Question

1. an observer at the top of a building sees a car on the road below. the angle of depression to the

car is 28°. if the car is about 50m from the building when it is seen, how tall is the building?

Explanation:

Step1: Relate the angle of depression to the angle in the right - triangle

The angle of depression is \(28^{\circ}\). In the right - triangle formed by the building (height \(h\)), the horizontal distance from the building to the car (\(x = 50m\)), and the line of sight from the top of the building to the car, the angle of depression is equal to the angle of elevation from the car to the top of the building. Let's call this angle \(\theta=28^{\circ}\).
We know that \(\tan\theta=\frac{h}{x}\) (where \(h\) is the height of the building and \(x\) is the horizontal distance from the building to the car).

Step2: Solve for the height \(h\)

Since \(\tan\theta=\frac{h}{x}\), and \(x = 50m\), \(\theta = 28^{\circ}\), we can solve for \(h\) by multiplying both sides of the equation by \(x\).
\(h=x\tan\theta\)
Substitute \(x = 50\) and \(\theta=28^{\circ}\) into the formula. We know that \(\tan(28^{\circ})\approx0.5317\)
\(h = 50\times\tan(28^{\circ})\)
\(h=50\times0.5317\)

Answer:

\(h = 26.585m\approx26.6m\)