Two towers stand on the same plane. From the foot of the second, the angle of elevation of the top of the first (height \(30\,\text{m}\)) is \(60^\circ\). From the foot of the first, the angle of elevation of the top of the second is \(30^\circ\). Find the distance between the towers and the height of the second.
Distance \(=10\sqrt3\,\text{m}\); height of second tower \(=10\,\text{m}.\)
Step 1: Understand the situation
There are two towers standing on the same horizontal ground.
Step 2: Use the first angle of elevation
From the foot of the second tower, the top of the first tower is seen at an angle of \(60^\circ\).
So, in the right triangle formed:
\[ \tan(60^\circ) = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{30}{D} \]
We know \(\tan(60^\circ) = \sqrt{3}\).
So, \(\sqrt{3} = \dfrac{30}{D}\).
Cross multiply: \(D = \dfrac{30}{\sqrt{3}}\).
Simplify: \(D = 10\sqrt{3}\,\text{m}\).
Step 3: Use the second angle of elevation
From the foot of the first tower, the top of the second tower is seen at an angle of \(30^\circ\).
So, in the right triangle formed:
\[ \tan(30^\circ) = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{H}{D} \]
We know \(\tan(30^\circ) = \dfrac{1}{\sqrt{3}}\).
So, \(\dfrac{1}{\sqrt{3}} = \dfrac{H}{10\sqrt{3}}\).
Cross multiply: \(H = \dfrac{10\sqrt{3}}{\sqrt{3}}\).
Simplify: \(H = 10\,\text{m}\).
Final Answer: