To find the height of the building above Xavier's eye level (x) and the distance to the top of the building (y), we can use trigonometric functions.
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Height of the Building above Xavier's Eye Level (x):
We can use the tangent function: \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \] Here, \(\theta = 52^\circ\), the opposite side is the height of the building (x), and the adjacent side is the distance to the building (123 meters).
Therefore: \[ \tan(52^\circ) = \frac{x}{123} \]
Rearranging gives: \[ x = 123 \cdot \tan(52^\circ) \]
Using a calculator: \[ \tan(52^\circ) \approx 1.2799 \] \[ x \approx 123 \cdot 1.2799 \approx 157.43 \text{ m} \]
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Distance from Xavier to the Top of the Building (y):
We can use the hypotenuse of the triangle formed: \[ y = \frac{x}{\sin(52^\circ)} \]
Using a calculator: \[ \sin(52^\circ) \approx 0.7880 \] \[ y^2 = 123^2 + x^2 = 123^2 + (157.43)^2 \] \[ = 15129 + 24840.4649 \approx 39969.4649 \] \[ y = \sqrt{39969.4649} \approx 199.92 \text{ m} , \text{ (rounding to two decimal places)} \]
Based on the calculations:
- The height of the building above Xavier's eye level: \( x \approx 157.43 , \text{m} \)
- The distance from Xavier to the top of the building: \( y \approx 199.79 , \text{m} \)
Thus, the correct option is: x=157.43 m and y=199.79 m.