Solution: Consider P as the man standing on the deck of the ship which is 20 m above the sea level and B is the bird Angle of elevation of the bird from P = 300 Angle of depression from P to the...

Solution: Width of the river PQ = 100 m B is the island and AB is the tree on it Angles of elevation from A to P and Q are 300 and 450 Consider AB = h PB = x BQ = 100 – x In right triangle APB tan θ...

Solution: Consider A as the man on the deck of a ship B and CE is the cliff AB = 16 m Angle of elevation from the top of the cliff = 450 Angle of depression at the base of the cliff = 300 Take CE =...

Solution: Consider A and D as the two positions of the aeroplane AB is the height and P is the point AB = 1 km Take AD = x and PB = y Angles of elevation from A and D at the point P are 600 and 300...

Solution: Consider PQ as the tower = h XQ = YR = y XY = 40 m PR = h – 40 In right triangle PXQ tan θ = PQ/XQ Substituting the values tan 600 = h/y So we get √3 = h/y y = h/√3 ….. (1) In right...

Solution: Consider AB as the boy and CD as the wall which is at a distance of 2.1 m AB = 1.54 m CD = 3.64 m BD = 2.1 m Construct AE parallel to BD ED = 1.54 m CE = 3.64 – 1.54 = 2.1 m AE = BD = 2.1...

 A boy 1.6 m tall is 20 m away from a tower and observes that the angle of elevation of the top of the tower is 600. Find the height of the tower. Solution: Consider AB as the boy and TR as the...

A 7 m long flagstaff is fixed on the top of a tower. From a point on the ground, the angles of elevation of the top and bottom of the flagstaff are 450 and 360 respectively. Find the height of the...

 An aeroplane 3000 m high passes vertically above another aeroplane at an instant when the angles of elevation of the two aeroplanes from the same point on the ground are 600 and 450 respectively....

The angle of elevation of the top of a tower from a point A (on the ground) is 300. On walking 50 m towards the tower, the angle of elevation is found to be 600. Calculate (i) the height of the...

 An aircraft is flying at a constant height with a speed of 360 km/h. From a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 450. After 20 seconds, the...

Solution: Consider TR as the tower where TR = h BR = x AB = 10 m Angles of elevation from the top of the tower at A and B are 300 and 450 In right triangle TAR tan θ = TR/AR Substituting the values...

The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 450. How much higher must the tower be raised so that its angle of elevation at the same point may...

Solution: Consider AB as the building where AB = 20 m CD as the monument where CD = x m Take the distance between the building and the monument as y In right triangle BCD tan θ = CD/BD Substituting...

Solution: Consider TR as the tower PL as the pole on the same level Ground PL = 20 m From the point P construct PQ parallel to LR ∠TPQ = 600 and ∠QPR = 300 Here ∠PRL = ∠QPR = 300 which are the...

A pole of height 5 m is fixed on the top of a tower. The angle of elevation of the top of the pole as observed from a point A on the ground is 600 and the angle of depression of the point A from the...

Solution: Consider AB as the height and CD as the building The angles of depression from A to C and D are 300 and 450 ∠ACE = 300 and ∠ADB = 450 CD = 8 m Take AB = h and BD = x From the point C...

Solution: Consider AB as the lamp post CD is the height of man BD is the distance of man from the foot of the lamp FD is the shadow of man Construct CE parallel to DB Take AB = x and CD = 1.8 m EB =...

 From a tower 126 m high, the angles of depression of two rocks which are in a horizontal line through the base of the tower are 160 and 120 20’. Find the distance between the rocks if they are on...

(i) The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 600 and 450 respectively. If the two ships are on the opposite sides of the light house,...

Solution: In triangle DBC tan 600 = 10/BC Substituting the values √3 = 10/BC BC = 10/√3 In triangle DBC tan 300 = 10/ (BC + AB) Substituting the values 1/√3 = 10/[10/√3 + AB] By further calculation...

Solution: Consider CD as the pillar of x m Angles of elevation of points A and B are 450 and 600 It is given that AB = 15 m Take AD = y DB = 15 – y In right triangle CAD tan θ = CD/AD Substituting...

As observed from the top of a 80 m tall light house, the angles of depression of two ships on the same side of the light house in horizontal line with its base are 300 and 400 respectively. Find the...

Solution: Consider the height of the first tower TR = x It is given that Height of the second tower PQ = 60 m Distance between the two towers QR = 140 m Construct PL parallel to QR LR = PQ = 60 m PL...

Solution: Consider CB = x and DB = y AB = 300 m In right triangle ACD tan θ = AB/CB Substituting the values tan 450 = 300/x 1 = 300/x So we get x = 300 m In right triangle ADB tan θ = AB/DB...

Solution: Consider CH as the height of the church A and B are two vehicles which make an angle of depression x0 and y0 from C Take AH = x and BH = y In a right triangle CBH tan x0 = CH/AH = 96/y...

 In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x0 where tan x = 2/5 and AF = 200 m. The elevation of T from B, where AB = 80 m, is y0. Calculate: (i) the height of...

Solution: Consider TR as the tower and P as the point on the ground such that tan θ = 5/12 tan α = ¾ PQ = 192 m Take TR = x and QR = y In right triangle TQR tan α = TR/QR = x\y So we get 3/4 = x/y y...

Solution: It is given that AB is a building CD = 60 m In triangle ABC tan 600 = AB/BC √3 = AB/BC So we get BC = AB/√3 ….. (1) In triangle ABD tan 300 = AB/BD 1/√3 = AB/ (BC + 60) By cross...

Solution: Consider A and B as the position of two consecutive kilometre stones Here AB = 1 km = 1000 m Take Distance BC = x m Distance AC = (1000 + x) m In right angled triangle BCD CD/BC = tan 450...

Solution: In the figure AB is the tower BD and BC are the shadow of the tower in two situations Consider BD = x m and AB = h m In triangle ABD tan 450 = h/x So we get 1 = h/x h = x ….. (1) In...

Solution: Consider TR as the tree and PR as the width of the river.   Take TR = X and PR = y In right triangle TPR tan θ = TR/PR Substituting the values tan 600 = x/y So we get √3 = x/y x = y...

Solution: Consider the distance between two planes = h m It is given that AD = 3125 m, ∠ACB = 600 and ∠ACD = 300 In triangle ACD tan 300 = AD/AC Substituting the values 1/√3 = 3125/AC AC = 3125√3...

Solution: Consider AB as the building and H as the helicopter hovering over it P is a point on the ground Angle of elevation of the top of building and helicopter are 300 and 600 We know that Height...

Solution: Consider CB as the cliff and AC as the pillar D as the boat which is 300 m away from the foot of the cliff BD = 300 m Angle of elevation of the top and foot of the pillar are 550 40’ and...

Solution: In the figure, AB is the tower and CD is an observer θ is the angle of observation It is given that AB = 22m CD = 1.5 m Distance BD = 20.5 m From the point C construct CE parallel tp DB AE...

Consider TR as the total height of the tree TP as the broken part which touches the ground at a distance of 6 m from the foot of the tree which makes an angle of 380 30’ with the ground Take PR = x...

Solution: Consider AB as the tower Take a man C stands at a distance x m from the foot of the tower cos θ = 0.53 We know that Height of the tower AB = 20 m cos θ = 0.53 So we get θ = 580 Let us take...

A bridge across a river makes an angle of 450 with the river bank. If the length of the bridge across the river is 200 metres, what is the breadth of the river. olution: Consider AB as the width of...

Consider AB as the pole and AC as the wire which makes an angle of 450 with the ground. Height of the pole AB = 10 m Consider x m as the length of wire AC We know that sin θ = AB/AC Substituting the...

Consider AB as the height of the kite A and AC as the string Angle of elevation of the kite = 260 32’ Take AB = x m and AC = 100 m We know that sin θ = AB/AC Substituting the values sin 260 32’ =...

Solution: Consider AB as the cliff whose height is 92 m So we get x = 92 × 2.7475 x = 252.7700 m Hence, the distance of the buoy from the foot of the cliff is 252.77 m.

Solution: Consider AB as the tower and P is at a distance of x m from B which is the foot of the tower. Height of the tower = 100 m Hence, the distance of P from the foot of the tower is 173.2 m....

Solution: Consider AB as the tree and BC as the width of the river   Hence Proved

Solution: Consider AB as the pole and CB as its shadow θ is the angle of elevation of the sun Take AB = x m and BC = x m We know that tan θ = AB/CB = x/x = 1 So we get Hence...

Solution: Consider AB as the wall and AC as the ladder whose foot C is 1.5 m away from B Hence, the height of wall is 2.6 m.

Consider BC as the tower and A as the point on the ground such that So we get Hence Proved  

Consider AB as the pole and OB as its shadow. It is given that Hence Proved