Question

when measured from a point on the ground that is a certain distance from the base of a cell phone tower, the angle of elevation to the top of the tower is (\theta = 45^{circ}) as shown below. the height of the cell phone tower is 100 feet. what is the distance, in feet, to the cell phone tower?
(100\tan45^{circ})
(\frac{100}{sin45^{circ}})
(100sin45^{circ})
(\frac{100}{\tan45^{circ}})

Explanation:

Step1: Recall the tangent function

In right - triangle \(ABC\) (\(\angle B = 90^{\circ}\)), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\theta = 45^{\circ}\), the opposite side \(BC = 100\) (height of the tower), and the adjacent side \(AB\) is the distance we want to find.
So, \(\tan45^{\circ}=\frac{BC}{AB}\).

Step2: Solve for \(AB\)

From \(\tan45^{\circ}=\frac{100}{AB}\), we can rewrite it as \(AB=\frac{100}{\tan45^{\circ}}\).

Answer:

\(\frac{100}{\tan45^{\circ}}\)