Angle of elevation problem

**fActor** · July 12th 2012, 03:52 AM

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.

**Prove It** · July 12th 2012, 04:11 AM

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?

**fActor** · July 12th 2012, 04:32 AM

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)

**Prove It** · July 12th 2012, 04:38 AM

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.

**fActor** · July 12th 2012, 04:53 AM

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!

**Soroban** · July 12th 2012, 05:19 AM

Hello, fActor

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

If you put the point on the building, where is the 35^o 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 62^o.
He moves 120 m away from $A$ (and away from the building) to $B.$
Now the angle of elevation to $T$ is only 35^o.
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 $h.$

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