Question

A pole 6 m high casts a shadow \(2\sqrt{3}\text{ m}\) long on the ground, then the Sun's angle of elevation is :

• A. \(60^{\circ}\)
• B. \(45^{\circ}\)
• C. \(30^{\circ}\)
• D. \(90^{\circ}\)

Hint:

Remember the standard ratios for \(30^{\circ}-60^{\circ}-90^{\circ}\) triangles:
If the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), the side opposite \(30^{\circ}\) is \(x\).
Here, the height is \(6\) and base is \(2\sqrt{3}\).
Dividing both by \(\sqrt{3}\) reveals the ratio of sides is \(\sqrt{3} : 1\).
Since the vertical side is larger (multiple of \(\sqrt{3}\)), the opposite angle must be \(60^{\circ}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question is from Trigonometry, specifically Heights and Distances.
We have a vertical pole of a given height and its shadow of a given length on the ground. We need to find the angle of elevation of the Sun.

Step 2: Key Formula or Approach:
A vertical pole forms a right-angled triangle with the ground and the line of sight to the Sun.
The height of the pole is the perpendicular side, and the length of the shadow is the base side of this triangle.
The tangent of the angle of elevation \(\theta\) is:
\[ \tan\theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{\text{Height of Pole}}{\text{Length of Shadow}} \]

Step 3: Detailed Explanation:
Given:
Height of the pole, \(h = 6\text{ m}\)
Length of the shadow, \(s = 2\sqrt{3}\text{ m}\)
Let the angle of elevation of the Sun be \(\theta\).
Using the trigonometric ratio:
\[ \tan\theta = \frac{h}{s} \]
Substitute the values:
\[ \tan\theta = \frac{6}{2\sqrt{3}} \]
Simplify the fraction:
\[ \tan\theta = \frac{3}{\sqrt{3}} \]
Rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{3}\):
\[ \tan\theta = \frac{3\sqrt{3}}{3} \]
\[ \tan\theta = \sqrt{3} \]
We know from standard trigonometric values that:
\[ \tan 60^{\circ} = \sqrt{3} \]
Since \(\theta\) is an acute angle, we have:
\[ \theta = 60^{\circ} \]

Step 4: Final Answer:
The Sun's angle of elevation is \(60^{\circ}\).