CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The angle of elevation of the top of a hill from the foot of a tower is $$60^{\circ}$$ and the angle of elevation of the top of the tower from the foot of the hill is $$30^{\circ}$$. If the tower is $$50$$m high, then find the height of the hill.


Solution

Let $$AD$$ be the height of tower and $$BC$$ be the height of the hill.

Given $$\angle CAB = 60^{\circ}, \angle ABD=30^{\circ}$$ and $$AD=50$$m.

Let $$BC=h$$ metres.

Now, in the right angled $$\displaystyle \bigtriangleup DAB, \tan 30^{\circ}=\frac{AD}{AB}$$

$$\displaystyle \Rightarrow AB=\frac{AD}{\tan30^{\circ}}$$

$$\therefore AB=50\sqrt{3}$$m                (1)

Also, in the right angled $$\displaystyle \bigtriangleup CAB, \tan 60^{\circ}=\frac{BC}{AB}$$

$$\Rightarrow BC=AB\tan 60^{\circ}$$

Thus, using (1) we get $$ h=BC=(50\sqrt{3})\sqrt{3}=150$$m

Hence, the height of the hill is $$150$$m.


1028086_622668_ans_b19609153e7541e5af751f9b4dee331f.jpg

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image