Aphotographer 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?

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

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

Thus the window is 46.82 m above the platform.

To find the answer you will plug in the first term for x and the second term for y in each ordered pair.

2(0)+3(1) = 3      Yes  3 is greater than -1
2(0)+3(-1) = -3    No -3 is  not greater than -1
2(-2)+3(1) = -1     No -1 is not greater than -1
2(-6)+3(0) = -12  no -12 is greater than -1
2(2)+3(-1) = 1      Yes  1 is  greater than -1

2x+3y>-1?
Select each correct answer.
(0,1)
(0,-1)
(-2,1)
(-6,0)
(2,-1)