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: Set up the right triangle.
The pole, its shadow and the line of sight to the sun form a right triangle with height $6$ m and base $2\sqrt{3}$ m.
Step 2: Find the hypotenuse using Pythagoras. \[ \text{Hypotenuse} = \sqrt{6^2 + (2\sqrt{3})^2} = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3} \text{ m} \]
Step 3: Use sine of the elevation angle instead of tangent. \[ \sin\theta = \frac{\text{height}}{\text{hypotenuse}} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]
Step 4: Identify the angle and conclude.
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$, we get $\theta = 60^{\circ}$. \[ \boxed{60^{\circ}} \]