Question

Xavier is looking up at a building that is 123 meters away from him, at an ang of elevation of 52° What is the building's height above Xavier's eye level, x. and the distance between Xavier and the top of the building, y? (1 point)
O a = 157.43 m and y = 98.26 m
O a = 199.70 m and y - 157.43 m
O z = 157.43 m and y - 199,79 m
O a = 96,1 m and y = 156.09 m

Answer 1

To find the building's height above Xavier's eye level, we can use the trigonometric function tangent.

From the given information, we know that the angle of elevation is 52° and the distance between Xavier and the building is 123 meters. Let x represent the height above Xavier's eye level.

Using the tangent function, we have:

tan(52°) = x/123

To find x, we can rearrange the equation:

x = tan(52°) * 123

Calculating this, we get x ≈ 98.26 m.

Now, to find the distance between Xavier and the top of the building (y), we can use the trigonometric function sine.

Using the angle of elevation (52°) and the height above Xavier's eye level (98.26 m), let y represent the distance.

Using the sine function, we have:

sin(52°) = y/123

To find y, we can rearrange the equation:

y = sin(52°) * 123

Calculating this, we get y ≈ 96.1 m.

Therefore, the correct answer is option O a = 157.43 m and y = 98.26 m.