Question:
A person on the top of a tower of \(100\sqrt{3}\) meters tall observes two points A and B on the opposite sides making angles of depression of \(30^\circ\) and \(60^\circ\) respectively. The distance between A and B in meters is:
A person on the top of a tower of \(100\sqrt{3}\) meters tall observes two points A and B on the opposite sides making angles of depression of \(30^\circ\) and \(60^\circ\) respectively. The distance between A and B in meters is:
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For angles of depression on opposite sides, compute both horizontal distances separately and then add them.
Updated On: Jun 12, 2026
- \(\frac{400}{\sqrt3}\)
- \(100\sqrt3\)
- \(400\sqrt3\)
- \(400\)
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The Correct Option is D
Solution and Explanation
Concept:
Use right triangle trigonometry with angles of depression.
Height of tower:
\[
h=100\sqrt3
\]
Step 1: Find distance to point A. \[ \tan 30^\circ = \frac{h}{x} \] \[ \frac{1}{\sqrt3}=\frac{100\sqrt3}{x} \] \[ x=300 \]
Step 2: Find distance to point B. \[ \tan 60^\circ = \frac{h}{y} \] \[ \sqrt3=\frac{100\sqrt3}{y} \] \[ y=100 \]
Step 3: Total distance between A and B. Since they are on opposite sides: \[ AB = x+y = 300+100 \] \[ =400 \] \[ \boxed{400} \]
Step 1: Find distance to point A. \[ \tan 30^\circ = \frac{h}{x} \] \[ \frac{1}{\sqrt3}=\frac{100\sqrt3}{x} \] \[ x=300 \]
Step 2: Find distance to point B. \[ \tan 60^\circ = \frac{h}{y} \] \[ \sqrt3=\frac{100\sqrt3}{y} \] \[ y=100 \]
Step 3: Total distance between A and B. Since they are on opposite sides: \[ AB = x+y = 300+100 \] \[ =400 \] \[ \boxed{400} \]
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