Question

From the top of the tower 60 meters high, the angle of depression of an object is 60°, then the distance of the object from the base of the tower is:

• A. \( \frac{60}{\sqrt{3}} \, \text{m} \)
• B. \( 60\sqrt{3} \, \text{m} \)
• C. \( 60 \, \text{m} \)
• D. \( 30\sqrt{3} \, \text{m} \)

Hint:

In right-angled triangles, use the trigonometric ratios like tangent to relate the angles and sides. The angle of depression from the top of the tower equals the angle of elevation from the object.

Answer 1

The Correct Option is A

We are given that the height of the tower is 60 meters and the angle of depression is 60°. We need to find the distance of the object from the base of the tower. Step 1: Represent the situation in a right triangle. - The height of the tower (opposite side) is 60 meters. - The angle of depression is 60°, so the angle of elevation from the object to the top of the tower is also 60° (alternate angles). Step 2: Use the tangent of the angle of elevation to find the distance \(d\) (adjacent side): \[ \tan 60^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{60}{d} \] Step 3: We know that \( \tan 60^\circ = \sqrt{3} \), so: \[ \sqrt{3} = \frac{60}{d} \] Step 4: Solve for \(d\): \[ d = \frac{60}{\sqrt{3}} \] Thus, the distance of the object from the base of the tower is \( \frac{60}{\sqrt{3}} \, \text{m} \).