Question

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 \(60^\circ\) and \(30^\circ\) respectively. Find the height of the poles and the distance of the point from the poles.

Hint:

For any equal height pole problem where angles of elevation are \(60^\circ\) and \(30^\circ\), the observation point always divides the total distance in a \(1 : 3\) ratio!
Since total width is 80 m, the segments are easily found as \(20\text{ m}\) and \(60\text{ m}\).

Answer 1

Correct Answer: 60

Step 1: Understanding the Question:
This question is from the chapter Some Applications of Trigonometry.
We have a road of width \(80\text{ m}\) with two vertical poles of equal height \(h\) at its two ends.
From an observation point on the road between the poles, the angles of elevation of the tops of the poles are \(60^\circ\) and \(30^\circ\).
We need to find the height \(h\) of the poles and the distances of the observation point from each pole.

Step 2: Key Formula or Approach:
1. Let \(h\) be the height of both poles.
2. Let the road width be \(80\text{ m}\). Let the distance of the observation point from the first pole be \(x\text{ m}\).
3. The distance from the second pole is \((80 - x)\text{ m}\).
4. Use trigonometric tangent ratios for both right-angled triangles: \[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \]

Step 3: Detailed Explanation:
1. Let the first pole be represented by \(AB\) and the second pole by \(CD\). Let their heights be \(AB = CD = h\).
2. Let the point on the road be \(P\). Let \(AP = x\) and \(PC = 80 - x\).
3. In right-angled triangle \(\triangle PAB\) (angle of elevation is \(60^\circ\)): \[ \tan 60^\circ = \frac{AB}{AP} \] \[ \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \text{--- (Equation 1)} \] 4. In right-angled triangle \(\triangle PCD\) (angle of elevation is \(30^\circ\)): \[ \tan 30^\circ = \frac{CD}{PC} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{80 - x} \implies 80 - x = h\sqrt{3} \quad \text{--- (Equation 2)} \] 5. Substitute the expression of \(h\) from Equation 1 into Equation 2: \[ 80 - x = (x\sqrt{3})\sqrt{3} \] \[ 80 - x = 3x \] \[ 80 = 4x \] \[ x = \frac{80}{4} = 20\text{ m} \] 6. Find the distance from the second pole: \[ 80 - x = 80 - 20 = 60\text{ m} \] 7. Calculate the height of the poles \(h\) using Equation 1: \[ h = x\sqrt{3} \] \[ h = 20\sqrt{3}\text{ m} \] Using \(\sqrt{3} \approx 1.732\): \[ h \approx 20 \times 1.732 = 34.64\text{ m} \]

Step 4: Final Answer:
The height of each pole is \(20\sqrt{3}\text{ m}\) (or approximately \(34.64\text{ m}\)).
The distances of the point from the poles are \(20\text{ m}\) and \(60\text{ m}\).