# Math Help - Angle of elevation problem

1. ## Angle of elevation problem

The angle of elevation at the top of a building at point A (on level ground) is 62 degrees. At a point 120m away from A, the angle of elevation is found to be 35 degrees. Find the height of the building.

2. ## Re: Angle of elevation problem

Originally Posted by fActor
The angle of elevation at the top of a building at point A (on level ground) is 62 degrees. At a point 120m away from A, the angle of elevation is found to be 35 degrees. Find the height of the building.
Have you started by drawing a diagram?

3. ## Re: Angle of elevation problem

Originally Posted by Prove It
Have you started by drawing a diagram?
Yes, but I'm unsure if the point away from A is supposed to lie on the building. I drew it on the building but didn't get the answer in the textbook (134m)

4. ## Re: Angle of elevation problem

Originally Posted by fActor
Yes, but I'm unsure if the point away from A is supposed to lie on the building. I drew it on the building but didn't get the answer in the textbook (134m)
I expect that the point away from A is on the ground.

5. ## Re: Angle of elevation problem

Originally Posted by Prove It
I expect that the point away from A is on the ground.
Oh yeah, you're right. I got the answer now. Thanks!

6. ## Re: Angle of elevation problem

Hello, fActor

From a point A on level ground the angle of elevation to the top of a building is 62o.
At a point 120m away from A, the angle of elevation is found to be 35o.
Find the height of the building.

If you put the point on the building, where is the 35o angle?

There is basically one diagram that satisfies the description.

Code:
                              * T
* * |
*   *   |
*     *     | h
*       *       |
* 35o     * 62o     |
* - - - - - * - - - - - *
B    120    A     x     :
The observer stands at $A.$
The angle of elevation to the top of the building $(T)$ is 62o.
He moves 120 m away from $A$ (and away from the building) to $B.$
Now the angle of elevation to $T$ is only 35o.
We want $h$, the height of the building.

Let $x$ be the distance from $A$ to the base of the building.

We have: . $\begin{Bmatrix}\tan62^o &=& \dfrac{h}{x} & [1] \\ \\[-2mm] \tan35^o &=& \dfrac{h}{x+120} & [2] \end{Bmatrix}$

From [1]: . $x \:=\:\frac{h}{\tan62^o}\;\;[3]$

From [2]: . $x \:=\:\frac{h}{\tan35^o} - 120 \;\;[4]$

Equate [3] and [4] and solve for $h.$