Trigonometry & Inverse Trigonometry
Heights And Distances
nta_abhyas_2025
Grade 11

Question:

A $150\left(\sqrt{3}+1\right)$ ft.

Step-by-Step Solution

Key Concept: Apply tangent ratios in two right triangles sharing a common height to find the total distance.
In $\triangle ABE$, $\tan 30° = \frac{h}{x}$, so $h = 100\sqrt{3}$ ft (equation 1). In $\triangle ABD$, $\tan 45° = \frac{h}{H} = 300$ ft (equation 2). Since $\frac{h}{\tan 30°} = \frac{h}{1/\sqrt{3}} = h\sqrt{3} = 300 + x$, and $h\sqrt{3} = 300 + x$ using equations (1) and (2), we get $x = 150(\sqrt{3}+1)$ ft.
Correct Answer: 150(√3+1)

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