Question

A surveyor stands at point A, which is due south of a tower OT of height h m. The angle of elevation of the top of the tower from A is 45 degrees. The surveyor then walks 100 m due east to point B, from where she measures the angle of elevation of the top of the tower to be 30 degrees.

P.S the answer says that the bearing is 130 degrees.

PLEASE PLEAS HELP ME I am really struggling to figure out how...

Answer 1

Solution.

Let bearing required is from base of the tower to point $B$ . Let $T$ is a top of the tower.

At first consider the triangle made by points $A, B$ and $T$ with $A = 45{}^{\circ}$ , $B = 30{}^{\circ}$ and $T = 105{}^{\circ}$ .

Then apply the law of sines to calculate $b$ (distance of $T$ to point $A$ ), also note that the distance from $A$ to $B$ is 100 meters:

$$\frac {b}{\sin 3 0 {}^ {\circ}} = \frac {1 0 0}{\sin 1 0 5 {}^ {\circ}}$$ $$b = \frac {1 0 0 \sin 3 0 {}^ {\circ}}{\sin 1 0 5 {}^ {\circ}} \approx 5 1. 7 6 m$$

Then consider the flat triangle $AOB$ where $O$ is the base of the tower. We know that the length of $AO$ is the height of the tower because triangle $AOT$ is an isosceles right triangle with two $45{}^{\circ}$ angles and one $90{}^{\circ}$ angle and $b$ is its hypotenuse, so

$$5 1. 7 6 ^ {2} = 2 h ^ {2}$$ $$2 6 7 9. 1 = 2 h ^ {2}$$ $$h ^ {2} = 1 3 3 9. 5 5 m ^ {2}$$ $$h = \sqrt {1 3 3 9 . 5 5} \approx 3 6. 6 m$$

In right triangle $AOB$ we have:

$$\tan \angle O = \frac {1 0 0}{3 6 . 6} = 2. 7 \Rightarrow \angle O = 7 0 {}^ {\circ}$$

Due south bearing is $180{}^{\circ}$ so $180{}^{\circ} - 70{}^{\circ} = 110{}^{\circ}$ bearing to $B$ .

Answer: $110{}^{\circ}$