Question

A student wants to know how far above the ground the top of a leaning flagpole is. At high noon, when the sun is almost directly overhead, the shadow cast by the pole is 5 ft long. The student holds a plumb bob with a string 3 ft long up to the flagpole and determines that the point of the plumb bob touches the ground 11 in. from the base of the flagpole. How far above the ground is the top of the pole?

Answer 1

Answer:

16.30 ft ≈ 16 ft 4 in

Step-by-step explanation:

Please see the image attached. According to the figure, AB is the leaning flag. When the sun is directly overhead at noon, the shadow of the flag is the line AC in the figure. The length of AC is 5ft. The length of the string of plumb bob is 3ft and this is represented by the line XY. Y is the point of the plumb bob and this is at a distance of 11 in ( = $\frac{11}{12} = 0.92$ft from the base of the flag, i.e, AY = 0.92ft.

Now, the angle between the AB and AC is taken as ∅. In traingle AXY, tan(∅) = $\frac{XY}{AY}$. In triangle ABC, for the same ∅, tan(∅) = $\frac{BC}{AC}$. Therefore,  tan(∅) = $\frac{XY}{AY} = \frac{BC}{AC}$. Here, XY = 3ft, AY = 0.92ft, AC = 5ft.

$\frac{XY}{AY} = \frac{BC}{AC} == \frac{3}{0.92} = \frac{BC}{5}$

Therefore, $BC = \frac{3}{0.92}$×5 = 16.30 ft ≈ 16 ft 4 in.