According to the given question, \[\left( x\text{ }+\text{ }5 \right)\text{ }\left( x\text{ }\text{ }5 \right)\text{ }=\text{ }24\] Or, \[{{x}^{2}}~\text{ }25\text{ }=\text{ }24\] Or,...
Solve: ![\[\left( \mathbf{i} \right)\text{ }\left( \mathbf{x}\text{ }+\text{ }\mathbf{5} \right)\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{5} \right)\text{ }=\text{ }\mathbf{24}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-85409799f13ae2131936b8199e40aef0_l3.png "Rendered by QuickLaTeX.com")
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Selina
Solve:
Solve: (x^2+1/x^2)-3(x-1/x)-2=0
Let \[\left( x\text{ }\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring on both sides \[{{\left( x\text{ }\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\] or,...
Solve: 2(x^2+1/x^2)-(x+1/x)=11
let \[\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring on both sides \[{{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\]...
Solve:9(x^2+1/x^2)-9(x+1/x)-52=0
Let, \[~\left( x\text{ }+\text{ }1/x \right)\text{ }=\text{ }y\text{ }\ldots .\text{ }\left( 1 \right)\] squaring both sides \[{{\left( x\text{ }+\text{ }1/x \right)}^{2}}~=\text{ }{{y}^{2}}\]...
Solve: x4 – 2×2 – 3 = 0
According to ques, \[{{x}^{4}}~\text{ }2{{x}^{2}}~\text{ }3\text{ }=\text{ }0\] Let’s take \[{{x}^{2}}~=\text{ }y\] Then, the equation becomes \[{{y}^{2}}~\text{ }2y\text{ }\text{ }3\text{ }=\text{...
Find the value of x, given that A^2 = B,
Solution: So, on comparison we get \[x\text{ }=\text{ }36.\]
If given matrix, find the matrix X such that: A + X = 2B + C
Solution: As per the given question,
Solve : 2×4 – 5×2 + 3 = 0
According to ques, \[~2{{x}^{4}}~\text{ }5{{x}^{2}}~+\text{ }3\text{ }=\text{ }0\] Let \[{{x}^{2}}~=\text{ }y\] The, \[2{{y}^{2}}~\text{ }5y\text{ }+\text{ }3\text{ }=\text{ }0\] \[2{{y}^{2}}~\text{...
If given matrix, find the matrix ‘X’ and matrix ‘Y’.
Solution: Now, On comparison, we get \[-28\text{ }-\text{ }3x\text{ }=\text{ }10\] \[3x\text{ }=\text{ }-38\] \[x\text{ }=\text{ }-38/3\] And, \[20\text{ }-\text{ }3y\text{ }=\text{ }-8\] \[3y\text{...
Find x and y, if:
Solution: According to the given question, On comparison, we get \[2x\text{ }+\text{ }3x\text{ }=\text{ }5\] And \[2y\text{ }+\text{ }4y\text{ }=\text{ }12\] \[5x\text{ }=\text{ }5\text{ }and\text{...
Solve: x + 4/x = -4; x ≠ 0
According to ques, \[~x\text{ }+\text{ }4/x\text{ }=\text{ }-4\] or, \[\left( {{x}^{2~}}+\text{ }4 \right)/\text{ }x\text{ }=\text{ }-4\] Or, \[{{x}^{2}}~+\text{ }4\text{ }=\text{ }-4x\] or,...
Solve: (i) A (BA) (ii) (AB) B.
Solution: \[\left( i \right)\text{ }A\text{ }\left( BA \right)\] \[\left( ii \right)\text{ }\left( AB \right)\text{ }B\]
Find the values of a, b and c.
Solution: According to the given ques, On comparison, we get \[a\text{ }+\text{ }1\text{ }=\text{ }5\Rightarrow a\text{ }=\text{ }4\] \[b\text{ }+\text{ }2\text{ }=\text{ }0\Rightarrow b\text{...
Solve: x2 – 11/4 x + 15/8 = 0
According to ques, \[~{{x}^{2}}~\text{ }11/4\text{ }x\text{ }+\text{ }15/8\text{ }=\text{ }0\] Taking L.C.M , \[\left( 8{{x}^{2}}~\text{ }22x\text{ }+\text{ }15 \right)/\text{ }8\text{ }=\text{ }0\]...
3A x M = 2B; find matrix M.
Solution: According to the given question, \[3A\text{ }x\text{ }M\text{ }=\text{ }2B\] Suppose the order of the \[matrix\text{ }M\text{ }be\text{ }\left( a\text{ }x\text{ }b \right)\] Now, we know...
Solve : a2x2 – b2 = 0
According to ques, \[{{a}^{2}}{{x}^{2}}~\text{ }{{b}^{2}}~=\text{ }0\] or, \[{{\left( ax \right)}^{2}}~\text{ }{{b}^{2}}~=\text{ }0\] Or, \[\left( ax\text{ }+\text{ }b \right)\left( ax\text{ }\text{...
Evaluate:
Solution: As per the given question,
Solve: (i) The order of the matrix X. (ii) The matrix X.
Solution: (i) Suppose, the order of the matrix be \[a\text{ }x\text{ }b\] We know that Hence, for product of matrices to be possible \[a\text{ }=\text{ }2\] And, form the order of the resultant...
Solve: (2x + 3)2 = 81
According to ques, \[{{\left( 2x\text{ }+\text{ }3 \right)}^{2}}~=\text{ }81\] Taking square root both sides, \[2x\text{ }+\text{ }3\text{ }=\text{ }\pm \text{ }9\] Or, \[2x\text{ }=\text{ }\pm...
If given matrix , find x and y, if: (i) x, y ∈ W (whole numbers) (ii) x, y ∈ Z (integers)
Solution: According to the given question, \[{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }25\] And, \[-2{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }-2\] \[\left( i \right)\text{ }x,\text{ }y~\in ~W\text{ }\left(...
Find x and y, if:
Solution: On comparison, we get \[3x\text{ }+\text{ }18\text{ }=\text{ }15\] And \[12x\text{ }+\text{ }77\text{ }=\text{ }10y\] \[3x\text{ }=\text{ }-3\] And \[y\text{ }=\text{ }\left( 12x\text{...
Solve each of the following equations: 2x/x-3+1/2x+3+3x+9/(x-3)(2x+3)=0; x
3,x -3/2
According to ques, \[4{{x}^{2}}~+\text{ }6x\text{ }+\text{ }x\text{ }\text{ }3\text{ }+\text{ }3x\text{ }+\text{ }9\text{ }=\text{ }0\] Or, \[4{{x}^{2}}~+\text{ }10x\text{ }+\text{ }6\text{ }=\text{...
Find x and y, if:
Solution: On comparison, we get \[6x\text{ }-\text{ }10\text{ }=\text{ }8\] And \[-2x\text{ }+\text{ }14\text{ }=\text{ }4y\] \[6x\text{ }=\text{ }18\] And \[y\text{ }=\text{ }\left( 14\text{...
If the given matrix. simplify: A2 + BC.
Solution: \[{{A}^{2}}~+\text{ }BC\]
If given matrix. Then show that: (i) A(B + C) = AB + AC (ii) (B – A)C = BC – AC.
Solution: \[\left( i \right)\text{ }A\left( B\text{ }+\text{ }C \right)\] \[AB\text{ }+\text{ }AC\] So, \[A\left( B\text{ }+\text{ }C \right)\text{ }=\text{ }AB\text{ }+\text{ }AC\] \[\left( ii...
If given matrix and A2 = I, find a and b.
Solution: \[{{A}^{2}}\] Given, \[~{{A}^{2~}}=\text{ }I\] On comparison, we get \[1\text{ }+\text{ }a\text{ }=\text{ }1\] \[a\text{ }=\text{ }0\] And, \[-1\text{ }+\text{ }b\text{ }=\text{ }0\]...
The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.
let the present ages of father and his son be \[~x\text{ }years\text{ }and\text{ }\left( 45\text{ }\text{ }x \right)\text{ }years\] hence five years ago, Father’s age \[=\text{ }\left( x\text{...
Find the matrix A, if B =given matrix and B2 = B + ½A.
Solution: \[{{B}^{2}}\] \[{{B}^{2}}~=\text{ }B\text{ }+\text{ }{\scriptscriptstyle 1\!/\!{ }_2}A\] \[{\scriptscriptstyle 1\!/\!{ }_2}A\text{ }=\text{ }{{B}^{2}}-\text{ }B\] \[A\text{ }=\text{...
Solve : (i) (A + B)^2 (ii) A2 + B2
Solution: According to the given ques, \[\left( i \right)\text{ }\left( A\text{ }+\text{ }B \right)\] \[So,\text{ }{{\left( A\text{ }+\text{ }B \right)}^{2}}~=\text{ }\left( A\text{ }+\text{ }B...
The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.
According to ques, \[S\text{ }=\text{ }n\left( n\text{ }+\text{ }1 \right)\] Also, \[S\text{ }=\text{ }420\] So, \[n\left( n\text{ }+\text{ }1 \right)\text{ }=\text{ }420\] Or, \[{{n}^{2}}~+\text{...
Solve: (i) AB (ii) A^2 – AB + 2B
Solution: \[\left( \mathbf{i} \right)\text{ }\mathbf{AB}\text{ }\] \[\left( ii \right)\text{ }{{\mathbf{A}}^{\mathbf{2}}}-\text{ }\mathbf{AB}\text{ }+\text{ }\mathbf{2B}\]
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.
Let the speed of the second train be \[~x\text{ }km/hr.\] Then, the speed of the first train is \[\left( x\text{ }+\text{ }5 \right)\text{ }km/hr\] Let O be the position of the railway station,...
Solve: (i) A – B (ii) A^2
Solution: \[\left( \mathbf{i} \right)\text{ }\mathbf{A}\text{ }-\text{ }\mathbf{B}\text{ }\] \[\text{ }\left( \mathbf{ii} \right)\text{ }{{\mathbf{A}}^{\mathbf{2}~}}\]
BA = M^2, find the values of a and b.
Solution: $BA$ \[{{M}^{2}}\] So, \[BA\text{ }={{M}^{2}}\] On comparison, we get \[-2b\text{ }=\text{ }-2\] \[b\text{ }=\text{ }1\] And, \[a\text{ }=\text{ }2\]
If the given matrix and I is a unit matrix of the same order as that of M; show that: M2 = 2M + 3I
Solution: \[{{M}^{2}}\] \[2M\text{ }+\text{ }3I\] Hence, \[{{M}^{2}}~=\text{ }2M\text{ }+\text{ }3I\]
A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Let the usual speed of the plane to be \[x\text{ }km/hr\] The distance to travel \[=\text{ }1500km\] since, Time = Distance/ Speed As the ques suggests, \[{{x}^{2}}~+\text{ }250x\text{ }\text{...
Find A2 + AC – 5B
Solution: \[{{A}^{2}}\] $AC$ $5B$ \[{{A}^{2}}~+\text{ }AC\text{ }-\text{ }5B\text{ }=\]
Is the following possible: A^2
Solution: \[{{A}^{2}}~=\text{ }A\text{ }x\text{ }A,\text{ }\]isn’t possible because the number of columns isn’t equal to its number of rows in matrix A.
Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.
let the original number of persons to be x. According to ques, Total money which was divided is \[=\text{ }Rs\text{ }6500\] Each person’s share is \[=\text{ }Rs\text{ }6500/x\] Then, as the question...
Is the following possible: (i) AB (ii) BA
Solution: \[\left( i \right)\text{ }AB\] \[\left( ii \right)\text{ }BA\]
Solve: (i) (AB) C (ii) A (BC)
Solution: According to the given ques, \[\left( i \right)\text{ }\left( AB \right)\] \[\left( AB \right)\text{ }C\] \[\left( ii \right)\text{ }BC\] \[A\text{ }\left( BC \right)\] So, \[\text{...
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for: (i) the onward journey; (ii) the return journey. If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
According to ques, Distance \[=\text{ }400\text{ }km\] Average speed of the airplane \[=\text{ }x\text{ }km/hr\] Also, speed while returning \[=\text{ }\left( x\text{ }+\text{ }40 \right)\text{...
Find x and y, if:
Solution: (i) On comparison, we get \[5x\text{ }-\text{ }2\text{ }=\text{ }8\] \[5x\text{ }=\text{ }10\] \[x\text{ }=\text{ }2\] And, \[20\text{ }+\text{ }3x\text{ }=\text{ }y\] \[20\text{ }+\text{...
A hotel bill for a number of people for overnight stay is Rs 4800. If there were 4 people more, the bill each person had to pay, would have reduced by Rs 200. Find the number of people staying overnight.
Let the number of people staying overnight as x. According to ques, total hotel bill \[~=\text{ }Rs\text{ }4800\] Now,hotel bill for each person \[=\text{ }Rs\text{ }4800/x\] therefore,...
If find x and y when x and y when A2 = B.
Solution: \[{{A}^{2}}~\] \[{{A}^{2}}~=\text{ }B\] On comparison, we get \[4x\text{ }=\text{ }16\] \[x\text{ }=\text{ }4\] And, \[1\text{ }=\text{ }-y\] \[y\text{ }=\text{ }-1\]
If the given matrix and I is a unit matrix of order 2×2, find: (i) A^2 (ii) B^2A
Solution: (i) \[{{A}^{2}}\] (ii) \[~{{B}^{2}}\] \[{{B}^{2}}A\]
A trader buys x articles for a total cost of Rs 600. (i) Write down the cost of one article in terms of x. If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less. (ii) Write down the equation in x for the above situation and solve it for x.
According to ques, Number of articles \[=\text{ }x\] And, the total cost of articles \[=\text{ }Rs\text{ }600\] Again, (i) Cost of one article \[=\text{ }Rs\text{ }600/x\] (ii) also,...
If the given matrix and I is a unit matrix of order 2×2, find: (i) AI (ii) IB
Solution: (i) AI = (ii) IB=
If given matrix and I is a unit matrix of order 2×2, find: (i) AB (ii) BA
Solution: According to the given question (i) (ii)
The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate: (iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it. (iv) Hence, find the speed of the train.
(iii) According to the question, \[4x\text{ }+\text{ }1728\text{ }=\text{ }{{x}^{2}}~+\text{ }16x\] Or, \[{{x}^{2}}~+\text{ }12x\text{ }\text{ }1728\text{ }=\text{ }0\] Or, \[{{x}^{2}}~+\text{...
Evaluate: if possible: If not possible, give reason.
Solution: The product of the given matrices isn’t possible as per the rule the number of columns in the first matrix isn’t equal to the number of rows in the second matrix.
Evaluate: if possible: If not possible, give reason.
Solution: \[=\text{ }\left[ 6\text{ }+\text{ }0 \right]\text{ }=\text{ }\left[ 6 \right]\] \[=\text{ }\left[ -2+2\text{ }3-8 \right]\text{ }=\text{ }\left[ 0\text{ }-5 \right]\]
The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate: (i) the time taken by the car to reach town B from A, in terms of x; (ii) the time taken by the train to reach town B from A, in terms of x.
According to ques, Speed of car = \[x\text{ }km/hr\] Speed of train = \[\left( x\text{ }+\text{ }16 \right)\text{ }km/hr\] Time = \[Distance/\text{ }Speed\] (i)Time taken by the car to reach town B...
(i) find the matrix 2A + B. (ii) find a matrix C such that:
(ii) Solution: (i) \[2A\text{ }+\text{ }B\] (ii)
Solve:
Solution: According to the given question, the matrix is
From given data below find (i) 2A – 3B + C (ii) A + 2C – B
Solution: \[\left( i \right)\text{ }2A\text{ }-\text{ }3B\text{ }+\text{ }C\] \[\left( ii \right)\text{ }A\text{ }+\text{ }2C\text{ }-\text{ }B\]
Find x and y if: (i) 3[4 x] + 2[y -3] = [10 0]
(ii) Solution: From L.H.S, we have \[3\left[ 4\text{ }x \right]\text{ }+\text{ }2\left[ y\text{ }-3 \right]\] \[=\text{ }\left[ 12\text{ }3x \right]\text{ }+\text{ }\left[ 2y\text{ }-6 \right]~\]...
Evaluate:
(I) (ii) Solution: According to the given ques, (i) (ii)
Evaluate: (i) 3[5 -2]
(ii) Solution: (i) \[3\left[ 5\text{ }-2 \right]\text{ }=\text{ }\left[ 3\times 5\text{ }3x-2 \right]\text{ }=\text{ }\left[ 15\text{ }-6 \right]\] (ii)
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152. Find its common ratio.
According to the given ques, Hence, the common ratio is \[3/5.\]
How many terms of the series 2 + 6 + 18 + ….. must be taken to make the sum equal to 728?
. According to the given question, G.P: \[2\text{ }+\text{ }6\text{ }+\text{ }18\text{ }+\text{ }\ldots ..\] Here, \[a\text{ }=\text{ }2\] And \[r\text{ }=\text{ }6/2\text{ }=\text{ }3\] Also given,...
Find the sum of G.P.: 3, 6, 12, …., 1536.
According to the given question, G.P: \[3,\text{ }6,\text{ }12,\text{ }\ldots .,\text{ }1536\] So, \[a\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }1536\] and \[r\text{ }=\text{ }6/3\text{ }=\text{...
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.
According to the given question, For a G.P., \[r\text{ }=\text{ }3,\text{ }l\text{ }=\text{ }486\] and \[{{S}_{n}}~=\text{ }728\] \[1458\text{ }-\text{ }a\text{ }=\text{ }728\text{ }x\text{ }2\text{...
The 4th term and the 7th term of a G.P. are 1/27 and 1/729 respectively. Find the sum of n terms of the G.P.
According to the given question, \[{{t}_{4}}~=\text{ }1/27\] and \[{{t}_{7~}}=\text{ }1/729\] We know that, \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4~}}=\text{...
A boy spends Rs.10 on first day, Rs.20 on second day, Rs.40 on third day and so on. Find how much, in all, will he spend in 12 days?
Amount spent on \[{{1}^{st}}~day\text{ }=\text{ }Rs\text{ }10\] Amount spent on \[{{2}^{nd}}~day\text{ }=\text{ }Rs\text{ }20\] Amount spent on \[{{3}^{rd}}~day\text{ }=\text{ }Rs\text{ }40\]...
The first term of a G.P. is 27 and its 8th term is 1/81. Find the sum of its first 10 terms.
\[First\text{ }term\text{ }\left( a \right)\text{ }of\text{ }a\text{ }G.P\text{ }=\text{ }27\] And, \[{{8}^{th}}~term\text{ }=\text{ }{{t}_{8}}~=\text{ }a{{r}^{8\text{ }-\text{ }1}}~=\text{ }1/81\]...
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?
According to the given question, G.P: \[1\text{ }+\text{ }4\text{ }+\text{ }16\text{ }+\text{ }64\text{ }+\text{ }\ldots \ldots ..\] Here, \[a\text{ }=\text{ }1\text{ }and\text{ }r\text{ }=\text{...
Find the sum of G.P.:
(i) (ii) Solution: (i) According to the given question Here, \[a\text{ }=\text{ }\left( x\text{ }+\text{ }y \right)/\text{ }\left( x\text{ }-\text{ }y \right)\] And \[~r\text{ }=\text{ }1/\left[...
Find the sum of G.P.: (i) 1 – 1/2 + 1/4 – 1/8 + …….. to 9 terms (ii) 1 – 1/3 + 1/32 – 1/33 + ……… to n terms
(i) According to the given question G.P: \[1\text{ }-\text{ }1/2\text{ }+\text{ }1/4\text{ }-\text{ }1/8\text{ }+\text{ }\ldots \ldots ..\text{ }to\text{ }9\text{ }terms\] Here, \[a\text{ }=\text{...
Find the sum of G.P.: (i) 1 + 3 + 9 + 27 + ………. to 12 terms (ii) 0.3 + 0.03 + 0.003 + 0.0003 +….. to 8 terms.
(i) According to the given question G.P: \[1\text{ }+\text{ }3\text{ }+\text{ }9\text{ }+\text{ }27\text{ }+\text{ }\ldots \ldots \ldots .\text{ }to\text{ }12\text{ }terms\] Here, \[a\text{ }=\text{...
If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that: (q – r) log a + (r – p) log b + (p – q) log c = 0
The first term of the G.P. be A and its common ratio be R. Hence, \[{{p}^{th}}~term\text{ }=\text{ }a\text{ }\Rightarrow \text{ }A{{R}^{p\text{ }-\text{ }1}}~=\text{ }a\] \[{{q}^{th}}~term\text{...
Find the G.P. 1/27, 1/9, 1/3, ……, 81; find the product of fourth term from the beginning and the fourth term from the end.
According to the given question, G.P. \[1/27,\text{ }1/9,\text{ }1/3,\text{ }\ldots \ldots ,\text{ }81\] Here, \[a\text{ }=\text{ }1/27,\] \[common\text{ }ratio\text{ }\left( r \right)\text{...
Find the third term from the end of the G.P. 2/27, 2/9, 2/3, ……., 162
Given series: \[2/27,\text{ }2/9,\text{ }2/3,\text{ }\ldots \ldots .,\text{ }162\] Here, \[a\text{ }=\text{ }2/27\] \[r\text{ }=\text{ }\left( 2/9 \right)\text{ }/\text{ }\left( 2/27 \right)\]...
Prove that : (v) 
Prove that : (iii) 
(iv) 
(iii) $\sin \left(28^{\circ}+\mathrm{A}\right)=\sin \left[90^{\circ}-\left(62^{\circ}-\mathrm{A}\right)\right]=\cos \left(62^{\circ}-\mathrm{A}\right)$ (iv)
Find the seventh term from the end of the series: √2, 2, 2√2, …… , 32
Given series: \[\surd 2,\text{ }2,\text{ }2\surd 2,\text{ }\ldots \ldots \text{ },\text{ }32\] Here, \[a\text{ }=\text{ }\surd 2\] \[r\text{ }=\text{ }2/\text{ }\surd 2\text{ }=\text{ }\surd 2\]...
Prove that: (i) 
(ii) 
(i) $\tan \left(55^{\circ}+x\right)=\tan \left[90^{\circ}-\left(35^{\circ}-x\right)\right]=\cot \left(35^{\circ}-x\right)$ (ii) $\sec \left(70^{\circ}-\theta\right)=\sec...
Evaluate: (vii) 
(viii) 
(vii) $3 \cos 80^{\circ} \operatorname{cosec} 10^{\circ}+2 \cos 59^{\circ} \operatorname{cosec} 31^{\circ}$ $=3 \cos (90-10)^{0} \operatorname{cosec} 10^{\circ}+2 \cos (90-31)^{0}...
The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term
According to the given question, Product of \[{{3}^{rd}}~and\text{ }{{8}^{th}}\] terms of a G.P. is \[243\] The general term of a G.P. First term \[a\] And Common ratio \[r\]is given by,...
Evaluate : (v) 
(vi) 
(v) $\sin 27^{\circ} \sin 63^{\circ}-\cos 63^{\circ} \cos 27^{\circ}$ $=\sin (90-63)^{0} \sin 63^{\circ}-\cos 63^{\circ} \cos (90-63)^{\circ}$ $=\cos 63^{\circ} \sin 63^{\circ}-\cos 63^{\circ} \sin...
Evaluate : (iii) 
(iv) 
(iii) (iv) $\cos 40^{\circ} \operatorname{cosec} 50^{\circ}+\sin 50^{\circ}$ sec $40^{\circ}$ $=\cos (90-50)^{0} \operatorname{cosec} 50^{\circ}+\sin (90-50)^{0} \sec 40^{\circ}$ $=\sin 50^{\circ}...
Evaluate:(i) 
(ii) 
(i) (ii) 1+1=2
If the first and the third terms of a G.P are 2 and 8 respectively, find its second term.
According to the given question, \[{{t}_{1}}~=\text{ }2\text{ }and\text{ }{{t}_{3}}~=\text{ }8\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{1}}~=\text{...
Fourth and seventh terms of a G.P. are 1/18 and -1/486 respectively. Find the G.P.
Given, \[{{t}_{4}}~=\text{ }1/18\text{ }and\text{ }{{t}_{7}}~=\text{ }-1/486\] General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So, \[{{t}_{4}}~=\text{ }a{{r}^{4\text{ }-\text{...
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
According to the given question, \[{{t}_{5}}~=\text{ }81\text{ }and\text{ }{{t}_{2}}~=\text{ }24\] We know that, General term is \[{{t}_{n}}~=\text{ }a{{r}^{n\text{ }-\text{ }1}}\] So,...
(i) If 
, show that: 
(ii) If 
, show that: 
(i) Since, $2 \sin \mathrm{A}-1=0$ Therefore, $\sin A=1 / 2$ since, $\sin 30^{\circ}=1 / 2$ Hence, $\mathrm{A}=30^{\circ}$ LHS = $\sin 3 A=\sin 3\left(30^{\circ}\right)=\sin 30^{\circ}=1$...
If tan A = n tan B and sin A = m sin B, prove that: cos^2 A = m^2 – 1/ n^2 – 1
$\tan A=n \tan B$ $n=\tan A / \tan B$ And, $\sin A=m \sin B$ $\mathrm{m}=\sin \mathrm{A} / \sin \mathrm{B}$ Substituting RHS in $\mathrm{m}$ and $\mathrm{n}$ $m^{2}-1 / n^{2}-1$...
If 
, show that: 
RHS, $\left(p^{2}-1\right) /\left(p^{2}+1\right)$ $=\frac{(\sec A+\tan A)^{2}-1}{(\sec A+\tan A)^{2}+1}$ $=\frac{\sec ^{2} A+\tan ^{2} A+2 \tan A \sec A-1}{\sec ^{2} A+\tan ^{2} A+2 \tan A \sec...
If 
and 
, show that: 
LHS = $a^{2} / x^{2}-b^{2} / y^{2}$ $=\frac{a^{2}}{a^{2} \cos ^{2} \theta}-\frac{b^{2}}{b^{2} \cot ^{2} \theta}$ $=\frac{1}{\cos ^{2} \theta}-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$...
If sin A + cos A = p and sec A + cosec A = q, then prove that: q(p^2 – 1) = 2p
LHS = $q\left(p^{2}-1\right)=(\sec A+\operatorname{cosec} A)\left[(\sin A+\cos A)^{2}-1\right]$ $=(\sec A+\operatorname{cosec} A)\left[\sin ^{2} A+\cos ^{2} A+2 \sin A \cos A-1\right]$ $=(\sec...
Which term of the G.P. :
Solution: In the given G.P. First term, \[a\text{ }=\text{ }-10\] Common ratio, \[r\text{ }=\text{ }\left( 5/\surd 3 \right)/\text{ }\left( -10 \right)\text{ }=\text{ }1/\left( -2\surd 3 \right)\]...
Prove: (xv) 
(xvi) 
(xv) LHS = $\sec ^{4} \mathrm{~A}\left(1-\sin ^{4} \mathrm{~A}\right)-2 \tan ^{2} \mathrm{~A}$ $=\sec ^{4} \mathrm{~A}\left(1-\sin ^{2} \mathrm{~A}\right)\left(1+\sin ^{2} \mathrm{~A}\right)-2 \tan...
Prove : (xvii) 
$(1+\tan A+\sec A)(1+\cot A-\operatorname{cosec} A)$ $=1+\cot A-\operatorname{cosec} A+\tan A+1-\sec A+\sec A+\operatorname{cosec} A-\operatorname{cosec} A \sec A$ $=2+\cos \mathrm{A} / \sin...
Prove : (xiii) 
(xiv) 
LHS = = RHS (xiv) LHS = = RHS
The mid-point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
According to the given question, The mid-point of \[\left( 2a,\text{ }4 \right)\text{ }and\text{ }\left( -2,\text{ }2b \right)\text{ }is\text{ }\left( 1,\text{ }2a\text{ }+\text{ }1 \right)\] By...
Prove : (xi) 
(xii) 
LHS = = RHS (xii) LHS = = RHS
Prove (ix) 
(x) 
LHS= = RHS (x) LHS =
Prove : (vii) 
(viii) 
(vii) LHS =$=(\sin A /(1-\cos A))-\cot A$ Since, $\cot A=\cos A / \sin A$ $=\left(\sin ^{2} A-\cos A+\cos ^{2} A\right) /(1-\cos A) \sin A$ $=(1-\cos A) /(1-\cos A) \sin A$ $=1 / \sin \mathrm{A}$...
Prove: (v) 
(vi) 
(v) LHS= cot A/ (1 – tan A) + tan A/ (1 – cot A) = RHS (vi) LHS= cos A/ (1 + sin A) + tan A = RHS
The mid-point of the line-segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
According to the given question, The mid-point of the line-segment joining \[\left( 4a,\text{ }2b\text{ }-\text{ }3 \right)\text{ }and\text{ }\left( -4,\text{ }3b \right)\text{ }is\text{ }\left(...
Prove: (iii) 
(iv) 
(iii) LHS= 1 – sin2 A/ (1 + cos A) = RHS (iv) LHS= (1 – cos A)/ sin A + sin A/ (1 – cos A) = RHS
Find the co-ordinates of the centroid of a triangle ABC whose vertices are: A (-1, 3), B (1, -1) and C (5, 1)
By the centroid of a triangle formula, we get the co- ordinates of the centroid of triangle \[ABC\]as \[=\text{ }\left( 5/3,\text{ }1 \right)\]
Use tables to find sine of: (i) 21° (ii) 34° 42′
(i) Taking LHS, $1 /(\cos A+\sin A)+1 /(\cos A-\sin A)$ (ii) Taking LHS, $\operatorname{cosec} A-\cot A$ $=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$ $=\frac{1-\cos A}{\sin A}$ $=\frac{1-\cos A}{\sin...
AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7), find (i) the length of radius AC (ii) the coordinates of B.
(i) \[Radius\text{ }AC\text{ }=\text{ }\surd [\text{ }{{\left( 3\text{ }+\text{ }2 \right)}^{2}}~+\text{ }{{\left( -7\text{ }-\text{ }5 \right)}^{2}}~]\] \[=\text{ }\surd [\text{ }({{5}^{2}}~+\text{...
Use trigonometrical tables to find tangent of: (iii) 17° 27′
(iii) $\tan 17^{\circ} 27^{\prime}=\tan \left(17^{\circ} 24^{\prime}+3^{\prime}\right)=0.3134+0.0010=0.3144$
Use trigonometrical tables to find tangent of: (i) 37° (ii) 42° 18′
(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...
If 
and 
, then prove that: 
\[\begin{array}{*{35}{l}} {{m}^{2}}~+\text{ }{{n}^{2}} \\ =\text{ }{{\left( x\text{ }cos\text{ }A\text{ }+\text{ }y-\text{ }sin\text{ }A \right)}^{2}}~+\text{ }{{\left( x\text{ }sin\text{ }A\text{...