Question

question 8 (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) 8.8 m
b) 12.6 m
c) 11.0 m
d) 9.5 m

Explanation:

Step1: Let the distance from Frank's new position to the base of the sign be \(x\)

We know that \(\tan19^{\circ}=\frac{h}{x}\), so \(x = \frac{h}{\tan19^{\circ}}\). Also, from the second - position (after walking \(20m\)), \(\tan52^{\circ}=\frac{h}{x - 20}\).

Step2: Substitute \(x=\frac{h}{\tan19^{\circ}}\) into \(\tan52^{\circ}=\frac{h}{x - 20}\)

We get \(\tan52^{\circ}=\frac{h}{\frac{h}{\tan19^{\circ}}-20}\).
Since \(\tan19^{\circ}\approx0.3443\) and \(\tan52^{\circ}\approx1.2799\), then \(1.2799=\frac{h}{\frac{h}{0.3443}-20}\).
Cross - multiply: \(1.2799(\frac{h}{0.3443}-20)=h\).
\(\frac{1.2799h}{0.3443}-25.598 = h\).
\(\frac{1.2799h}{0.3443}-h=25.598\).
\(h(\frac{1.2799}{0.3443}-1)=25.598\).
\(\frac{1.2799 - 0.3443}{0.3443}h=25.598\).
\(\frac{0.9356}{0.3443}h=25.598\).
\(h=\frac{25.598\times0.3443}{0.9356}\).
\(h\approx9.5\)

Answer:

d) \(9.5m\)