} 54^{\circ}\right)=\cos 24^{\circ} 77^{\circ}=\cos 25^{\circ} 17^{\circ}=\cos \left(25^{\circ}...

### Use tables to find cosine of: (iii) 26° 32’ (iv) 65° 41’

(iii) $\cos 26^{\circ} 32^{\prime}=\cos \left(26^{\circ} 30^{\prime}+2^{\prime}\right)=0.8949-0.0003=0.8946$ (iv) $\cos 65^{\circ} 41^{\prime}=\cos \left(65^{\circ}...

### In what ratio is the line joining A (0, 3) and B (4, -1) divided by the x-axis? Write the co-ordinates of the point where AB intersects the x-axis.

Suppose, that the line segment \[AB\]intersects the \[x-axis\]by point \[P\text{ }\left( x,\text{ }0 \right)\]in the ratio \[k:\text{ }1.\] \[0\text{ }=\text{ }\left( -k\text{ }+\text{ }3...

### Use tables to find cosine of: (i) 2° 4’ (ii) 8° 12’

(i) $\cos 2^{\circ} 4^{\prime}=0.9994-0.0001=0.9993$ (ii) $\cos 8^{\circ} 12^{\prime}=\cos 0.9898$

### Use tables to find sine of: (v) 10° 20′ + 20° 45′

(v) $\sin \left(10^{\circ} 20^{\prime}+20^{\circ} 45^{\prime}\right)=\sin 30^{\circ} 65^{\prime}=\sin 31^{\circ} 5^{\prime}=0.5150+0.0012=0.5162$

### A line segment joining A (-1, 5/3) and B (a, 5) is divided in the ratio 1: 3 at P, point where the line segment AB intersects the y-axis. (i) Calculate the value of ‘a’. (ii) Calculate the co-ordinates of ‘P’.

As, the line segment \[AB\] intersects the \[y-axis\text{ }at\text{ }point\text{ }P\], suppose the co-ordinates of \[point\text{ }P\] be \[\left( 0,\text{ }y \right).\] And, \[P\text{...

### Use tables to find sine of: (iii) 47° 32′ (iv) 62° 57′

(iii) $\sin 47^{\circ} 32^{\prime}=\sin \left(47^{\circ} 30^{\prime}+2^{\prime}\right)=0.7373+0.0004=0.7377$ (iv) $\sin 62^{\circ} 57^{\prime}=\sin \left(62^{\circ}...

### Use tables to find sine of: (i) 21° (ii) 34° 42′

(i) $\sin 21^{\circ}=0.3584$ (ii) $\sin 34^{\circ} 42^{\prime}=0.5693$

### Find the co-ordinates of points of trisection of the line segment joining the point (6, -9) and the origin.

Suppose \[P\text{ }and\text{ }Q\] to be the points of trisection of the line segment joining \[A\text{ }\left( 6,\text{ }-9 \right)\text{ }and\text{ }B\text{ }\left( 0,\text{ }0 \right).\] So,...

### A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP: PB = 3: 5 and AQ: QC = 3: 5. Show that: PQ = 3/8 BC.

According to the given question, Point \[P\text{ }lies\text{ }on\text{ }AB\] such that \[AP:\text{ }PB\text{ }=\text{ }3:\text{ }5.\] So, the co-ordinates of point P are \[=\text{ }\left(...

### A triangle ABC is right angled at B; find the value of (sec A. cosec C – tan A. cot C)/ sin B

Since, ABC is a right angled triangle right angled at B Therefore, \[\begin{array}{*{35}{l}} A\text{ }+\text{ }C\text{ }=\text{ }{{90}^{o}} \\ \left( sec\text{ }A.\text{ }cosec\text{ }C\text{...

### A (20, 0) and B (10, -20) are two fixed points. Find the co-ordinates of a point P in AB such that: 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.

According to the given question, \[3PB\text{ }=\text{ }AB\] So, \[AB/PB\text{ }=\text{ }3/1\] \[\left( AB\text{ }-\text{ }PB \right)/\text{ }PB\text{ }=\text{ }\left( 3\text{ }-\text{ }1...

### Evaluate: (ix)

(ix) \[14\text{ }sin\text{ }{{30}^{o}}~+\text{ }6\text{ }cos\text{ }{{60}^{o}}~-\text{ }5\text{ }tan\text{ }{{45}^{o}}\] \[=\text{ }14\text{ }\left( 1/2 \right)\text{ }+\text{ }6\text{ }\left( 1/2...

### Given a triangle ABC in which A = (4, -4), B = (0, 5) and C = (5, 10). A point P lies on BC such that BP: PC = 3: 2. Find the length of line segment AP.

According to the given question, \[BP:\text{ }PC\text{ }=\text{ }3:\text{ }2\] Then by section formula, the co-ordinates of \[point\text{ }P\]are: \[=\text{ }\left( 15/5,\text{ }40/5 \right)\]...

### Evaluate: (vii) (viii)

(vii) = 1 – 2 = -1 (viii)

### Evaluate: (v) (vi)

\[\begin{array}{*{35}{l}} v)\text{ }cosec\text{ }\left( {{65}^{o}}~+\text{ }A \right)\text{ }-\text{ }sec\text{ }\left( {{25}^{o}}~-\text{ }A \right) \\ =\text{ }cosec\text{ }\left[...

### Evaluate: (iii) (iv)

\[\begin{array}{*{35}{l}} \left( iii \right)\text{ }sin\text{ }{{80}^{o}}/\text{ }cos\text{ }{{10}^{o}}~+\text{ }sin\text{ }{{59}^{o}}~sec\text{ }{{31}^{o}} \\ =\text{ }sin\text{ }{{\left( 90\text{...

### Evaluate: (i) (ii) cosec

(i) (ii) \[\begin{array}{*{35}{l}} 3\text{ }cos\text{ }{{80}^{o}}~cosec\text{ }{{10}^{o}}~+\text{ }2\text{ }cos\text{ }{{59}^{o}}~cosec\text{ }{{31}^{o}} \\ =\text{ }3\text{ }cos\text{ }{{\left(...

### For triangle ABC, show that: (i) sin (A + B)/ 2 = cos C/2 (ii) tan (B + C)/ 2 = cot A/2

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \quad$ [Angle sum property of a triangle] $(\angle \mathrm{A}+\angle \mathrm{B}) / 2=90^{\circ}-\angle \mathrm{C} / 2$ $\sin ((A+B)...

### Show that: (i) (ii)

\[\begin{array}{*{35}{l}} =\text{ }sin\text{ }A\text{ }cos\text{ }A\text{ }\text{ }si{{n}^{3}}~A\text{ }cos\text{ }A\text{ }\text{ }co{{s}^{3}}~A\text{ }sin\text{ }A \\ =\text{ }sin\text{ }A\text{...

### Express each of the following in terms of angles between 0°and 45°: (iii) cos 74°+ sec 67°

\[\begin{array}{*{35}{l}} \left( iii \right)\text{ }cos\text{ }74{}^\circ +\text{ }sec\text{ }67{}^\circ \\ =\text{ }cos~\left( 90\text{ }-\text{ }16 \right){}^\circ +\text{ }sec\text{ }\left(...

### Express each of the following in terms of angles between 0°and 45°: (i) sin 59°+ tan 63° (ii) cosec 68°+ cot 72°

\[\begin{array}{*{35}{l}} \left( i \right)\text{ }sin\text{ }59{}^\circ +\text{ }tan\text{ }63{}^\circ \\ =\text{ }sin\text{ }\left( 90\text{ }-\text{ }31 \right){}^\circ +\text{ }tan\text{...

### Show that: (iii) sin26/ sec64 + cos 26/ cosec 64 = 1

### Show that: (i) tan10 tan15 tan75 tan80 = 1 (ii) sin42 sec48 + cos42 cosec48 = 2

(i) \[sin\text{ }{{42}^{o}}~sec\text{ }{{48}^{o}}~+\text{ }cos\text{ }{{42}^{o}}~cosec\text{ }{{48}^{o}}\] \[~=\text{ }sin\text{ }{{42}^{o}}~sec\text{ }({{90}^{o}}~-\text{ }{{42}^{o}})\text{...

### One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1).

We know that, The mid-point of any diameter of a circle is it’s centre. Suppose assume the required co-ordinates of the other end of mid-point to be \[\left( x,\text{ }y \right).\] \[2\text{...

### Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.

Solution: According to the given question, \[AB\text{ }=\text{ }BC\text{ }=\text{ }CD\] Thus, \[B\]is the mid-point of \[AC.\] Suppose the co-ordinates of point \[A\text{ }be\text{ }\left( x,\text{...

### (-5, 2), (3, -6) and (7, 4) are the vertices of a triangle. Find the length of its median through the vertex (3, -6).

Suppose \[A\text{ }\left( -5,\text{ }2 \right),\text{ }B\text{ }\left( 3,\text{ }-6 \right)\text{ }and\text{ }C\text{ }\left( 7,\text{ }4 \right)\] be the vertices of the given, triangle. And...

### In the given figure, P (4, 2) is mid-point of line segment AB. Find the co-ordinates of A and B.

Solution: Suppose point A lies on x-axis, hence its co-ordinates can be \[\left( x,\text{ }0 \right)\] And, Point B lies on y-axis, hence its co-ordinates can be \[\left( 0,\text{ }y \right)\]...

### P (-3, 2) is the mid-point of line segment AB as shown in the given figure. Find the co-ordinates of points A and B.

We know that, Point A lies on y-axis, so its co-ordinates are \[\left( 0,\text{ }y \right).\] Point B lies on x-axis, so its co-ordinates are taken to be \[\left( x,\text{ }0 \right).\] \[P\text{...

### Given M is the mid-point of AB, find the co-ordinates of: (i) A; if M = (1, 7) and B = (-5, 10) (ii) B; if A = (3, -1) and M = (-1, 3).

(i) Suppose that the co-ordinates of \[A\text{ }to\text{ }be\text{ }\left( x,\text{ }y \right).\] Thus, \[\left( 1,\text{ }7 \right)\text{ }=\text{ }\left( x-5/2,\text{ }y+10/2 \right)\] \[1\text{...

### A (5, 3), B (-1, 1) and C (7, -3) are the vertices of triangle ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = ½ BC.

According to the given question, \[L\text{ }is\text{ }the\text{ }mid-point\text{ }of\text{ }AB\] And \[M\text{ }is\text{ }the\text{ }mid-point\text{ }of\text{ }AC.\] Co-ordinates of L are,...

### Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.

According to the According to the given question, question, \[Mid-point\text{ }of\text{ }AB\text{ }=\text{ }\left( 2,\text{ }3 \right)\] Thus, \[\left( 3+x/2,\text{ }5+y/2 \right)\text{ }=\text{...

### Find the mid-point of the line segment joining the points: (i) (-6, 7) and (3, 5) (ii) (5, -3) and (-1, 7)

(i) Suppose \[A\text{ }\left( -6,\text{ }7 \right)\text{ }and\text{ }B\text{ }\left( 3,\text{ }5 \right)\] Thus, the mid-point of \[AB\text{ }=\text{ }\left( -6+3/2,\text{ }7+5/2 \right)\text{...

### Use ruler and compasses only for this question. (i) Construct the locus of points inside the triangle which are equidistant from B and C. (ii) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.

Steps of construction: Draw line \[BC\text{ }=\text{ }6\text{ }cm\] and construct angle \[CBX\text{ }=\text{ }{{60}^{o}}\]. Cut off \[AB\text{ }=\text{ }3.5\]. Join \[AC\], triangle \[ABC\]is the...

### Use ruler and compasses only for this question. (i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60degree (ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.

Steps of construction: (i) Draw line \[BC\text{ }=\text{ }6\text{ }cm\] and construct angle \[CBX\text{ }=\text{ }{{60}^{o}}\]. Cut off \[AB\text{ }=\text{ }3.5\]. Join \[AC\], triangle \[ABC\]is...

### In triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that: i) point A is equidistant from all the three sides of the triangle. ii) AM bisects angle LMN.

Construction: \[Join\text{ }AM\] Proof: (i) Since,\[A\] lies on bisector of \[\angle N\] So, \[A\]is equidistant from \[MN\text{ }and\text{ }LN.\] Now again, as \[A\] lies on the bisector...

### In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.

Construction: From \[P\], draw \[PL\bot AB~and~PM\bot BC\] Proof: In \[\vartriangle PLB\text{ }and\text{ }\vartriangle PMB\], \[\angle PLB\text{ }=\angle PMB\text{ }[Each\text{ }{{90}^{o}}]\]...

### Construct a triangle ABC in which angle ABC = 75o, AB = 5cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P; prove that P is equidistant from B and C; and also from AC and BC.

Steps of Construction: i) Draw a line segment \[BC\text{ }=\text{ }6.4\text{ }cm\] ii) At\[B\], draw a ray \[BX\]making an angle of \[{{75}^{o}}\] with \[BC\]and cut off \[BA\text{ }=\text{ }5\text{...

### Construct a right angled triangle PQR, in which ∠Q = 90degree, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meet PR at point T. Prove that T is equidistant from PQ and QR.

Steps of Construction: i) Draw a line segment \[QR\text{ }=\text{ }4.5\text{ }cm\] ii) At \[Q,\]draw a ray \[QX\]of an angle of \[{{90}^{o}}\] iii) With\[centre\text{ }R\text{ }and\text{...

### In each of the given figures: PA = PB and QA = QB Prove in each case, that PQ (produced, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.

(i) (ii) SOLUTION: Construction: Join \[PQ\]which meets \[AB\text{ }in\text{ }D.\] Proof: \[As\text{ }P\text{ }is\text{ }equidistant\text{ }from\text{ }A\text{ }and\text{ }B.\] So, \[P\]lies on the...

### Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.

Given: In triangle ABC, \[AB\text{ }=\text{ }4.2\text{ }cm,\text{ }BC\text{ }=\text{ }6.3\text{ }cm\text{ }and\text{ }AC\text{ }=\text{ }5\text{ }cm\] Steps of Construction: i) Draw a line segment...

### ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

We assume that \[ABCD\]be the given cyclic quadrilateral. \[PA\text{ }=\text{ }PD\text{ }\left[ Given \right]\] So, \[\angle PAD\text{ }=\angle PDA\text{ }\ldots \ldots \text{ }\left( 1 \right)\]...

### Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.

Solution: Join \[AC,\text{ }PQ\text{ }and\text{ }BD.\] As \[ACQP\]is a cyclic quadrilateral \[\angle CAP\text{ }+\angle PQC\text{ }=\text{ }{{180}^{o}}~\ldots \ldots .\text{ }\left( i \right)\]...

### Given: AX bisects angle BAC and PQ is perpendicular bisector of AC which meets AX at point Y. Prove: i) X is equidistant from AB and AC. ii) Y is equidistant from A and C

solution: Draw \[XL\bot AC\text{ }and\text{ }XM\bot AB.\] Now, join \[YC\] Proof: (i) In \[\vartriangle AXL\text{ }and\text{ }\vartriangle AXM,\] \[\angle XAL\text{ }=\text{ }\angle XAM\text{...

### In the figure given alongside, AB || CD and O is the center of the circle. If ∠ADC = 25o; find the angle AEB. Give reasons in support of your answer.

Solution: Join \[AC\text{ }and\text{ }BD.\] Hence, we get \[\angle CAD\text{ }=\text{ }{{90}^{o}}~and\angle CBD\text{ }=\text{ }{{90}^{o}}\] [Angle is a semicircle is a right angle] And, \[AB\text{...

### Given: CP is the bisector of angle C of ∆ABC. Prove: P is equidistant from AC and BC.

solution: Construction: \[From\text{ }P,\text{ }draw\text{ }PL\bot AC\text{ }and\text{ }PM\bot CB\] Proof: In \[\vartriangle LPC\text{ }and\text{ }\vartriangle MPC,\] \[\angle PLC\text{ }=\text{...

### In the figure given RS is a diameter of the circle. NM is parallel to RS and angle MRS = 29degree Calculate: (i) ∠RNM; (ii) ∠NRM.

Solution: (i) Join \[RN\text{ }and\text{ }MS\] \[\angle RMS\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] By angle sum property of \[\vartriangle RMS\] \[\angle RMS\text{...

### Given: PQ is a perpendicular bisector of side AB of the triangle ABC. Prove: Q is equidistant from A and B.

solution: Construction: Join AQ Proof: In \[\vartriangle AQP\text{ }and\text{ }\vartriangle BQP,\] \[AP\text{ }=\text{ }BP\text{ }\left[ Given \right]\] \[\angle QPA\text{ }=\text{ }\angle QPB\text{...

### Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the center O is equal to twice the angle APC

Solution: To prove: \[\angle AOC\text{ }+\angle BOD\text{ }=\text{ }2\angle APC\] \[OA,\text{ }OB,\text{ }OC\text{ }and\text{ }OD\] are joined. \[AD\] is joined. Now, it’s seen that \[\angle...

### The figure given below, shows a circle with centre O. Given: ∠AOC = a and ∠ABC = b. (i) Find the relationship between a and b (ii) Find the measure of angle OAB, if OABC is a parallelogram.

Solution: (i) According to the given question, it’s clear that \[\angle ABC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }Reflex\text{ }(\angle COA)\] [Angle at the centre is double the...

### Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

Solution: According to the given question, Let \[O\text{ }and\text{ }O\]be the centres of two intersecting circles, where points of the intersection are \[P\text{ }and\text{ }Q\text{ }and\text{...

### In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.

Solution: According to the given question, \[AB\text{ }=\text{ }AC\] So, \[\angle B\text{ }=\angle C\text{ }\ldots \text{ }\left( 1 \right)\] [Angles opposite to equal sides are equal] And,...

### Prove that: (i) the parallelogram, inscribed in a circle, is a rectangle. (ii) the rhombus, inscribed in a circle, is a square.

Solution: (i) Let’s expect that \[ABCD\]is a parallelogram which is inscribed in a circle. Hence, we know that \[\angle BAD\text{ }=\angle BCD\] [Opposite angles of a parallelogram are equal] And...

### ABCD is a parallelogram. A circle

Solution: Now, \[\angle BAD\text{ }+\angle BFE\text{ }=\text{ }{{96}^{o}}~+\text{ }{{84}^{o}}~=\text{ }{{180}^{o}}\] But these two are interior angles on the same side of a pair of lines \[AD\text{...

### In the following figure, (i) if ∠BAD = 96degree, find ∠BCD and ∠BFE. (ii) Prove that AD is parallel to FE.

Solution: \[ABCD\]is a cyclic quadrilateral (i) Hence, \[\angle BAD\text{ }+\angle BCD\text{ }=\text{ }{{180}^{o}}\] [Pair of opposite angles in a cyclic quadrilateral are supplementary] \[\angle...

### In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.

Solution: Join \[OB.\] Then,\[\angle OBA\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle is a right angle] Which means, \[OB\]is perpendicular to \[AE.\] Now, we know that the perpendicular...

### In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80degree and ∠CDE = 40degree. Find the number of degrees in: (i) ∠DCE; (ii) ∠ABC.

Solution: (i) According to the given question, We know that \[\angle DCE\text{ }=\text{ }{{90}^{o}}~\angle CDE\] \[=\text{ }{{90}^{o}}-\text{ }{{40}^{o}}~=\text{ }{{50}^{o}}\] Hence, \[\angle...

### In ABCD is a cyclic quadrilateral in which ∠DAC = 27degree, ∠DBA = 50degree and ∠ADB = 33degree. Calculate ∠CAB.

Solution: In quad. \[ABCD,\] \[\angle DAB\text{ }+\angle DCB\text{ }=\text{ }{{180}^{o}}\] \[{{27}^{o}}~+\angle CAB\text{ }+\text{ }{{83}^{o}}~=\text{ }{{180}^{o}}\] Hence, \[\angle CAB\text{...

### In ABCD is a cyclic quadrilateral in which ∠DAC = 27degree, ∠DBA = 50degree and ∠ADB = 33degree. Calculate (i) ∠DBC, (ii) ∠DCB

Solution: (i) According to the given question, We know that \[\angle DBC\text{ }=\angle DAC\text{ }=\text{ }{{27}^{o}}\] [Angles subtended by the same chord on the circle are equal] (ii)...

### In the following figure, O is the centre of the circle; ∠AOB = 60degree and ∠BDC = 100degree, find ∠OBC.

Solution: According to the given question, we get \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB\] \[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{...

### In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110degree, find ∠BDC.

Solution: Join \[AD\] So, we get \[\angle ADC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{110}^{o}}~=\text{...

### ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130degree, find ∠BAC.

Solution: According to the given question \[\angle ACB\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is 90o] Also, \[\angle ABC\text{ }=\text{ }{{180}^{o}}-\angle ADC\] \[=\text{...

### Given: ∠CAB = 75degree and ∠CBA = 50degree. Find the value of ∠DAB + ∠ABD.

Solution: According to the given question, \[\angle CAB\text{ }=\text{ }{{75}^{o}}~and\angle CBA\text{ }=\text{ }{{50}^{o}}\] In \[\vartriangle ABC\], by angle sum property we have \[\angle...

### In the figure given below, O is the centre of the circle and triangle ABC is equilateral. Find: (i) ∠ADB, (ii) ∠AEB

Solution: (i) According to the given question, \[\angle ACB\text{ }and\angle ADB\]are in the same segment, So, \[\angle ADB\text{ }=\angle ACB\text{ }=\text{ }{{60}^{o}}\] (ii) Now, join \[OA\text{...

### In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75degree; ∠ABD = 58degree and ∠ADC = 77degree. Find ∠BCA.

Solution: According to the given ques, \[\angle BCA\text{ }=\angle ADB\text{ }=\text{ }{{47}^{o}}\] [Angles subtended by the same chord on the circle are equal]

### In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD = 75Degree; ∠ABD = 58degree and ∠ADC = 77degree. Find: (i) ∠BDC, (ii) ∠BCD,

Solution: (i) According to the given question, By angle sum property of triangle ABD, \[\angle ADB\text{ }=\text{ }{{180}^{o}}-\text{ }{{75}^{o}}-\text{ }{{58}^{o}}~=\text{ }{{47}^{o}}\] Hence,...

### Calculate: ∠ACB.

Solution: According to the given question By angle sum property of a triangle we have \[\angle ACB\text{ }=\text{ }{{180}^{o}}-\text{ }{{49}^{o}}-\text{ }{{43}^{o}}~=\text{ }{{88}^{o}}\]

### Calculate: (i) ∠CDB, (ii) ∠ABC,

Solution: According to the given question We get, (i) \[\angle CDB\text{ }=\angle BAC\text{ }=\text{ }{{49}^{o}}\] (ii) \[\angle ABC\text{ }=\angle ADC\text{ }=\text{ }{{43}^{o}}\] [Angles subtended...

### In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠OAB, (ii) ∠CBA.

Solution: According to the given question, (i) In \[\vartriangle AOB,\]we have \[OA\text{ }=\text{ }OB\text{ }\left( radii \right)\] Thus, \[\angle OBA\text{ }=\angle OAB\] By angle sum property of...

### In the figure, given below, O is the centre of the circle. If ∠AOB = 140degree and ∠OAC = 50degree; Find (i) ∠ACB, (ii) ∠OBC,

Solution: According to the given question, Given, \[\angle AOB\text{ }=\text{ }{{140}^{o}}~\]and \[\angle OAC\text{ }=\text{ }{{50}^{o}}\] (i) So, \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle...

### In the figure, given below, find: ∠ABC. Show steps of your working.

Solution: The sum of angles in a quadrilateral is \[{{360}^{o}}\] Thus, \[\angle ADC\text{ }+\angle DAB\text{ }+\angle BCD\text{ }+\angle ABC\text{ }=\text{ }{{360}^{o}}\] \[{{75}^{o~}}+\text{...

### In the figure, given below, find: (i) ∠BCD, (ii) ∠ADC, Show steps of your working.

Solution: According to the given question, it’s clear that In cyclic quadrilateral \[ABCD,\text{ }DC\text{ }||\text{ }AB\] And given, \[\angle DAB\text{ }=\text{ }{{105}^{o}}\] (i) Hence, \[\angle...

### In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and O2 are the centers of two circles.

Solution: According to the given question, it’s clear that \[\angle DBA\text{ }=\angle CBA\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] So, adding both \[\angle DBA\text{...

### In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

(i) (ii) SOLUTION: (i) According to the given question, \[\angle AOB\text{ }=\text{ }2\angle AOB\text{ }=\text{ }2\text{ }x\text{ }{{50}^{o}}~=\text{ }{{100}^{o}}\] [Angle at the center is double...

### In each of the following figures, O is the center of the circle. Find the values of a, b, c and d.

SOLUTION: (i) In the following figure, \[\angle BAD\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle] Therefore, \[\angle BDA\text{ }=\text{ }{{90}^{o}}-\text{ }{{35}^{o}}~=\text{ }{{55}^{o}}\]...

### In each of the following figures, O is the centre of the circle. Find the values of a, b and c.

(i) (ii) Solution: (i)In the following figure, \[b\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{130}^{o}}\] [Angle at the center is double the angle at the circumference...

### Given O is the centre of the circle and ∠AOB = 70o. Calculate the value of: (i) ∠OCA, (ii) ∠OAC.

Solution: \[\angle AOB\text{ }=\text{ }2\angle ACB\] [Angle at the center is double the angle at the circumference subtend by the same chord] \[\angle ACB\text{ }=\text{ }{{70}^{o}}/\text{ }2\text{...

### In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45° (i) Prove that AC is a diameter of the circle. (ii) Find ∠ACB.

Solution: (i) In \[\vartriangle ABD,\] \[\angle DAB\text{ }+\angle ABD\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}\] \[{{65}^{o}}~+\text{ }{{70}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}\]...

### In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30Degree and 40Degree respectively. Find ∠AOC Show your steps of working.

Solution: Let’s join \[AC.\] And, let \[\angle OAC\text{ }=\angle OCA\text{ }=\text{ }x\] [Angles opposite to equal sides are equal] Thus, \[\angle AOC\text{ }=\text{ }{{180}^{o}}-\text{ }2x\] Also,...

### From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that: i) ∠AOP = ∠BOP ii) OP is the ⊥ bisector of chord AB.

According to the given question, i) In \[\vartriangle AOP\text{ }and\text{ }\vartriangle BOP\], we have \[AP\text{ }=\text{ }BP\] \[\left[ Tangents\text{ }from\text{ }P\text{ }to\text{ }the\text{...

### Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if – i) they touch each other externally. ii) they touch each other internally.

According to the given question \[Radius\text{ }of\text{ }bigger\text{ }circle\text{ }=\text{ }6.3\text{ }cm\]and \[smaller\text{ }circle\text{ }radius\text{ }=\text{ }3.6\text{ }cm\] i) The two...

### In the figure, if AB = AC then prove that BQ = CQ.

Solution: From point \[A\], we know that \[AP\text{ }and\text{ }AR\]are the tangents to the circle So, we have \[AP\text{ }=\text{ }AR\] Similarly, we also have \[BP\text{ }=\text{ }BQ\text{...

### From the given figure prove that: AP + BQ + CR = BP + CQ + AR. Also, show that AP + BQ + CR = ½ x perimeter of triangle ABC.

Solution: From point \[B,\text{ }BQ\text{ }and\text{ }BP\]are the tangents to the circle We have, \[BQ\text{ }=\text{ }BP\text{ }\ldots \ldots \ldots ..\left( 1 \right)\] Similarly, we also get...

### If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.

Solution: Suppose a circle touch the sides \[AB,\text{ }BC,\text{ }CD\text{ }and\text{ }DA\]of parallelogram \[ABCD\text{ }at\text{ }P,\text{ }Q,\text{ }R\text{ }and\text{ }S\]respectively. Now,...

### If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.

Solution: Suppose, a circle touch the sides \[AB,\text{ }BC,\text{ }CD\text{ }and\text{ }DA\] of quadrilateral \[ABCD\text{ }at\text{ }P,\text{ }Q,\text{ }R\text{ }and\text{ }S\]respectively. As,...

### Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.

Suppose, \[ABC\]be the triangle formed when centres of 3 circles are joined. Given, \[AB\text{ }=\text{ }6\text{ }cm,\text{ }AC\text{ }=\text{ }8\text{ }cm\text{ }and\text{ }BC\text{ }=\text{...

### Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.

Solution: Given, \[OS\text{ }=\text{ }5\text{ }cm\text{ }and\text{ }OT\text{ }=\text{ }3\text{ }cm\] In right triangle\[OST\], we have \[S{{T}^{2}}~=\text{ }O{{S}^{2}}~\text{ }O{{T}^{2}}\] \[=\text{...

### Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.

Solution: Suppose, \[Q\]be the point on the common tangent from which, two tangents \[QA\text{ }and\text{ }QP\]are drawn to the circle with \[centre\text{ }O.\] So, \[QA\text{ }=\text{ }QP\text{...

### Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Solution: Suppose, \[Q\]be the point from which, \[QA\text{ }and\text{ }QP\]are two tangents with centre \[O\] So, \[QA\text{ }=\text{ }QP\text{ }\ldots ..\left( a \right)\] Similarly, from point...

### In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.

Solution: Given, \[AB\text{ }=\text{ }15\text{ }cm,\text{ }AC\text{ }=\text{ }7.5\text{ }cm\] Suppose,the radius of the circle to be \[r.\] So, \[AO\text{ }=\text{ }AC\text{ }+\text{ }OC\text{...

### The radius of a circle is 8cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10cm from its centre.

Given, a circle with \[centre\text{ }O\]and \[radius\text{ }8\text{ }cm.\] An external point \[P\]is the point, from where a tangent is drawn to meet the circle at \[T.\] \[OP\text{ }=\text{...

### If and , prove that

Taking LHS, $\mathrm{m}^{2}-\mathrm{n}^{2}$ $=(a \sec A+b \tan A)^{2}-(a \tan A+b \sec A)^{2}$ $=a^{2} \sec ^{2} A+b^{2} \tan ^{2} A+2 a b \sec A \tan A-a^{2} \tan ^{2} A-b^{2} \sec ^{2} A-2 a b...

### Prove: (ix)

### Prove: (vii) (viii)

(vii) (viii)

### Prove: (v) (vi)

(v) \[\begin{array}{*{35}{l}} 2\text{ }si{{n}^{2}}~A\text{ }+\text{ }co{{s}^{2}}~A \\ =\text{ }2\text{ }si{{n}^{2}}~A\text{ }+\text{ }{{\left( 1\text{ }-\text{ }si{{n}^{2}}~A \right)}^{2}} \\...

### Prove: (iii) (iv)

(iii) (iv)

### Prove: (i). cosA/1-tanA+ sinA/1-cotA = sinA + cosA (ii)

(i) (ii)

### Prove: 1/ 1 – sin A + 1/ 1 + sin A = 2 sec^2 A

### Prove: 1/ 1 + cos A + 1/ 1 – cos A = 2 cosec^2A

### Prove:

\[\begin{array}{*{35}{l}} RHS\text{ }=\text{ }ta{{n}^{2}}~A\text{ }+\text{ }co{{t}^{2}}~A\text{ }+\text{ }2\text{ }=\text{ }ta{{n}^{2}}~A\text{ }+\text{ }co{{t}^{2}}~A\text{ }+\text{ }2\text{...

### Prove:

\[\begin{array}{*{35}{l}} LHS, \\ {{\left( sin\text{ }A\text{ }+\text{ }cosec\text{ }A \right)}^{2}}~+\text{ }{{\left( cos\text{ }A\text{ }+\text{ }sec\text{ }A \right)}^{2}} \\ =\text{...

### Prove: sec A – tan A/ sec A + tan A = 1 – 2 secA tanA + 2 tan^2 A

= 1 + tan2 A + tan2 A – 2 sec A tan A = 1 – 2 sec A tan A + 2 tan2 A = RHS

### Prove: cosec A + cot A = 1/ cosec A – cot A

cosec A + cot A

### Prove: 1/ sec A + tan A = sec A – tan A

### Prove: (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1

(cosec A – sin A)(sec A – cos A)(tan A + cot A)

### Prove:

\[\begin{array}{*{35}{l}} {{\left( cos\text{ }A\text{ }+\text{ }sin\text{ }A \right)}^{2}}~+\text{ }{{\left( cosA\text{ }-\text{ }sin\text{ }A \right)}^{2}} \\ =\text{ }cos2\text{ }A\text{ }+\text{...

### Prove:

\[\begin{array}{*{35}{l}} \left( sec\text{ }A\text{ }-\text{ }cos\text{ }A \right)\left( sec\text{ }A\text{ }+\text{ }cos\text{ }A \right) \\ =\text{ }\left( se{{c}^{2}}~A\text{ }-\text{...

### Prove:

\[\begin{array}{*{35}{l}} \left( cosec\text{ }A\text{ }+\text{ }sin\text{ }A \right)\text{ }\left( cosec\text{ }A\text{ }-\text{ }sin\text{ }A \right) \\ =\text{ }cose{{c}^{2}}~A\text{ }-\text{...

### Prove:

cot2 A – cos2 A

### Prove:

tan2 A – sin2 A

### Prove:

### Prove

### Prove: cosec A (1 + cos A) (cosec A – cot A) = 1

### Prove: sec A (1 – sin A) (sec A + tan A) = 1

sec A (1 – sin A) (sec A + tan A)

### In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.

Let’s expect the point \[P\text{ }\left( x,\text{ }0 \right)\text{ }on\text{ }x-axis\] divides the line segment joining \[A\text{ }\left( 4,\text{ }3 \right)\text{ }and\text{ }B\text{ }\left(...

### Prove: cosec^4 A – cosec^2 A = cot^4 A + cot^2 A

\[\begin{array}{*{35}{l}} cose{{c}^{4}}~A\text{ }-\text{ }cose{{c}^{2}}~A \\ =\text{ }cose{{c}^{2}}~A\left( cose{{c}^{2}}~A\text{ }-\text{ }1 \right) \\ =\text{ }\left( 1\text{ }+\text{...

### Prove: (1 – tan A)^2 + (1 + tan A)^2 = 2sec^2 A

\[\begin{array}{*{35}{l}} {} \\ {{\left( 1\text{ }-\text{ }tan\text{ }A \right)}^{2}}~+\text{ }{{\left( 1\text{ }+\text{ }tan\text{ }A \right)}^{2}} \\ =\text{ }\left( 1\text{ }+\text{...

### Prove the following : sin^4 A – cos^4 A = 2 sin^2 A – 1

\[\begin{array}{*{35}{l}} \mathbf{L}.\mathbf{H}.\mathbf{S}, \\ si{{n}^{4}}~A\text{ }-\text{ }co{{s}^{4}}~A \\ =\text{ }{{\left( si{{n}^{2~}}A \right)}^{2}}~-\text{ }{{\left( co{{s}^{2}}~A...

### Prove the following: tan A – cot A = 1 – 2 cos^2 A/ sin A cos A

### Prove the following : 1/ tan A + cot A = cos A sin A

### Prove the following identities: 1 + sin A/ 1 – sin A = cosec A + 1/ cosec A – 1

### Find the seventh term of the G.P: 1, √3, 3, 3 √3, …..

First term is \[\left( a \right)\text{ }=\text{ }1\] And, common ratio\[\left( r \right)\text{ }=\text{ }\surd 3/1\text{ }=\text{ }\surd 3\] We know that, the general term is \[{{t}_{n}}~=\text{...

### Prove the following identities: sec A – 1/ sec A + 1 = 1 – cos A/ 1 + cos A

### Find the value of ‘m’, if

and

leave the same remainder when each is divided by

.

Let f(x) = \[\mathbf{m}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\] and g(x) = \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{mx}\text{ }+\text{...

### Factorise

completely using factor theorem.

Let f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{6}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{11x}\text{ }+\text{ }\mathbf{6}\] For \[x\text{ }=\text{ }-1\], the value of f(x) is...

### If

is a factor of

and

, find the values of a and b.

Let f(x) = \[{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{ax}\text{ }+\text{ }\mathbf{b}\] Given, \[\left( x\text{ }\text{ }2 \right)\]is a factor of f(x). Then, remainder = \[f\left( 2...

### If

and

are factors of

, find the values of a and b. And then, factorise the given expression completely.

Let’s take f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\left( \mathbf{a}\text{ }+\text{ }\mathbf{1} \right){{\mathbf{x}}^{\mathbf{2}}}~\text{ }\left( \mathbf{b}\text{ }\text{ }\mathbf{2}...

### What should be subtracted from

, so that the resulting expression has

as a factor?

Let’s assume the required number to be k. And let f(x) = \[3{{x}^{3}}~\text{ }8{{x}^{2}}~+\text{ }4x\text{ }\text{ }3\text{ }\text{ }k\] From the question, we have \[f\left( -2 \right)\text{...

### When

is divided by

, the remainder is

. Find the value of m.

Let us consider f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{mx}\text{ }+\text{ }\mathbf{4}\] From the question, we have \[f\left( 2...

### Using Remainder Theorem, factorise:

completely.

Let f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{37x}\text{ }+\text{ }\mathbf{26}\] According to remainder theorem, we know that For \[x\text{...

### Show that

is a factor of

. Hence, completely factorise the given expression.

Let us consider f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{14x}\text{ }\text{ }\mathbf{8}\] Then, for \[x\text{ }=\text{ }1\] \[f\left( 1...

### Given that

and

are factors of f(x) =

; calculate the values of a and b. Hence, find all the factors of f(x).

Let \[f\left( x \right)\text{ }=\text{ }{{x}^{3}}~+\text{ }3{{x}^{2}}~+\text{ }ax\text{ }+\text{ }b\] As, \[\left( x\text{ }\text{ }2 \right)\] is a factor of f(x), so \[f\left( 2 \right)\text{...

### Factorise the expression . Hence, find all possible values of for which 0

Given: f(x) = 2x3 – 7x2 – 3x + 18 By hit and trial method For x = 2, the value of f(x) will be f(2) = 2(2)3 – 7(2)2 – 3(2) + 18 = 16 – 28 – 6 + 18 = 0 As f(2) = 0, (x – 2) is a factor of f(x). Now,...

### Using the Remainder Theorem, factorise the expression

. Hence, solve the equation

.

Let’s take f(x) = \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{10}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{x}\text{ }\text{ }\mathbf{6}\] For \[x\text{ }=\text{ }-1\], the value of...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{4x}\text{ }\text{ }\mathbf{4}\] For \[x\text{ }=\text{ }-1\], the value of f(x) will be...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = \[\mathbf{4}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{7}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{36x}\text{ }\text{ }\mathbf{63}\] For \[x\text{ }=\text{ }3\], the value of f(x)...

### Using the Remainder Theorem, factorise each of the following completely.

Given: f(x) = 2x3 + x2 – 13x + 6 By hit and trial we put x=-2 For x = -2, the value of f(x) will be f(-2) = 3(-2)3 + 2(-2)2 – 23(-2) – 30 = -24 + 8 + 46 – 30 = -54 + 54 = 0 As f(-2) = 0, so (x + 2)...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }+\text{ }\mathbf{6}\] For \[x\text{ }=\text{ }2\], the value of f(x) will...

### Using the Remainder Theorem, factorise each of the following completely.

Let f(x) = \[\mathbf{3}{{\mathbf{x}}^{\mathbf{3}~}}+\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }\mathbf{19x}\text{ }+\text{ }\mathbf{6}\] For x=2, the value of f(x) will be...

### Using the Factor Theorem, show that:

is a factor of

. Hence, factorise the expression

completely.

Here, f(x) = \[3{{x}^{3}}~+\text{ }2{{x}^{2}}~\text{ }3x\text{ }\text{ }2\] So, \[3x\text{ }+\text{ }2\text{ }=\text{ }0~\Rightarrow x\text{ }=\text{ }-2/3\] Thus, remainder = \[f\left( -2/3...

### Using the Factor Theorem, show that:

is a factor of

. Hence, factorise the expression

completely

Given, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{28x}\text{ }\text{ }\mathbf{15}\] So, \[x\text{ }+\text{ }5\text{ }=\text{...

### Using the Factor Theorem, show that:

is a factor of

. Hence, factorise the expression

completely.

Given, f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~\text{ }\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{9x}\text{ }+\text{ }\mathbf{18}\] So, \[x\text{ }\text{ }2\text{ }=\text{ }0~\Rightarrow...

### Find the values of and so that and both are factors of .

Given: f(x) = x3 + (3m + 1) x2 + nx – 18 Given, (x – 1) and (x + 2) are the factors of f(x). We know that when a polynomial f (x) is divided by (x – a), the remaining is f from the remainder theorem...

### Find the value of , if is a factor of .

Given, f(x) = 2x5 – 6x4 – 2ax3 + 6ax2 + 4ax + 8 and x – 2 is a factor of f(x). We know that when a polynomial f (x) is divided by (x – a), the remaining is f from the remainder theorem (a). So, x –...

### Find the value of k, if

is a factor of

.

Let us consider f(x) = \[\left( \mathbf{3k}\text{ }+\text{ }\mathbf{2} \right){{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\left( \mathbf{k}\text{ }\text{ }\mathbf{1} \right)\] Now, \[2x\text{ }+\text{...

### Find the values of constants a and when and both are the factors of expression

Given expression, f(x) = x3 + ax2 + bx – 12 Given, x – 2 and x + 3 both are the factors of f(x) So, according to remainder theorem, f(2) and f(-3) both should be equal to zero. When we put the...

### Find the value of k, if

is a factor of expression

.

Given, \[3x\text{ }\text{ }4\]is a factor of g(x) = \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{2x}\text{ }\text{ }\mathbf{k}\]. So, \[f\left( 4/3 \right)\text{ }=\text{ }0\]...

### If

is a factor of

, find the value of a

Given, \[\mathbf{2x}\text{ }+\text{ }\mathbf{1}\]is a factor of f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{ax}\text{ }\text{ }\mathbf{3}\]. So, \[f\left( -1/2 \right)\text{...

### Use the Remainder Theorem to find which of the following is a factor of

. (i)

(ii)

(iii)

According to remainder theorem, when a polynomial f (x) is divided by x – a, then the remainder is f(a). Here, f(x) = \[\mathbf{2}{{\mathbf{x}}^{\mathbf{3}}}~+\text{...

### Show that: is a factor of

We know that when a polynomial f (x) is divided by (x – a), the remaining is f from the remainder theorem (a). Given: f(x) = 3x2 – x – 2 f(-2/3) = 3(-2/3)2 – (-2/3) – 2 = 4/3 + 2/3 – 2 = 2 – 2 = 0...

### Show that:

is a factor of

(x – a) is a factor of a polynomial f(x) if the remainder, when f(x) is divided by (x – a), is \[0\], i.e., if f(a) = \[0\]. Given, f(x) = \[\mathbf{5}{{\mathbf{x}}^{\mathbf{2}}}~+\text{...

### Find, in each case, the remainder when:

is divided by

.

From the question, f(x) = \[{{\mathbf{x}}^{\mathbf{4}}}~+\text{ }\mathbf{1}\]is divided by \[\mathbf{x}\text{ }+\text{ }\mathbf{1}\] So, remainder = \[f\left( -1 \right)\text{ }=\text{ }{{\left( -1...

### Find, in each case, the remainder when:

is divided by

.

from the question, f(x) = \[{{\mathbf{x}}^{\mathbf{3}}}~+\text{ }\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{12x}\text{ }+\text{ }\mathbf{4}\] is divided by \[\mathbf{x}\text{ }\text{...

### Find, in each case, the remainder when: is divided by

We know that when a polynomial f (x) is divided by (x – a), the remaining is f from the remainder theorem (a). Given, f(x) = x4 – 3x2 + 2x + 1 is divided by x – 1 So, remainder = f(1) = (1)4 –...

### The age of the father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.

We should expect the current age of the child to be x years. Thus, the current age of the dad = \[2x2\text{ }years\] Eight years henceforth, Child's age = \[\left( x\text{ }+\text{ }8 \right)\] a...

### One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

ACCORDING TO QUES, How about we think about the current age of the child to be x years. Along these lines, the current age of the man \[=\text{ }x2\text{ }years\] One year prior, Child's age...

### The ages of two sisters are 11 years and 14 years. In how many years’ time will the product of their ages be 304?

Given, the ages of two sisters are 11 years and 14 years. Leave x alone the quantity of years some other time when their result of their ages become 304. In this way, \[\left( 11\text{ }+\text{ }x...

### The product of the digits of a two digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.

How about we accept the ten's and unit's digit of the necessary number to be x and y individually. Then, at that point, from the inquiry we have \[x\text{ }\times \text{ }y\text{ }=\text{ }24\]...

### A stone is thrown vertically downwards and the formula d = 16t2 + 4t gives the distance, d metres, that it falls in t seconds. How long does it take to fall 420 metres?

As indicated by the inquiry, \[16t2\text{ }+\text{ }4t\text{ }=\text{ }420\] \[4t2\text{ }+\text{ }t\text{ }\text{ }105\text{ }=\text{ }0\] \[4t2\text{ }\text{ }20t\text{ }+\text{ }21t\text{ }\text{...

### The sum S of n successive odd numbers starting from 3 is given by the relation: S = n(n + 2). Determine n, if the sum is 168.

From the inquiry, we have \[n\left( n\text{ }+\text{ }2 \right)\text{ }=\text{ }168\] \[n2\text{ }+\text{ }2n\text{ }\text{ }168\text{ }=\text{ }0\] \[n2\text{ }+\text{ }14n\text{ }\text{ }12n\text{...

### A girl goes to her friend’s house, which is at a distance of 12 km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find ‘x’.

Given, The young lady covers a distance of 6 km at a speed x km/hr. Along these lines, the time taken to cover initial 6 km \[=\text{ }6/x\text{ }hr\] [Since, Time = Distance/Speed] Likewise given,...

### A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.

We should accept x km/h to be the first speed of the vehicle. We realize that, Time = Distance/Speed From the inquiry, The time taken by the vehicle to finish 400 km = \[400/x\text{ }hrs\]...

### If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.

How about we think about the first speed of the plane to be x km/hr. Presently, the time taken to cover a distance of \[1200\text{ }km\text{ }=\text{ }1200/x\text{ }hrs\] [Since, Time =...

### The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr. (i) Find the time taken by each train to cover 300 km. (ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.

(i) Given, Speed of the conventional train \[=\text{ }x\text{ }km/hr\] Speed of the express train \[=\text{ }\left( x\text{ }+\text{ }25 \right)\text{ }km/hr\] Distance \[=\text{ }300\text{ }km\] We...

### The perimeter of a rectangle is 104 m and its area is 640 m2. Find its length and breadth.

We should take the length and the expansiveness of the square shape be x m and y m. Thus, the edge \[=\text{ }2\left( x\text{ }+\text{ }y \right)\text{ }m\] \[104\text{ }=\text{ }2\left( x\text{...

### The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of the rectangle.

How about we think about the more limited side of the square shape to be x m. Then, at that point, the length of the opposite side \[=\text{ }\left( x\text{ }+\text{ }30 \right)\text{ }m\] Length of...

### The hypotenuse of a right-angled triangle exceeds one side by 1 cm and the other side by 18 cm; find the lengths of the sides of the triangle.

Leave the hypotenuse of the right triangle alone x cm. From the inquiry, we have Length of one side \[=\text{ }\left( x\text{ }\text{ }1 \right)\text{ }cm\] Length of opposite side \[=\text{ }\left(...

### The sides of a right-angled triangle are (x – 1) cm, 3x cm and (3x + 1) cm. Find: (i) the value of x, (ii) the lengths of its sides, (iii) its area.

Given, The more drawn out side = \[Hypotenuse\text{ }=\text{ }\left( 3x\text{ }+\text{ }1 \right)\text{ }cm\] Furthermore, the lengths of other different sides are\[\left( x\text{ }\text{ }1...

### The hypotenuse of a right-angled triangle is 26 cm and the sum of other two sides is 34 cm. Find the lengths of its sides

Given, a right triangle \[Hypotenuse\text{ }=\text{ }26\text{ }cm\] and the amount of other different sides is \[34\text{ }cm.\] Presently, let believe the other different sides to be \[x\text{...

### The sides of a right-angled triangle containing the right angle are 4x cm and (2x – 1) cm. If the area of the triangle is 30 cm2; calculate the lengths of its sides.

Given, the space of triangle \[=\text{ }30\text{ }cm2\] As, x can't be negative, just \[x\text{ }=\text{ }3\] is substantial. Thus, we have \[AB\text{ }=\text{ }4\text{ }\times \text{ }3\text{...

### The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.

Leave the two back to back certain even numbers alone taken as \[x\text{ }and\text{ }x\text{ }+\text{ }2.\] Now, \[x2\text{ }+\text{ }\left( x\text{ }+\text{ }2 \right)2\text{ }=\text{ }52\]...

### The sum of the squares of two positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.

How about we expect the two numbers to be x and y, y being the bigger of the two numbers. Then, at that point, from the inquiry \[x2\text{ }+\text{ }y2\text{ }=\text{ }208\text{ }\ldots \text{...

### Divide 15 into two parts such that the sum of their reciprocals is 3/10

We should expect the two sections to be \[x\text{ }and\text{ }15\text{ }\text{ }x.\] now, \[150\text{ }=\text{ }45x\text{ }\text{ }3x2\] \[3x2\text{ }\text{ }45x\text{ }+\text{ }150\text{ }=\text{...

### Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is 7/10.

How about we believe the two normal numbers to be\[x\text{ }and\text{ }x\text{ }+\text{ }3\] . (As they vary by 3) now, \[20x\text{ }+\text{ }30\text{ }=\text{ }7x2\text{ }+\text{ }21x\] \[7x2\text{...

### The sum of a number and its reciprocal is 4.25. Find the number.

Leave the number alone x. In this way, its complementary is 1/x now, \[4x2\text{ }\text{ }17x\text{ }+\text{ }4\text{ }=\text{ }0\] \[4x2\text{ }\text{ }16x\text{ }\text{ }x\text{ }+\text{...

### Find the two natural numbers which differ by 5 and the sum of whose squares is 97.

We should expect the two normal numbers to be\[x\text{ }and\text{ }x\text{ }+\text{ }5\] . (As given they contrast by 5) So from the inquiry, \[x2\text{ }+\text{ }\left( x\text{ }+\text{ }5...

### The sum of the squares of two consecutive natural numbers is 41. Find the numbers.

Allow us to take the two successive regular numbers as x and x + 1. So from the inquiry, \[x2\text{ }+\text{ }\left( x\text{ }+\text{ }1 \right)2\text{ }=\text{ }41\] \[2x2\text{ }+\text{ }2x\text{...

### The product of two consecutive integers is 56. Find the integers.

let two continuous numbers to be $$ \[x\text{ }and\text{ }x\text{ }+\text{ }1.\] So from the inquiry, \[x\left( x\text{ }+\text{ }1 \right)\text{ }=\text{ }56\] \[x2\text{ }+\text{ }x\text{...

### If (4a + 9b) (4c – 9d) = (4a – 9b) (4c + 9d), prove that: a: b = c: d.

Applying componendo and dividendo: \[\begin{array}{*{35}{l}} 8a/18b\text{ }=\text{ }8c/18d \\ a/b\text{ }=\text{ }c/d \\ \end{array}\]

### If x= (fig 1) ,find the value of:

fig 1: SOLUTION: \[\begin{array}{*{35}{l}} x\text{ }=\text{ }2ab/\text{ }\left( a\text{ }+\text{ }b \right) \\ x/a\text{ }=\text{ }2b/\left( a\text{ }+\text{ }b \right) \\ \end{array}\] Applying...

### If x and y be unequal and x: y is the duplicate ratio of x + z and y + z, prove that z is mean proportional between x and y.

\[\begin{array}{*{35}{l}} x\left( {{y}^{2}}~+\text{ }{{z}^{2}}~+\text{ }2yz \right)\text{ }=\text{ }y\left( {{x}^{2}}~+\text{ }{{z}^{2}}~+\text{ }2xz \right) \\ x{{y}^{2}}~+\text{...

### Find two numbers such that the mean proportional between them is 14 and third proportional to them is 112.

Let the required numbers be a and b. Given, 14 is the mean proportional between a and b. \[\begin{array}{*{35}{l}} a:\text{ }14\text{ }=\text{ }14:\text{ }b \\ ab\text{ }=\text{ }196 \\ a\text{...

### Find the (iii) mean proportional to (x – y) and (x^3 – x^2y).

(iii) Let the mean proportional to \[\left( x\text{ }-\text{ }y \right)\text{ }and\text{ }\left( {{x}^{3}}~\text{ }-{{x}^{2}}y \right)\] be n. \[\left( x\text{ }-\text{ }y \right)\text{ }and\text{...

### Find the: (i) fourth proportional to 2xy, x2 and y2. (ii) third proportional to a2 – b2 and a + b.

(i) Let the fourth proportional to 2xy, x2 and y2 be n. \[\begin{array}{*{35}{l}} 2xy:\text{ }{{x}^{2}}~=\text{ }{{y}^{2}}:\text{ }n \\ 2xy~\times ~n\text{ }=\text{ }{{x}^{2}}~\times ~{{y}^{2}} \\...

### If 15(2×2 – y2) = 7xy, find x: y; if x and y both are positive.

15(2x2 – y2) = 7xy Let the substitution as \[\begin{array}{*{35}{l}} x/y\text{ }=\text{ }a \\ 2a\text{ }-\text{ }1/a\text{ }=\text{ }7/15 \\ \left( 2{{a}^{2}}~-\text{ }1 \right)/\text{ }a\text{...

### A woman reduces her weight in the ratio 7: 5. What does her weight become if originally it was 84 kg?

Let the woman’s reduced weight as x. the original weight = 84 kg So, we have \[\begin{array}{*{35}{l}} 84:\text{ }x\text{ }=\text{ }7:\text{ }5 \\ 84/x\text{ }=\text{ }7/5 \\ 84\text{ * }5\text{...

### What quantity must be added to each term of the ratio x: y so that it may become equal to c: d?

Let the required quantity which has to be added be p. \[\begin{array}{*{35}{l}} dx\text{ }+\text{ }pd\text{ }=\text{ }cy\text{ }+\text{ }cp \\ pd\text{ }-\text{ }cp\text{ }=\text{ }cy-\text{...

### Find the value of x, if: (iii) (3x – 7): (4x + 3) is the sub-triplicate ratio of 8: 27.

(iii) (3x – 7): (4x + 3) is the sub-triplicate ratio of 8: 27 And the sub-triplicate ratio of \[\begin{array}{*{35}{l}} 8:\text{ }27\text{ }=\text{ }2:\text{ }3 \\ \left( 3x\text{ }-\text{ }7...

### Find the value of x, if: (i) (2x + 3): (5x – 38) is the duplicate ratio of √5: √6. (ii) (2x + 1): (3x + 13) is the sub-duplicate ratio of 9: 25.

(i) (2x + 3): (5x – 38) is the duplicate ratio of √5: √6 And, the duplicate ratio of √5: √6 = 5: 6, \[\begin{array}{*{35}{l}} \left( 2x\text{ }+\text{ }3 \right)/\text{ }\left( 5x\text{ }-\text{ }38...

### Find the : (v) reciprocal ratio of 3: 5 (vi) ratio compounded of the duplicate ratio of 5: 6, the reciprocal ratio of 25: 42 and the sub-duplicate ratio of 36: 49.

(v) Reciprocal ratio of \[3:\text{ }5\text{ }=\text{ }5:\text{ }3\] (vi) Duplicate ratio of \[5:\text{ }6\text{ }=\text{ }25:\text{ }36\] Reciprocal ratio of \[25:\text{ }42\text{ }=\text{...

### Find the: (iii) sub-duplicate ratio of 9x^2a^4 : 25y^6b^2 (iv) sub-triplicate ratio of 216: 343

(iv) Sub-triplicate ratio of \[216:\text{ }343\text{ }=~{{\left( 216 \right)}^{1/3}}:\text{ }{{\left( 343 \right)}^{1/3}}~=\text{ }6:\text{ }7\] (v) Reciprocal ratio of \[3:\text{ }5\text{ }=\text{...

### Find the: (i) duplicate ratio of 2√2: 3√5 (ii) triplicate ratio of 2a: 3b

(i) Duplicate ratio of \[2\surd 2:\text{ }3\surd 5\text{ }=\text{ }{{\left( 2\surd 2 \right)}^{2}}:\text{ }{{\left( 3\surd 5 \right)}^{2}}~=\text{ }8:\text{ }45\] (ii) Triplicate ratio of...

### If (3x – 4y): (2x – 3y) = (5x – 6y): (4x – 5y), find x: y.

Since, \[\left( 3x\text{ }-\text{ }4y \right):\text{ }\left( 2x\text{ }-\text{ }3y \right)\text{ }=\text{ }\left( 5x\text{ }-\text{ }6y \right):\text{ }\left( 4x\text{ }-\text{ }5y \right)\] This...

### Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image: (a)A” of A under reflection in the y-axis. (b) B” of B under reflection in the line AA”.

(a) \[A''\text{ }=\text{ }Image\text{ }of\text{ }A\text{ }under\text{ }appearance\]in the y-hub = (- 3, 4) (b) \[B''\text{ }=\text{ }Image\text{ }of\text{ }B\text{ }under\text{ }appearance\]in the...

### If 5x + 6y: 8x + 5y = 8: 9, find x: y.

Given, On cross multiplying, we get \[\begin{array}{*{35}{l}} 45x\text{ }+\text{ }54y\text{ }=\text{ }64x\text{ }+\text{ }40y \\ 14y\text{ }=\text{ }19x \\ {} \\ x/y\text{ }=\text{ }14/19 \\...

### If a: b = 3: 5, find: (10a + 3b): (5a + 2b)

\[\begin{array}{*{35}{l}} a/b\text{ }=\text{ }3/5 \\ \left( 10a\text{ }+\text{ }3b \right)/\text{ }\left( 5a\text{ }+\text{ }2b \right) \\ ~ \\ \end{array}\]

### Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image: (a) A’ of A under reflection in the x-axis. (b) B’ of B under reflection in the line AA’.

(a) \[A'\text{ }=\text{ }Image\text{ }of\text{ }A\text{ }under\text{ }appearance\] in the \[x-hub\text{ }=\text{ }\left( 3,\text{ }-\text{ }4 \right)\] (b) \[B'\text{ }=\text{ }Image\text{ }of\text{...

### If (a + b + c + d) (a – b – c + d) = (a + b – c – d) (a – b + c – d), prove that a: b = c: d.

Rewriting the given, we have applying componendo and dividendo: Applying componendo and dividendo :