Chapter 9.5, Problem 47PS

Solution

a.

To state The distance between the person and the cliff if a building is constructed on the cliff that is 300 meter high. A person standing on level ground below the cliff observes that the angle of elevation to the top of the building is ${72}^{\circ }$ , and the angle of elevation to the top of the cliff is ${63}^{\circ }$ .

The resultant answer is 152.9 meters.

Given information:

The cliff is 300 meter high. A person standing on level ground below the cliff observes that the angle of elevation to the top of the building is ${72}^{\circ }$ , and the angle of elevation to the top of the cliff is ${63}^{\circ }$ .

Explanation:

First draw the diagram using the given information:

  

The distance x can be found using the right triangle ACD . Write a trigonometric equation involving x .

  $\tan {63}^{\circ }=\frac{300}{x}$

Now solve for x ,

  $\begin{array}{l}x=\frac{300}{\tan {63}^{\circ }}\\ x\approx 152.9\end{array}$

Therefore, the distance between the cliff and the person is about 152.9 meters.

b.

To state: Two different methods that can be used to find out the height of the building. Then use one of the methods to find the height of the building.

The resultant answer is170.58 meters.

Given information:

The given statement says to describe two different methods that can be used to find out the height of the building.

Explanation:

The first method to find the height of the building is by writing a trigonometric equation that involves h using the right triangle ABD .

By substituting the respective values and solving the equations, the value of h can be found.

The second method is by using the law of sines. The height h of the building makes one of the sides of the triangle ABC .

Using the law of sines, it can be written as $\frac{h}{\sin A}=\frac{AC}{\sin B}$ , where angles $A\text{ and }B$ , and side AC can be determined using the triangle $ABC\text{ and }ABD$ .

Now use the any of the methods to find the height of the building,

Let’s use the first method to find h ,

  $\tan {72}^{\circ }=\frac{BD}{AD}$

Now replace $BD\text{ with 30}0+h\text{ and }AD\text{ with }152.9$ ,

  $\tan {72}^{\circ }=\frac{300+h}{152.9}$

Now solve for $\text{30}0+h$ ,

  $\begin{array}{c}\tan {72}^{\circ }\left(152.9\right)=300+h\\ 470.58\approx 300+h\\ 170.58\approx h\end{array}$

Therefore, the height of the building is about 170.58 meters.