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.
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.
The observer stands at $\displaystyle A.$Code:* T * * | * * | * * | h * * | * 35o * 62o | * - - - - - * - - - - - * B 120 A x :
The angle of elevation to the top of the building $\displaystyle (T)$ is 62^{o}.
He moves 120 m away from $\displaystyle A$ (and away from the building) to $\displaystyle B.$
Now the angle of elevation to $\displaystyle T$ is only 35^{o}.
We want $\displaystyle h$, the height of the building.
Let $\displaystyle x$ be the distance from $\displaystyle A$ to the base of the building.
We have: .$\displaystyle \begin{Bmatrix}\tan62^o &=& \dfrac{h}{x} & [1] \\ \\[-2mm] \tan35^o &=& \dfrac{h}{x+120} & [2] \end{Bmatrix}$
From [1]: .$\displaystyle x \:=\:\frac{h}{\tan62^o}\;\;[3]$
From [2]: .$\displaystyle x \:=\:\frac{h}{\tan35^o} - 120 \;\;[4]$
Equate [3] and [4] and solve for $\displaystyle h.$