Question

A water tower 30 m tall is located at the top of a hill. From a distance of D = 110 m down the hill it is observed that the angle formed between the top and base of the tower is 8°. Find the angle of inclination of the hill. (Round your answer to one decimal place.)

Answer 1

Answer:

$\displaystyle \theta=40.1^o$

Step-by-step explanation:

Application of Right Triangles

The right triangles have an internal angle of 90°, we can take advantage of it because the fundamental trigonometric functions can be expressed to relate angles and lengths in a right triangle.  

Please refer to the image below to understand the upcoming relations and variables. The lower triangle has an angle \theta and h_h and D are the opposite and adjacent legs respectively, then:

$\displaystyle tan\theta =\frac{h_h}{D}$

Where h_h is the height of the tower. We can solve:

$\displaystyle h_h=D\ tan\theta$

For the big triangle:

$\displaystyle h_t+h_h=D\ tan(\theta +\alpha)$

Where $\alpha$ is 8° and $h_h$ is the height of the hill. Knowing that:

$\displaystyle h_h=D\ tan \theta$

We replace it into the above equation

$\displaystyle h_t+D\ tan \theta = D\ tan(\theta +\alpha)$

We have an equation in $tan\theta$, but we need to expand the tangent of a sum of angles:

$\displaystyle h_t+D\ tan \theta = D\ \frac{tan\theta +tan\alpha}{1-tan\theta \ tan\alpha }$

Rearranging

$\displaystyle (h_t+D\ tan\theta)(1-tan\theta \ tan\alpha)=D\ tan\theta +D\ tan\alpha$  

Multiplying

$\displaystyle h_t-h_t\ tan \theta \ tan\alpha +D\ tan\theta -D\ tan^2\theta \ tan\alpha =D\ tan\theta +D\ tan\alpha$

Simplifying, we have a second-degree equation for tan\theta

$\displaystyle -D\ tan^2\theta \ tan\alpha -h_t\ tan\alpha \ tan\theta +h_t-D\ tan\alpha =0$

Using the known values D=110, ht=30, $\alpha=8^o$

$\displaystyle -15.46\ tan^2\theta -4.22\ tan\theta +14.54=0$

Solving the equation, we get two answers:

$\displaystyle tan\theta=-1.12$

This solution is not feasible, since the angle cannot exceed 90° or go below 0°, thus the other answer

$\displaystyle tan\theta=0.843$

is the correct option. Computing the angle of inclination of the hill:

$\boxed{\displaystyle \theta=40.1^o}$