Question

1. from the top of a cliff at the edge of a lake, the angle of depression to a buoy on the lake is 33°. if the buoy is 80 meters from the base of the cliff, how high is the cliff? round to the nearest hundredth. (see yellow hint below.) 123.19 m 51.95 m 67.09 m 43.57 m

Explanation:

Step1: Use the tangent function

The angle of depression is \(33^{\circ}\), and the adjacent side of the right - triangle (distance from the base of the cliff to the buoy) is \(x = 80\) m. Let the height of the cliff be \(h\). Since the angle of depression is equal to the angle of elevation in such a geometric setup, we use the formula \(\tan\theta=\frac{opposite}{adjacent}\). Here, \(\theta = 33^{\circ}\), \(opposite=h\), and \(adjacent = 80\) m. So, \(\tan(33^{\circ})=\frac{h}{80}\).

Step2: Solve for \(h\)

We know that \(h = 80\times\tan(33^{\circ})\). Using a calculator, \(\tan(33^{\circ})\approx0.6494\). Then \(h=80\times0.6494 = 51.952\approx51.95\) m.

Answer:

\(51.95\) m