Question

Two poles of equal heights are standing opposite to each other on either side of the road which is 90 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 30\(^\circ\) and 60\(^\circ\) respectively. Find the height of the poles and the distances of the point from the poles. [Use \(\sqrt{3}\) = 1.732]

Hint:

The pole with the larger angle of elevation (\(60^\circ\)) will always be closer to the observation point than the pole with the smaller angle (\(30^\circ\)). This is a great conceptual sanity check!

Answer 1

Step 1: Set up the two right triangles.
Let the common pole height be $h$, and let $x$ be the distance of the observation point from the pole seen at $60^\circ$, so $(90-x)$ is its distance from the pole seen at $30^\circ$.
Step 2: Form a ratio directly instead of solving a linear equation for h.
Since $h = x\tan 60^\circ = (90-x)\tan 30^\circ$, dividing gives:
\[ \frac{x}{90-x} = \frac{\tan 30^\circ}{\tan 60^\circ} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3} \]
Step 3: Solve the ratio for x.
$x:(90-x) = 1:3$ means the 90 m splits into 4 equal parts, so $x = \frac{90}{4} = 22.5$ m and $90-x = 67.5$ m.
Step 4: Find the height.
\[ h = x\tan 60^\circ = 22.5 \times 1.732 = 38.97 \text{ m} \]
\[ \boxed{h = 38.97 \text{ m}, \text{ distances } 22.5 \text{ m and } 67.5 \text{ m}} \]