Question

A photographer points a camera at a window in a nearby building forming an angle of 42° with the camera platform. if the camera is 52 m from the building, how high above the platform is the window, to the nearest hundredth of a meter?

Answer 1

Answer:

46.82 m

Step-by-step explanation:

Given : A photographer points a camera at a window in a nearby building forming an angle of 42° with the camera platform.

To Find: if the camera is 52 m from the building, how high above the platform is the window, to the nearest hundredth of a meter?

Solution:

Refer the attached figure.

A photographer points a camera at a window in a nearby building forming an angle of 42° with the camera platform i.e. ∠ACB = 42°

The camera is 52 m from the building i.e. BC = 52 m

Now we are supposed to find how high above the platform is the window i.e. AB

In ΔABC

Using trigonometric ratio

$Tan \theta = \frac{Perpendicular}{base}$

$Tan 42^{\circ} = \frac{AB}{BC}$

$Tan 42^{\circ} = \frac{AB}{52}$

$0.9004 = \frac{AB}{52}$

$0.9004 \times 52 =AB$

$46.8208 =AB$

Thus the window is 46.82 m above the platform.

Answer 2

the base is 52 ft and the angle is 42

 to find the height

 multiply 52 by the tangent of 42

52 x tan(42) = 46.821,

round to nearest hundredth = 46.82 feet