Question

A surveyor measures the angle of elevation to a point on a mountain that is 4 miles away to be 18 degrees. The vertical change in elevation from the point where the surveyor is standing to the point on the mountain is _____ miles. (Round your answer to the nearest hundredth of a mile.)

Answer 1

Answer on Question #57925 – Math – Trigonometry

Question

A surveyor measures the angle of elevation to a point on a mountain that is 4 miles away to be 18 degrees. The vertical change in elevation from the point where the surveyor is standing to the point on the mountain is ______ miles. (Round your answer to the nearest hundredth of a mile.)

Solution

$$\angle A = 18{}^\circ$$ $$AB = 4 \text{ miles}, BC - ?$$ $$\tan \angle A = \frac{BC}{AB}$$ $$\tan \angle A = \tan(18{}^\circ) = 0.3249 = BC/4$$

The value of $\tan(18{}^\circ) = \frac{\sin(18{}^\circ)}{\cos(18{}^\circ)}$ can be found from

$\sin(18{}^\circ) = \frac{\sqrt{5} - 1}{4} \approx 0.30902, \quad \cos(18{}^\circ) = \frac{\sqrt{10 + 2\sqrt{5}}}{4} \approx 0.95106$ (details can be found at http://www.math-only-math.com/exact-value-of-cos-18-degree.html)

Thus,

$$\frac{BC}{4} = 0.3249.$$ $$h = BC = 4 \cdot 0.3249 = 1.2996 \approx 1.3 \text{ (miles)}$$

**Answer**: $BC \approx 1.3$ miles.

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