Astudent 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?

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 ( = 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(∅) = . In triangle ABC, for the same ∅, tan(∅) = . Therefore,  tan(∅) = . Here, XY = 3ft, AY = 0.92ft, AC = 5ft.

Therefore, ×5 = 16.30 ft ≈ 16 ft 4 in.

they are simplified

step-by-step explanation:

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet