Question

the angle of depression of one side of a lake, measured from a balloon 2500 feet above the lake is 43°. the angle of depression to the opposite side of the lake is 27°. find the width of the lake.

Explanation:

Step1: Define variables and use tangent

Let the horizontal distance to the near side of the lake be \( x \) and to the far side be \( y \). The height of the balloon is 2500 feet. For the angle of depression \( 43^\circ \), the angle with the horizontal is also \( 43^\circ \) (alternate interior angles). So, \( \tan(43^\circ)=\frac{2500}{x} \), so \( x = \frac{2500}{\tan(43^\circ)} \).

Step2: Calculate for the other angle

For the angle of depression \( 27^\circ \), the angle with the horizontal is \( 27^\circ \). So, \( \tan(27^\circ)=\frac{2500}{y} \), so \( y = \frac{2500}{\tan(27^\circ)} \).

Step3: Find the width of the lake

The width of the lake \( w = y - x \). Calculate \( \tan(43^\circ)\approx0.9325 \), \( \tan(27^\circ)\approx0.5095 \). Then \( x=\frac{2500}{0.9325}\approx2681 \), \( y=\frac{2500}{0.5095}\approx4907 \). Then \( w = 4907 - 2681 = 2226 \) (approximate, using more precise values for tangent may vary slightly).

Answer:

The width of the lake is approximately \(\boldsymbol{2226}\) feet (or more precisely, using calculator - derived tangent values: \( \frac{2500}{\tan(27^\circ)}-\frac{2500}{\tan(43^\circ)}\approx\frac{2500}{0.5095}-\frac{2500}{0.9325}\approx4907 - 2681 = 2226 \) feet).