Trigonometry & Inverse Trigonometry
Heights And Distances
nta_abhyas_2025
Grade 11

Question:

(D) $10\sqrt{3}(2 + \sqrt{3})$

Step-by-Step Solution

Key Concept: Use simultaneous equations from two right triangles with different angles of elevation to find the height of the tower
In the figure $AB$ is the vertical tower and $C, D$ are two points in right angled $\triangle ABD$. From equation (1): $\frac{h}{\tan 15°} = h - \sqrt{3}x$. In right angled $\triangle BCE$: $\frac{10}{\tan 15°} = x - \frac{10}{\sqrt{3}}$ (equation 2). Solving these simultaneously gives $h = 10\sqrt{3}(2 + \sqrt{3})$ meters.
Correct Answer: 4

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