Deshaun is looking up at a flag that is 50 feet away from him at an angle of elevation from ground level of 35° What is the flagpole’s height, x, and the distance between Deshaun and the top of the flagpole, y?(1 point) Responses x=35.01 ft. and y=35.70 ft. x equals 35.01 ft. and x equals 35.70 ft. x=61.04 ft. and y=35.01 ft. x equals 61.04 ft. and x equals 35.01 ft. x=71.41 ft. and y=87.17 ft. x equals 71.41 ft. and x equals 87.17 ft. x=35.01 ft. and y=61.04 ft.

To find the height of the flagpole \( x \) and the distance \( y \) from Deshaun to the top of the flagpole, we can use trigonometry.

1. Since Deshaun is looking up at the flag at an angle of elevation of \( 35^\circ \) and the distance from him to the base of the flagpole is \( 50 \) feet, we can use the tangent function to find the height of the flagpole \( x \):

\[
\tan(35^\circ) = \frac{x}{50}
\]

Rearranging this gives:

\[
x = 50 \cdot \tan(35^\circ)
\]

Calculating \( \tan(35^\circ) \):

\[
x \approx 50 \cdot 0.7002 \approx 35.01 \text{ ft}
\]

2. To find the distance \( y \) from Deshaun to the top of the flagpole, we can use the sine function. The distance \( y \) can be calculated as follows:

\[
y = \frac{x}{\sin(35^\circ)}
\]

Using \( x \approx 35.01 \text{ ft} \):

\[
y = \frac{35.01}{\sin(35^\circ)}
\]

Calculating \( \sin(35^\circ) \):

\[
y \approx \frac{35.01}{0.5736} \approx 61.04 \text{ ft}
\]

Thus, we have:

- Height of the flagpole \( x \approx 35.01 \text{ ft} \)
- Distance from Deshaun to the top of the flagpole \( y \approx 61.04 \text{ ft} \)

The correct answer based on the calculations is:

**x = 35.01 ft. and y = 61.04 ft.**