To find the height of the building above Xavier's eye level (x) and the distance between Xavier and the top of the building (y), we can use trigonometry.
- Finding x (the height above Xavier’s eye level):
- We can use the tangent function, which relates the angle of elevation to the height (opposite side) and the distance from the building (adjacent side): \[ \tan(\theta) = \frac{x}{d} \] where \( \theta = 52^\circ \) and \( d = 123 \) m.
Rearranging gives us: \[ x = d \cdot \tan(\theta) \] \[ x = 123 \cdot \tan(52^\circ) \]
Calculating this: \[ \tan(52^\circ) \approx 1.2799 \] \[ x = 123 \cdot 1.2799 \approx 157.43 \text{ m} \]
- Finding y (the total distance from Xavier to the top of the building):
- We can use the hypotenuse with the cosine function: \[ \cos(\theta) = \frac{d}{y} \] Rearranging gives us: \[ y = \frac{d}{\cos(\theta)} \] \[ y = \frac{123}{\cos(52^\circ)} \] Calculating this: \[ \cos(52^\circ) \approx 0.6157 \] \[ y \approx \frac{123}{0.6157} \approx 199.79 \text{ m} \]
Thus,
- The height above Xavier's eye level: \( x \approx 157.43 \text{ m} \)
- The distance to the top of the building: \( y \approx 199.79 \text{ m} \)
So the correct response is: x = 157.43 m and y = 199.79 m.
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