**1. If sun’s elevation is 45****°. ****Find the length of the shadow of a pole of height 8 m.**

In rt. Δ PQR,

∴ tan 45° =

1 =

**2. A pole 8 metre high cast a shadow 8 m on the ground. Find the sun’s elevation.**

In rt. Δ PQR,

∴ PQ = 8 m

QR = 8 m

Let

∴ tan

=

** **

**3. If the angle of elevation of the top of a hill from two points at distances p and q from the base and in the same straight line with it are complementary. Find the height of the hill.**

Let PQ = h be the height of the hill,

QR = b, QS = a, such that

In rt. Δ PQS, we have

h = a tan

In rt. Δ PQR, we have

h = b cot

Multiplying (1) and (2), we get

**4. From the figure, find the angle of depression of point Z from the point R.**

**5. Find the angle of elevation of the moon when the shadow of a tree h metres high is 3–√h metres long.**

The height of the pole be h m and length of its shadow is

Let

Consider rt.agl Δ ABC

tan

= tan

** 6. ****A ladder 14 metres long just reaches the top of a wall. If the ladder makes an angle of 45° with the wall, find the height of the wall.**

Here, length of the ladder is 14m and angle of elevation is 90°-45°=45°.

Let h m be the height of the wall

Consider rt.agl Δ PQR

h =

∴ The length of the wall is

** **

** 7. ****An observer 1.8 m tall is 20.8 m away from a tower 22.6 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.**

Here, PQ be the observer and RS be the tower of height 24 m.

PQ = TS = 1.8 m

RT = RS – TS

= 22.6 – 1.8

= 20.8 m

Now, Consider rt.agl ΔPTR, we have

1 = tan

tan

**8.From a balloon vertically above a straight road, the angles of depression of two cars at an instant is found to be 45° and 60°. If the car are 100 m apart, find the height of the balloon.**

Let the height of the balloon at P be h metres. Let A and B be the two cars. Thus, AB = 100m.

∠ PAQ= 45° and ∠ PBQ=60°

Now, Consider rt.agl ΔAQP,

Now, Consider rt.agl ΔPBQ,

h =

h =

h =

h =

=

** **

** 9. ****The shadow of a tower on a level surface is 40 m long when moon’s elevation is 30° then when it is 60°.Find the height of the tower.**

XY=40 m,

Now, Consider rt.agl ΔRYS

Consider rt.agl ΔRXQ

3h = 40

2h = 40

h = 20

∴ The height of the tower is 20

**10. From the top of a tree h m high, the angle of depression of two objects, which are in the line with foot of the tree are γ and δ (δ> γ). Find the distance between the two objects.**

Let RS be the tree of height h m, X and Y are two objects such that

Now, Consider rt.agl ΔYSR

Now, Consider rt.agl ΔXSR

XY = h (cot

**11. The angle of elevation of the top of a tower from two distinct points x and y from its foot are complementary. Prove that the height of the tower is xy−−√.**

Let h m be the height of the tower PQ. X and Y are two distinct points from its foot, such that PX = x and PY = y

Now, Consider rt.agl ΔXQP, we obtain

h = x tan

Consider rt.agl ΔYQP, we obtain

h = y

Multiply (I) and (ii), we get

h =

∴ The height of the tower is

**12. The angle of elevation of the top of a tower from certain point at 30°. If the observer moves 10 m towards the tower, the angle of elevation of the top increases by 15° . Find the height of the tower.**

Let h m be the height of the tower RS and P, Q are two points 10 m apaet, such that

Now, Consider rt.agl ΔQRS

*h* m

Now, Consider rt.agl ΔPRS

(

∴ The height of the tower is 5(

**13. The angle of elevation of the top of a vertical tower from a point on the groung is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.**

Let XY be the vertical tower of height *h* m.

C and D are two points 10 m apart such that,

CD = ZY = 10 m

XZ = XY – ZY = (h – 10) m

Consider rt.agl ΔDYX

Consider rt.agl ΔCZX

XZ = CZ

H – 10 =

h

h

h =

=

= 5(3 +

**14. The angle of elevation of the top of a tower 27 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and the height of the other tower.**

Let us assume that PQ and RS be the two tower of height h m and 27 m respectively.

Consider rt.agl ΔRQS

Consider rt.agl ΔPSQ

From (i) and (ii), we have

From (ii), we get QS = 9

∴ The distance between two towers is 9

**15. The lower window of a house is at a height of 3 m above the ground and its upper window is 6 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30° ****respectively. Find the height of the balloon from the ground.**

Let P and Q be the positions of two windows of a house, such that PR = 3 m, QR = 9 m. Let U be the position of balloon of height h m above the ground.

Consider rt.agl ΔQTU

h – 9 = QT

Consider rt.agl ΔUPS

h – 3 = PS

Also, QT = PS …..(iii)

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

∴ height of the balloon above the ground is 12 m.