Chapter 4.8, Problem 14E

Solution

To find: The height of the top of Long’s Peak from Otis’s elevation.

The height of the top of Long’s Peak from Otis’s elevation is approximately 3290.53 mt.

Given information:

Given, while hiking on a level path toward Colorado’s front range, Otis Evans determines that the angle of elevation to the top of Long’s Peak is 30°. Moving 1000 ft closer to the mountain, Otis determines the angle of elevation to be 35°.

Calculation:

The triangle representing the data is:

  

Using definitions of trigonometric functions for the triangle CBD:

  $\begin{array}{l}\tan D=\frac{\text{opposite side}}{\text{adjacent side}}\\ \tan \left(35°\right)=\frac{h}{x}\\ x=\frac{h}{\tan \left(35°\right)}\end{array}$

Also for the triangle ABC,

  $\begin{array}{l}\tan A=\frac{\text{opposite side}}{\text{adjacent side}}\\ \tan \left(30°\right)=\frac{h}{1000+x}\\ h=\left(1000+x\right)\tan \left(30°\right)\end{array}$

Plugging the value of x :

  $\begin{array}{l}h=\left(1000+\frac{h}{\tan \left(35°\right)}\right)\tan \left(30°\right)\\ h=\left(1000+\frac{h}{\tan \left(35°\right)}\right)\left(\frac{1}{\sqrt{3}}\right)\\ h=\frac{1000}{\sqrt{3}}+\frac{h}{\sqrt{3}\tan \left(35°\right)}\\ h-\frac{h}{\sqrt{3}\tan \left(35°\right)}=\frac{1000}{\sqrt{3}}\\ h\left(\frac{\sqrt{3}\tan \left(35°\right)-1}{\sqrt{3}\tan \left(35°\right)}\right)=\frac{1000}{\sqrt{3}}\\ h=\frac{1000\tan \left(35°\right)}{\sqrt{3}\tan \left(35°\right)-1}\\ h\approx 3290.53\end{array}$

The height of the top of Long’s Peak from Otis’s elevation is approximately 3290.53 mt.

Conclusion:

The height of the top of Long’s Peak from Otis’s elevation is approximately 3290.53 mt.