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. 10.3 ft 17.6 ft 30.2 ft 97.2 ft

Explanation:

Step1: Find initial distance

We use the tangent function $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Initially, $\theta = 37^{\circ}$ and the height of the building (opposite side) $h = 130$ ft. Let the initial distance from the building be $x_1$. So, $\tan37^{\circ}=\frac{130}{x_1}$, and $x_1=\frac{130}{\tan37^{\circ}}$. Since $\tan37^{\circ}\approx0.7536$, $x_1=\frac{130}{0.7536}\approx172.5$ ft.

Step2: Find new - distance

After walking forward, the angle changes to $40^{\circ}$. Let the new distance from the building be $x_2$. Using the tangent function again, $\tan40^{\circ}=\frac{130}{x_2}$, and $x_2 = \frac{130}{\tan40^{\circ}}$. Since $\tan40^{\circ}\approx0.8391$, $x_2=\frac{130}{0.8391}\approx154.9$ ft.

Step3: Calculate the distance walked

The distance she walked forward $d=x_1 - x_2$. So, $d=172.5-154.9 = 17.6$ ft.

Answer:

C. 17.6 ft