9. A 40-foot tall monument sits on top of a hill. Pete is standing at a point on the hill and observes the top of the monument at an angle of elevation of \( 50^{\circ} \) and the bottom of the monument at an angle of elevation of \( 22^{\circ} \). Find the distance Pete must climb to reach the monument.

Let the horizontal distance from Pete to the point at the base of the monument (on the hill) be \( d \) and let the vertical difference (along a vertical line) from Pete’s position to that base be \( h_1 \). Then from the geometry of the situation: 1. From the angle of elevation of the base of the monument, we have \[ \tan(22^\circ)=\frac{h_1}{d}\quad\Longrightarrow\quad h_1=d\tan(22^\circ). \] 2. The monument is 40 feet tall. When Pete looks up to the top, the total vertical rise from his eye level is \[ h_1+40. \] The angle of elevation to the top is given as \( 50^\circ \); thus, \[ \tan(50^\circ)=\frac{h_1+40}{d}. \] 3. Substitute \( h_1=d\tan(22^\circ) \) into the second equation: \[ \tan(50^\circ)=\frac{d\tan(22^\circ)+40}{d}=\tan(22^\circ)+\frac{40}{d}. \] Rearranging to solve for \( d \): \[ \tan(50^\circ)-\tan(22^\circ)=\frac{40}{d}\quad\Longrightarrow\quad d=\frac{40}{\tan(50^\circ)-\tan(22^\circ)}. \] 4. The computed \( d \) is the horizontal distance. However, note that Pete is standing on the hill, so the hill itself rises at the same angle as the elevation of the base, namely \( 22^\circ \). The distance Pete must climb along the hill (i.e. the slope distance) is the hypotenuse of the right triangle formed by \( d \) (horizontal) and \( h_1=d\tan(22^\circ) \) (vertical). Thus, the distance along the hill is \[ s=\frac{d}{\cos(22^\circ)}. \] 5. Combining the results, we have \[ s=\frac{40}{\tan(50^\circ)-\tan(22^\circ)}\cdot\frac{1}{\cos(22^\circ)}. \] That is, \[ s=\frac{40\,\sec(22^\circ)}{\tan(50^\circ)-\tan(22^\circ)}\, \text{feet}. \] 6. Approximating using a calculator: - \( \tan(50^\circ)\approx 1.1918 \) - \( \tan(22^\circ)\approx 0.4040 \) - \( \cos(22^\circ)\approx 0.9272 \) so that \( \sec(22^\circ)=\frac{1}{\cos(22^\circ)}\approx 1.0783 \) Then, \[ d\approx \frac{40}{1.1918-0.4040}=\frac{40}{0.7878}\approx 50.79\,\text{feet}, \] and \[ s\approx \frac{50.79}{0.9272}\approx 54.80\,\text{feet}. \] Thus, the distance Pete must climb along the hill to reach the monument is approximately \( 54.8 \) feet.