To determine the height of the building above Xavier's eye level (x) and the distance from Xavier to the top of the building (y), we can use trigonometric functions.
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Find the height (x): We can use the tangent function since we have an angle of elevation and the adjacent side (the distance from Xavier to the building): \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \implies \tan(52^\circ) = \frac{x}{123} \] Rearranging for x: \[ x = 123 \cdot \tan(52^\circ) \] Now calculating \( \tan(52^\circ) \): \[ \tan(52^\circ) \approx 1.2799 \] Therefore: \[ x \approx 123 \cdot 1.2799 \approx 157.43 \text{ m} \]
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Find the distance to the top of the building (y): We can use the hypotenuse of the right triangle formed by Xavier, the building, and the ground: \[ y = \frac{x}{\sin(52^\circ)} = \frac{157.43}{\sin(52^\circ)} \] Now calculating \( \sin(52^\circ) \): \[ \sin(52^\circ) \approx 0.7880 \] Therefore: \[ y \approx \frac{157.43}{0.7880} \approx 199.79 \text{ m} \]
So the correct answers are: \( x = 157.43 \) m and \( y = 199.79 \) m.
Therefore, the answer is: x = 157.43 m and y = 199.79 m.
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