Chapter 8.2, Problem 43AYU

To determine

Find the height of the tree.

Answer to Problem 43AYU

  $\boxed{39.4ft}$

Explanation of Solution

Given information:

Pat needs to determine the height of a tree before cutting down to be sure that it will not fall on a nearly fence. The angle of elevation of the tree from one position on a flat path from the tree is $30°$ , and from a second position $40ft$ farther along this path it is $20°$ . What is the height of the tree?

Calculation:

  

  $\angle ADB=20°,\angle ACB=30°andCD=40ft$

By using the property of triangle, we get

  $\angle ACD+30°=180°$

  $\angle ACD=150°\mathrm{.......}\left(2\right)$

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

  $\angle DAC=180°-\left(150°+20°\right)$

  $\angle DAC=180°-\left(\angle ACD+\angle CDA\right)$ Using $\left(1\right)and\left(2\right)$

  $\angle DAC=10°\mathrm{.....}\left(3\right)$

In the triangle $\angle ACD,$ using low of sine, we get

  $\frac{\sin A}{CD}=\frac{\sin C}{DA}=\frac{\sin D}{AC}$

Putting the values,

  $\begin{array}{l}\frac{\sin A}{CD}=\frac{\sin C}{DA}=\frac{\sin D}{AC}\\ \frac{\sin 10°}{40}=\frac{\sin 150°}{DA}=\frac{\sin 20°}{AC}\\ now,\\ \frac{\sin 10°}{40}=\frac{\sin 20°}{AC}\\ AC=40\times \frac{\sin 20°}{\sin 10°}\\ AC=40\times \frac{0.34202}{0.17365}\\ AC=78.78459ft\end{array}$

In the right triangle $\Delta ABC$ using trigonometric ratio for acute angle for right triangle,

  $\begin{array}{l}\sin \left(30°\right)=\frac{AB}{AC}\\ \sin \left(30°\right)=\frac{AB}{78.78459}\\ AB=78.78459\times 0.5\\ AB=39.392295ft\end{array}$

Hence, the height of the tree $\boxed{39.4ft}$ .