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diana works in a building that is 130 feet tall. she is outside, lookin…

Question

diana works in a building that is 130 feet tall. she is outside, looking up at the building at an angle of 37° from her feet to the top of the building. if diana walks forward and her angle looking to the top of the building changes to 40°, how much closer is she to the building? round the answer to the nearest tenth of a foot. 30.2 ft 97.2 ft 17.6 ft 10.3 ft

Explanation:

Step1: Find initial distance

We use the tangent function $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. The height of the building (opposite side) is $h = 130$ feet. When $\theta = 37^{\circ}$, let the initial distance be $d_1$. So $\tan37^{\circ}=\frac{130}{d_1}$, and $d_1=\frac{130}{\tan37^{\circ}}$. Since $\tan37^{\circ}\approx0.7536$, $d_1=\frac{130}{0.7536}\approx172.5$ feet.

Step2: Find final distance

When $\theta = 40^{\circ}$, let the final - distance be $d_2$. Using $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, we have $\tan40^{\circ}=\frac{130}{d_2}$, and $d_2=\frac{130}{\tan40^{\circ}}$. Since $\tan40^{\circ}\approx0.8391$, $d_2=\frac{130}{0.8391}\approx154.9$ feet.

Step3: Calculate the difference

The difference in distances $\Delta d=d_1 - d_2$. $\Delta d\approx172.5-154.9 = 17.6$ feet.

Answer:

$17.6$ ft