Trigonometry & Inverse Trigonometry
Height and Distance
Grade 11
Question:
<p>A balloon is observed simultaneously from three points A, B and C on a straight road directly under it. The angular elevation at B is twice and at C is thrice that of A. If the distance between A and B is 200 m and the distance between B and C is 100 m, then the height of the balloon is given by</p>
<p>(a) 50 m</p>
<p>(b) \(50\sqrt{3}\) m</p>
<p>(c) \(50\sqrt{2}\) m</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the angle of elevation relationships combined with distance constraints to set up equations in terms of height.
<p>Let the height of the balloon be h. If the angles of elevation at A, B, C are \(\alpha, 2\alpha, 3\alpha\) respectively, then:
\[\tan\alpha = \frac{h}{x_A}, \quad \tan 2\alpha = \frac{h}{x_B}, \quad \tan 3\alpha = \frac{h}{x_C}\]
where \(x_A, x_B, x_C\) are horizontal distances. Using the given distances AB = 200 m and BC = 100 m along with angle relationships, we solve to get \(h = 50\sqrt{3}\) m.</p>
Correct Answer: B