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A building casts a shadow. The building is 125 feet tall and the shadow is 13 feet in length from the base of the building. Usi...Question
A building casts a shadow. The building is 125 feet tall and the shadow is 13 feet in length from the base of the building. Using the inverse of tangent, what is the approximate angle of elevation from the end of the shadow’s length to the top of the building?
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Answered by
GPT-4o mini
To find the angle of elevation from the end of the shadow to the top of the building, we can use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right triangle.
Here, the height of the building (125 feet) is the opposite side, and the length of the shadow (13 feet) is the adjacent side.
We can use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{125}{13}
\]
To find the angle \(\theta\), we need to take the inverse tangent (arctan) of the ratio:
\[
\theta = \tan^{-1}\left(\frac{125}{13}\right)
\]
Now we can calculate this value:
\[
\frac{125}{13} \approx 9.615384615
\]
Now using this value to find the angle:
\[
\theta \approx \tan^{-1}(9.615384615)
\]
Using a calculator:
\[
\theta \approx 82.73^\circ
\]
Thus, the approximate angle of elevation from the end of the shadow to the top of the building is:
\[
\boxed{82.73^\circ}
\]
Here, the height of the building (125 feet) is the opposite side, and the length of the shadow (13 feet) is the adjacent side.
We can use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{125}{13}
\]
To find the angle \(\theta\), we need to take the inverse tangent (arctan) of the ratio:
\[
\theta = \tan^{-1}\left(\frac{125}{13}\right)
\]
Now we can calculate this value:
\[
\frac{125}{13} \approx 9.615384615
\]
Now using this value to find the angle:
\[
\theta \approx \tan^{-1}(9.615384615)
\]
Using a calculator:
\[
\theta \approx 82.73^\circ
\]
Thus, the approximate angle of elevation from the end of the shadow to the top of the building is:
\[
\boxed{82.73^\circ}
\]
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