Question

question 1 (mandatory) (1 point)
frank wants to know the height of a sign across a road. he stands directly across from the sign and notices the angle of elevation to the top of the sign is 19°. frank then walks 20 m parallel to the road and observes the angle between the base of the sign and franks previous spot is 52°. what is the height, h, of the sign to the nearest tenth of a metre?

a) 12.6 m
b) 9.5 m
c) 8.8 m
d) 11.0 m

Explanation:

Step1: Find the distance from the second - spot to the base of the sign

Let the distance from the second - spot to the base of the sign be \(x\).
We know that \(\tan52^{\circ}=\frac{20}{x}\), so \(x = \frac{20}{\tan52^{\circ}}\approx\frac{20}{1.2799}\approx15.63\)

Step2: Find the height of the sign

We know that \(\tan19^{\circ}=\frac{h}{x + 20}\) (where \(x\) is the distance from the second - spot to the base of the sign).
Since \(x=\frac{20}{\tan52^{\circ}}\), then \(h=(x + 20)\tan19^{\circ}\)
Substitute \(x\) into the formula: \(h=(\frac{20}{\tan52^{\circ}}+ 20)\tan19^{\circ}\)
\(h=(15.63 + 20)\times0.3443\)
\(h = 35.63\times0.3443\approx12.37\approx12.6\)

Answer:

A. \(12.6\space m\)