Question

1. (4 - 43) review: geometry when she was younger, mary had to look up at a 68° angle to see into her father’s eyes whenever she was standing 15 inches away. how high above the flat ground were her father’s eyes if mary’s eyes were 32 inches above the ground? draw a sketch of this situation and use your trig table to help you solve.

Explanation:

Step1: Set up a right - triangle model

We have a right - triangle where the adjacent side to the angle of elevation is the horizontal distance between Mary and her father ($x = 15$ inches) and the angle of elevation $\theta=68^{\circ}$. We want to find the opposite side $y$ (the vertical distance from Mary's eyes to her father's eyes). We use the tangent function $\tan\theta=\frac{y}{x}$.

Step2: Solve for the vertical distance between their eyes

We know that $\tan\theta=\tan(68^{\circ})$, and from the trigonometric table, $\tan(68^{\circ})\approx2.475$. Since $\tan\theta=\frac{y}{x}$ and $x = 15$ inches, then $y=x\tan\theta=15\times\tan(68^{\circ})\approx15\times2.475 = 37.125$ inches.

Step3: Find the height of father's eyes above the ground

Mary's eyes are 32 inches above the ground. The height $h$ of her father's eyes above the ground is the sum of the height of Mary's eyes above the ground and the vertical distance between their eyes. So $h=32 + y=32+37.125=69.125$ inches.

Answer:

$69.125$ inches