To solve for 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 some trigonometry.
- Finding the height of the building above Xavier's eye level (x):
Given:
- The distance from Xavier to the building (adjacent side, a) = 123 meters
- The angle of elevation (θ) = 52 degrees
We can use the tangent function, which relates the angle to the opposite and adjacent sides:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]
Here, the opposite side represents the height above eye level (x), so we can write:
\[ \tan(52^\circ) = \frac{x}{123} \]
To isolate x, we rearrange the equation:
\[ x = 123 \cdot \tan(52^\circ) \]
Now, we compute \( \tan(52^\circ) \):
Using a calculator:
\[ \tan(52^\circ) \approx 1.2799 \]
Now plug this value in:
\[ x = 123 \cdot 1.2799 \approx 157.4 \text{ meters} \]
So, the height of the building above Xavier's eye level (x) is approximately 157.4 meters.
- Finding the distance between Xavier and the top of the building (y):
To find y, we can use the Pythagorean theorem since we now have one side (adjacent) and the opposite side (height).
\[ y^2 = 123^2 + x^2 \]
Substituting x:
\[ y^2 = 123^2 + 157.4^2 \]
Calculating:
\[ 123^2 = 15129 \] \[ 157.4^2 \approx 24855.76 \]
Now add them:
\[ y^2 = 15129 + 24855.76 \approx 39984.76 \]
Taking the square root of both sides:
\[ y \approx \sqrt{39984.76} \approx 199.96 \text{ meters} \]
Thus, the distance between Xavier and the top of the building (y) is approximately 200.0 meters.
In summary:
- Height above eye level (x): 157.4 meters
- Distance to the top of the building (y): 200.0 meters
1 answer