Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
Step-by-Step Solution
Key Concept: Use the definition of tangent in right‑angled triangles: \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\). Let the distances from the point to the two poles be \(x\) and \(80-x\). Set up two equations using \(\tan 60^{\circ}\) and \(\tan 30^{\circ}\), then solve for \(x\) and the common height \(h\).
1. Introduce variables
Let the pole whose top subtends an angle of \(60^{\circ}\) be Pole A and the other be Pole B.
Let the distance of the point \(P\) from Pole A be \(x\) metres.
Hence the distance of \(P\) from Pole B is \(80 - x\) metres (since the road width is 80 m).
2. Apply the tangent definition
For Pole A (\(60^{\circ}\)):
$$\tan 60^{\circ}=\frac{h}{x}\quad\Rightarrow\quad \sqrt{3}=\frac{h}{x}\quad\Rightarrow\quad h=\sqrt{3}\,x$$
For Pole B (\(30^{\circ}\)):
$$\tan 30^{\circ}=\frac{h}{80-x}\quad\Rightarrow\quad \frac{1}{\sqrt{3}}=\frac{h}{80-x}\quad\Rightarrow\quad h=\frac{80-x}{\sqrt{3}}$$
3. Equate the two expressions for \(h\)
$$\sqrt{3}\,x = \frac{80-x}{\sqrt{3}}$$
Multiply both sides by \(\sqrt{3}\):
$$3x = 80 - x$$
$$4x = 80$$
$$x = 20\text{ m}$$
Hence the distance from \(P\) to Pole B is \(80 - x = 60\) m.
4. Find the height of the poles
Using \(h = \sqrt{3}\,x\):
$$h = \sqrt{3}\times 20 = 20\sqrt{3}\text{ m} \approx 34.6\text{ m}$$
5. Answer
• Height of each pole = \(20\sqrt{3}\) m (≈ 34.6 m).
• Distance of the point from the pole with \(60^{\circ}\) elevation = 20 m.
• Distance of the point from the pole with \(30^{\circ}\) elevation = 60 m.
Correct Answer: Height of each pole = $20\sqrt{3}\,$ m (≈ 34.6 m).\nDistance of the point from the pole subtending $60^{\circ}$ = 20 m.\nDistance of the point from the pole subtending $30^{\circ}$ = 60 m.