Answer:
![\displaystyle \theta=40.1^o](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctheta%3D40.1%5Eo)
Step-by-step explanation:
<u>Application of Right Triangles
</u>
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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20tan%5Ctheta%20%3D%5Cfrac%7Bh_h%7D%7BD%7D)
Where h_h is the height of the tower. We can solve:
![\displaystyle h_h=D\ tan\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h_h%3DD%5C%20tan%5Ctheta)
For the big triangle:
![\displaystyle h_t+h_h=D\ tan(\theta +\alpha)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h_t%2Bh_h%3DD%5C%20tan%28%5Ctheta%20%2B%5Calpha%29)
Where
is 8° and
is the height of the hill. Knowing that:
![\displaystyle h_h=D\ tan \theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h_h%3DD%5C%20tan%20%5Ctheta)
We replace it into the above equation
![\displaystyle h_t+D\ tan \theta = D\ tan(\theta +\alpha)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h_t%2BD%5C%20tan%20%5Ctheta%20%3D%20D%5C%20tan%28%5Ctheta%20%2B%5Calpha%29)
We have an equation in
, 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 }](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h_t%2BD%5C%20tan%20%5Ctheta%20%3D%20D%5C%20%5Cfrac%7Btan%5Ctheta%20%2Btan%5Calpha%7D%7B1-tan%5Ctheta%20%5C%20tan%5Calpha%20%7D)
Rearranging
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h_t-h_t%5C%20tan%20%5Ctheta%20%5C%20tan%5Calpha%20%2BD%5C%20tan%5Ctheta%20-D%5C%20tan%5E2%5Ctheta%20%5C%20tan%5Calpha%20%3DD%5C%20tan%5Ctheta%20%2BD%5C%20tan%5Calpha)
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-D%5C%20tan%5E2%5Ctheta%20%5C%20tan%5Calpha%20-h_t%5C%20tan%5Calpha%20%5C%20tan%5Ctheta%20%2Bh_t-D%5C%20tan%5Calpha%20%3D0)
Using the known values D=110, ht=30, ![\alpha=8^o](https://tex.z-dn.net/?f=%5Calpha%3D8%5Eo)
![\displaystyle -15.46\ tan^2\theta -4.22\ tan\theta +14.54=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20-15.46%5C%20tan%5E2%5Ctheta%20-4.22%5C%20tan%5Ctheta%20%2B14.54%3D0)
Solving the equation, we get two answers:
![\displaystyle tan\theta=-1.12](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20tan%5Ctheta%3D-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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20tan%5Ctheta%3D0.843)
is the correct option. Computing the angle of inclination of the hill:
![\boxed{\displaystyle \theta=40.1^o}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cdisplaystyle%20%5Ctheta%3D40.1%5Eo%7D)