Question

1. two buildings are built besides each other. a video camera is mounted on the top of the shorter, 120 - m tall building. when this camera makes an angle of depression of 42° it views the bottom of the other building. if the camera makes an angle of elevation of 28° it looks at the top of the other building. determine the height of the taller building.a = 5t = 1+5

Explanation:

Step1: Find the horizontal distance between buildings

Let the horizontal distance between the two buildings be $x$. The angle of depression to the bottom of the other building is $42^{\circ}$. Since the angle of depression is equal to the angle of elevation from the bottom of the other building to the top of the shorter building. We know that $\tan42^{\circ}=\frac{120}{x}$. So, $x = \frac{120}{\tan42^{\circ}}$.

Step2: Find the height above 120m of the taller building

Let the height above 120m of the taller building be $h$. The angle of elevation to the top of the taller building from the top of the shorter building is $28^{\circ}$. Using the tangent - function, $\tan28^{\circ}=\frac{h}{x}$. Substitute $x=\frac{120}{\tan42^{\circ}}$ into this equation, we get $h = x\tan28^{\circ}=\frac{120\tan28^{\circ}}{\tan42^{\circ}}$.

Step3: Calculate the height of the taller building

The height $H$ of the taller building is $H=120 + h$. First, calculate $\tan28^{\circ}\approx0.5317$ and $\tan42^{\circ}\approx0.9004$. Then $h=\frac{120\times0.5317}{0.9004}=\frac{63.804}{0.9004}\approx70.86$. So, $H = 120+70.86=190.86$ m.

Answer:

$190.86$ m