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**Q.9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.****Sol. **Given: height of the tower AB = 50 m

∠ACB = 60°, ∠DBC = 30°

Let the height of the building

CD = x m

Hence, height of the building = 50/3 m**Q.10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles.****Sol. **Let AB = CD = h m [Height of the poles]

Given: BC = 80 m [Width of the road]

Let CE = x m

∴ BE = (80 - x) m

In ΔCDE,

In ΔABE,

... (ii)

From equation (i) and (ii), we get

Substituting h in equation (i),

Hence, position of the point is at a distance of 60 m from pole CD and 20 m from pole AB.**Q.11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see figure). Find the height of the tower and the width of the canal.**

**Sol. **Let the height of the tower AB = h m and BC be the width of the canal.

Given: ∠ACB = 60° and ∠ADB = 30°

⇒

⇒

⇒

Hence, the height of the tower = 10√3 m and width of the canal = 10 m.**Q.12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.****Sol. **Let height of the tower AB = (h + 7) m

Given: CD = 7m (height of the building),

∠ACE = 60°, and ∠ECB = 45°

⇒ ∠CBD = 45º

⇒

⇒

**Q.13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.****Sol. **Given: height of the lighthouse = 75 m

Let C and D are the positions of two ships.

We have ∠XAD = ∠ADB = 30°

and ∠XAC = ∠ACB = 45°

⇒

Hence, the distance between two ships is 54.75 m.**Q.14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the balloon during the interval.**

**Sol**. Let the first position of the balloon is A and after some time it will reach to the point D. The vertical height ED = AB = (88.2 - 1.2) m = 87 m.

Distance travelled by the balloon from A to D is BE.

So, BE = CE - CB**Q.15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.****Sol. **Let the height of the tower AB = h m

Given: ∠XAD = ∠ADB = 30°

and ∠XAC - ∠ACB = 60°

Let the speed of the car = x m/sec

From equation (i) and (ii), we get

Hence, time taken from C to B = 3 sec.**Q.16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.****Sol. **Let the height of the tower AB = h m

We have PB = 4 m, QB = 9 m

Let ∠AQB = θ [Both are complementary angles]

Then ∠APB = 90° - θ

⇒

⇒ h = 4 cot θ .......(i)

h = 9 tan θ ....(ii)

From equation (i) and (ii), we get

Hence, the height of the tower is 6 m.

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