**Problem 1 :**

A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45° . The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30° . Determine the speed at which the bird flies.(√3 = 1.732)

**Solution :**

Let "x" be the speed at which the bird flies.

Time = distance / speed

2 = distance / x

Distance covered = 2x

In triangle ABC,

tan θ = Opposite side / Adjacent side

tan 30 = AB / BC

1/√3 = 80/ BC

BC = 80√3 ----(1)

tan 45 = AB / CD

1 = 80/(BC + CD)

1 = 80/(BC + 2x)

BC + 2x = 80

BC = 80 - 2x ------(2)

(1) = (2)

80√3 = 80 - 2x

By dividing the entire equation by 2, we get

40√3 = 40 - x

x = 40 - 40√3

x = 40(1 - 1.732)

x = 40 (-0.732)

x = -29.28

Because, the speed can not be negative, speed of the bird is 29.28 m/s

**Problem 2 :**

An aeroplane is flying parallel to the Earth’s surface at a speed of 175 m/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth’s surface is 37° at a given point. After what period of time does the angle of elevation increase to 53° ? (tan 53° = 1.3270, tan 37° = 0.7536)

**Solution :**

Let "x" be the required time.

In triangle DBC,

tan 53 = DC/BC

1.3270 = 600/BC

BC = 600/1.320 ----(1)

In triangle ADC,

tan 37 = DC/AC

0.7536 = 600/(175x + BC)

175x + BC = 600/0.7536

BC = (600/0.7536) - 175x----(2)

(1) = (2)

600/1.3270 = (600/0.7536) - 175x

175x = (600/0.7536) - (600/1.3270)

175x = 600(1/0.7536 - 1/1.3270)

175x = 600(1.32 - 0.75)

175x = 600(0.57)

175x = 342

x = 342/175

x ≈ 1.95 seconds

**Problem 3 :**

A bird is flying from A towards B at an angle of 35° , a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away.

(i) How far is B to the North of A?

(ii) How far is B to the West of A?

(iii) How far is C to the North of B?

(iv) How far is C to the East of B?

(sin 55° = 0.8192, cos 55° = 0.5736, sin 42° = 0.6691, cos 42° = 0.7431)

**Solution :**

(i) To find how far is B to the North of A, we should find the length of AC.

sin θ = Opposite side/hypotenuse

cos θ = Adjacent side/hypotenuse

sin 55 = AC/30

0.8192 = AC/30

AC = 0.8192(30)

AC = 24.58 km

(ii) To find how far is B to the west of A, we should find the length of BC.

cos 55 = BC/30

0.5736 = BC/30

BC = 0.5736(30)

BC = 17.21 km

(iii) To find how far is C to the north of B, we should find the value of BD.

sin 42 = BD/32

0.6691 = BD/32

BD = 0.6691(32)

BD = 21.41

(iv) To find how far is C to the east of B, we should find the length of DE.

cos 42 = DE/32

0.7341 = DE/32

DE = 0.7341(32)

DE = 23.49 km

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