Chapter 8.2, Problem 52AYU

To determine

How far is the ranger from the tower at this time?

Answer to Problem 52AYU

  $\boxed{110.01ft}$

Explanation of Solution

Given information:

A forest ranger is walking on a path inclined at $5°$ to the horizontal directly towards a 100-foot-tall fire observation tower. The angle of elevation from the path to the top of the tower is $40°$ . How far is the ranger from the tower at this time?

  

Calculation:

Let us consider the fact that the forest ranger is walking on a path BC inclined at $5°$ to the horizontal line $CD$ towards $100feet$ tall observation tower $AB$ (base at $B$ ). The angle of elevation from the path (from the position of the forest ranger $C$ ) to the top of the tower is $40°$ .The information can be figured as:

  

In the right triangle $\Delta ACD$ , we have

  $\angle ACD=40°+5°$

  $=45°$

By the property of interior angles of triangle , we get

  $\angle ACD+\angle CDA+\angle DAC=180°$

  $\angle DAC=180°-\left(\angle ACD+\angle CDA\right)$

  $=180°-\left(45°-90°\right)$

  $=45°$

In the right triangles $\Delta ABC$ by the property of interior angles of a triangle, we get

  $\angle ABC+\angle BCA+\angle CAB=180°$

  $\angle ABC=180°-\left(\angle BCA+\angle CAB\right)$

  $=180°-\left(40°+45°\right)$

  $=95°$

In the right triangle $\Delta ABC$ by Law of Sines, we get

  $\frac{\sin \left(\angle ABC\right)}{AC}=\frac{\sin \left(\angle BCA\right)}{CB}=\frac{\sin \left(\angle CAB\right)}{BA}$

  $\frac{\sin 95°}{AC}=\frac{\sin 40°}{CB}=\frac{\sin 45°}{100}$

Taking last two, we get

  $\frac{\sin 40°}{CB}=\frac{\sin 45°}{100}$

  $CB=100\times \frac{\sin 45°}{\sin 40°}$

  $CB\approx 110.006ft$

Hence, forest ranger is approximately $\boxed{110.01ft}$ from the tower.