Chapter 10.6, Problem 41P

Gayle stood at the edge of a 120-ft deep canyon. She is approximately 5 ft tall and when she looked across the canyon to the far corner, her line of sight made a 22° angle as shown in the figure. How wide was the canyon?

To determine

To calculate: The width of the canyon

Answer to Problem 41P

The canyon was $50.5\text{ feet}$ wide.

Explanation of Solution

Given information:

Gayle stood at the edge of 120 feet deep canyon. She is approximately 5 feet tall. When she looked across the canyon her line of sight made ${22}^{\circ }$ angle.

Consider the diagram given in the question

Formula used:

  $\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

Calculation:

According to the diagram Gayle make a right angled triangle.

For a right angled triangle, assume hypotenuse as $AC$ , base as $BC$ and perpendicular as $AB$

Now,

The depth of the canyon $=20$ feet

Height of Gayle $=5$ feet

Therefore, the perpendicular $\left(AB\right)=125\text{ feet}$

Assume the width of the canyon $\left(BC\right)$ as $\text{'}x\text{'}$

Hence, the width can be calculated as,

  $\begin{array}{l}\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}\\ \tan {22}^{\circ }=\frac{BC}{AB}\\ \tan {22}^{\circ }=\frac{x}{125}\\ x=125\times \tan {22}^{\circ }\\ x=125\times 0.404\\ x=50.5\text{ feet}\end{array}$

Therefore, the canyon was $50.5\text{ feet}$ wide.