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Trigonometry
Question

From a window \( 32.0 ft \) above the street, the ...

From a window \( 32.0 ft \) above the street, the angle of elevation to the top of the building across the street is \( 50.0 ^ { \circ } \) and the angle of depression to the base of this building is \( 16.0 ^ { \circ } \) . 

Find the height of the building across the street. 

The height of the building across the street is _____ft.

(Round to the nearest whole number as needed.) 

Answer

The height (h)(h) of the building across the street is:

\(h= h_1+ 32\)ft              (Eq. 1)

Use the tangent of the angle of elevation (\(θ_e= 50° \)) to find \(h_1\), where the opposite side is\(h_1\) and the adjacent side is x:

\(tanθ_e= \frac{opposite \space  side}{adjacent\space  side}\)

\(tan50° = \frac{h_1}{x}\)

\(x \space tan50° = h_1\)

\(h_1= x\space tan50° \)    (Cross multiply x)       (Eq. 2)

Substituting Eq. 2 in Eq. 1:

\(h= h_1+ 32\) ft            (Eq. 1)

\(h= x\space tan50° + 30\)   ft (Eq. 3)

Now we need to know the length of x, so use the tangent of the angle of depression \((θ_d= 20\degree)\) to look for it. Using the height of the window of 32 ft as the opposite side and x as the adjacent side, solve for x

\(tanθ_d= \frac{opposite \space  side}{adjacent\space  side}\)

\(tan20\degree= \frac{32}{x}\)

\(x\space tan20\degree= 32\)

\(x= \frac{32}{tan20\degree}\)

x=\(\frac{32}{0.3639702343}\)

x=87.9192774144

The value of x remains unrounded because this is not yet the final answer. Substituting this in Eq. 3, the height of the building is:

\(h= x\space tan50\degree+ 32 ft\)

\(h= 87.9192774144\space tan50\degree+ 32\) ft

=104.77810+32 ft

=136.7781

≈136.8 ft

Thus, the building across the street is 136.8 feet high.

Solution
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