Question

The angle of elevation of the top of a rock fromt he top and foot of a 60m high tower are 45 and 60 respectively find the height of the rock

Answer 1

Answer:

The height of rock is 142 m (approx)

Step-by-step explanation:

In the figure attached below, AB represents the rock and let its height be 'h' m and CD represents 60 m high tower.

Then EBCD forms a rectangle thus, EB= CD = 60 m

Let the distance between foot of rock and tower is 'x' m then BD = EC = x

The angle of elevation of the top of a rock from the top and foot of tower are 45 and 60 respectively.

In ΔAEC ,

$\tan 45^\circ=\frac{AE}{EC}$

$\tan 45^\circ=\frac{h-60}{x}$

We know tan 45°= 1

$1=\frac{h-60}{x}$

$x=h-60$ .............(1)

Also , In ΔABD,

$\tan 60^\circ=\frac{AB}{BD}$

$\tan 60^\circ=\frac{h}{x}$

we know, tan 60° =√3

$\sqrt{3}=\frac{h}{x}$

$x\sqrt{3}={h}$  ...........(2)

From (1) and (2) , we get,

$(h-60)\sqrt{3}={h}$

$60\sqrt{3}=\sqrt{3}h-h$

$60\sqrt{3}=(\sqrt{3}-1)h$

$\frac{60\sqrt{3}}{(\sqrt{3}-1)} =h$

We know value of √3 = 1.732

$\frac{60 \times 1.732}{0.732} =h$

$h=141.96721311$

Thus, the height of rock is 142 m (approx).

 

Answer 2

Answer:

147.2

Step-by-step explanation: