Question

The angle of elevation of the top of a rock from the top and foot of a 60m high tower are 45° and 60° respectively. Find the height of the rock. (√3=1.732) ​

Answer 1

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}tan45∘=ECAE​
\tan 45^\circ=\frac{h-60}{x}tan45∘=xh−60​
We know tan 45°= 1
1=\frac{h-60}{x}1=xh−60​
x=h-60x=h−60 .............(1)
Also , In ΔABD,
\tan 60^\circ=\frac{AB}{BD}tan60∘=BDAB​
\tan 60^\circ=\frac{h}{x}tan60∘=xh​
we know, tan 60° =√3
\sqrt{3}=\frac{h}{x}3​=xh​
x\sqrt{3}={h}x3​=h  ...........(2)
From (1) and (2) , we get,
(h-60)\sqrt{3}={h}(h−60)3​=h
60\sqrt{3}=\sqrt{3}h-h603​=3​h−h
60\sqrt{3}=(\sqrt{3}-1)h603​=(3​−1)h
\frac{60\sqrt{3}}{(\sqrt{3}-1)} =h(3​−1)603​​=h
We know value of √3 = 1.732
\frac{60 \times 1.732}{0.732} =h0.73260×1.732​=h
h=141.96721311h=141.96721311
Thus, the height of rock is 142 m (approx)

Answer 2

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