Chapter 14.4, Problem 16PPSE

Solution

To find: The height of tower above the top of a hill.

The height of the tower above the hill is $329.3$ ft.

Given data: The distance from a point on the earth to the base of tower is 2760 ft. the angle of elevation of the base of the tower is ${26}^{\text{ο}}$ and the angle of top of the tower is ${32}^{\text{ο}}$ .

Method/Formula used:

In a right angle triangle ABC shown in Fig. 1 the trigonometric ratio cos and tan of angle $\theta$ are given by

  

  $\begin{array}{l}\cos \theta =\frac{x}{l}\\ \tan \theta =\frac{h}{x}\end{array}$

Calculations:

The tower at the top of a hill whose base B is at distance 2760 ft from a point E on the ground where a person is standing, the angle of elevation of the base of tower B and the angle of elevation of top T of the tower is shown in Fig 2 here.

  

Suppose a vertical line TBfrom top and base of the tower meet ground at point H.

Then in right angle triangle BEH ,

  $\cos {26}^{\text{ο}}=\frac{EH}{EB}$ … (1)

Substitute 2670 for EBand0 .8988 for $\cos {26}^{\text{ο}}$ ,

  $\begin{array}{c}0.8988=\frac{EH}{2670}\\ \Rightarrow EH=2670\times 0.8988\\ =2399.80\\ EH\approx 2400\end{array}$

In right angle triangle BEH ,

  $\begin{array}{c}\tan {26}^{\text{ο}}=\frac{BH}{EH}\\ \Rightarrow BH=EH\tan {26}^{\text{ο}}\end{array}\}$ … (2)

  $\begin{array}{c}\tan {32}^{\text{ο}}=\frac{TH}{EH}\\ \Rightarrow TH=EH\tan {32}^{\text{ο}}\end{array}\}$ … (3)

From result (2) and (3),

  $TH-BH=EH\tan {32}^{\text{ο}}-EH\tan {26}^{\text{ο}}$ …. (4)

But $TH-BH$ equals BT the height of the tower, therefore from result (4)

  $\begin{array}{c}BT=EH\tan {32}^{\text{ο}}-EH\tan {26}^{\text{ο}}\\ =EH\left(\tan {32}^{\text{ο}}-\tan {26}^{\text{ο}}\right)\end{array}$

Substitute 2400 for EH , 0.6249 for $\tan {32}^{\text{ο}}$ , and 0.4877 for $\tan {26}^{\text{ο}}$

  $\begin{array}{c}BT=EH\tan {32}^{\text{ο}}-EH\tan {26}^{\text{ο}}\\ =2400\left(0.6249-\text{0}\text{.4977}\right)\\ =2400\left(0.1372\right)\\ =329.28\end{array}$

Therefore, the height of the tower above the hill is $329.3$ ft.