Chapter 8.2, Problem 54AYU

To determine

Find the height of the aircraft as per time.

Answer to Problem 54AYU

  $\boxed{710.97ft}$

Explanation of Solution

Given information:

Two sensors are spaced $700$ feet apart along the approach to a small airport. When an aircraft is nearing the airport, the angle of elevation from the first sensor to the aircraft is $20°$ , and from the second sensor to the aircraft it is $15°$ . Determine how high the aircraft is at this time.

  

Calculation:

Consider the following figure;

  

By using the property of triangle, we get

  $\angle A{S}_2{S}_1+20°=180°$

  $\angle A{S}_2{S}_1=180°-20°$

  $\angle A{S}_2{S}_1=160°\mathrm{.......}\left(1\right)$

Also, $\angle A{S}_2{S}_1+\angle S_2{S}_{1}A+\angle S_{1}A{S}_2=180°$

  $\angle S_{1}A{S}_2=180°-\left(\angle A{S}_2{S}_1+\angle S_2{S}_{1}A\right)$

  $\angle S_{1}A{S}_2=180°-\left(160°+15°\right)$

  $\angle S_{1}A{S}_2=5°\mathrm{......}\left(2\right)$

By using Law of Sines in the triangle $\Delta A{S}_1{S}_2$ , we get

  $\frac{\sin A}{S_1{S}_2}=\frac{\sin S_1}{S_{2}A}=\frac{\sin S_2}{A{S}_1}$

Putting the values from (1), (2) and (3), we get

  $\frac{\sin 5°}{700}=\frac{\sin 15°}{S_{2}A}=\frac{\sin 160°}{A{S}_1}\mathrm{......}\left(3\right)$

Taking first two in (3), we get

  $\frac{\sin 5°}{700}=\frac{\sin 15°}{S_{2}A}$

  $S_{2}A=700\times \frac{\sin 15°}{\sin 5°}$

  $S_{2}A\approx 700\times \frac{0.25882}{0.08716}$

  $S_{2}A\approx 2078.7385feet\mathrm{....}\left(4\right)$

In the right the triangle $\Delta AB{S}_2$ , using trigonometric ratio (function) of acute angle foe right triangle, we get

  $\sin \left(20°\right)=\frac{AB}{S_{2}A}$

  $\sin \left(30°\right)\approx \frac{AB}{2078.7385}$

  $AB\approx 2078.7385\times \sin \left(20°\right)$

  $AB\approx 2078.7385\times 0.34202$

  $AB\approx 710.97044feet$

Hence , the height of aircraft is approximately $\boxed{710.97}$ feet.