Login/Sign Up
Home » The horizontal distance between two towers is 60 meters. The angle of depression of the top of the first tower when seen from the top of the second tower is . If the height of the second tower is 90 meters. Find the height of the first tower.
The horizontal distance between two towers is 60 meters. The angle of depression of the top of the first tower when seen from the top of the second tower is 30^{\circ}. If the height of the second tower is 90 meters. Find the height of the first tower.
CBSE | Class 10 | Maths | RS Aggarwal | Some Applications Of Trigonometry
The horizontal distance between two towers is 60 meters. The angle of depression of the top of the first tower when seen from the top of the second tower is 30^{\circ}. If the height of the second tower is 90 meters. Find the height of the first tower.

Let \mathrm{DE} be the first tower and \mathrm{AB} be the second tower.

=>  \mathrm{AB}=90 \mathrm{~m} and \mathrm{AD}=60 \mathrm{~m} such that \mathrm{CE}=60 \mathrm{~m} and \angle \mathrm{BEC}=30^{\circ}.

Let D E=h m such that A C=h m and B C=(90-h) m .

In the right \triangle \mathrm{BCE}, we have:

\frac{\mathrm{BC}}{\mathrm{CE}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}

\Rightarrow \frac{(90-\mathrm{h})}{60}=\frac{1}{\sqrt{3}}

\Rightarrow(90-\mathrm{h}) \sqrt{3}=60

\Rightarrow \mathrm{h} \sqrt{3}=90 \sqrt{3}-60

\Rightarrow \mathrm{h}=90-\frac{60}{\sqrt{3}}=90-34.64=55.36 \mathrm{~m}

\therefore Height of the first tower =\mathrm{DE}=\mathrm{h}=55.36 \mathrm{~m}

i

More Material