Question

Sung sees the top of a 32 meter tall building by looking up at an angle of 75 degrees. To the nearest tenth of a meter, how far is Sung from the base of the building? *

Answer 1

9.0 m

Explanation:

Given:-

- The top of the building is at a height of P = 32 m

- The sight elevation from Sung to top is, θ = 75°

Find:-

To the nearest tenth of a meter, how far is Sung from the base of the building? *

Solution:-

- We can construct a right angle triangle from between Sung, foot of the building and top of the building.

- We will denote the distance between the building and Sung to be "B".

- We will apply trigonometric ratio of tangent which relates the distance of building from Sung "B" to the height of the building "H" with and angle of elevation "θ".

tan ( θ ) = P / B

B = P / tan ( θ )

- Pug in the values and solve for "B":

B = 32 / tan ( 75° )

B = 8.57 m

- The distance "B" to the nearest tenth would be = 9 m

Answer 2

31m

Explanation:

Sung is looking at the top of the 32m tall building at an angle of 75°

So, we need to calculate the distance of Sung to the base of the building.

So we call the unknown variable x

So to find x,

We solve.

x= (Sin 75°) * 32

x= 0.9659 * 32

x= 30.9088m

So, to the nearest tenth of a metre,

x= 31m.