Stadium Pole?

Question

You want to measure the height of a stadium light pole. However, it is on the other side of a fence and you don't want to test the trespassing laws. Instead, you stand at point A and measure the angle to the top of the pole and find it to be 40 degrees. You then walk 50 feet from point A in a direction directly away from the pole and measure the angle to the top again. This time it measures 28 degrees. How tall is the pole?

Please show each step!

Thank you!

Answer 1

The tangent of an angle is defined to be the ratio of the opposite leg of a right triangle to the adjacent leg of the right triangle.

First, using A for the angle, the pole, y, for the opposite side and x for the distance from point A to the pole write an equation.

Tan 40 = y/x

Then do the same from point B

Tan 28 = y/(x + 50)

Solve the first equation for y and substitute into the second

x tan 40 = y

tan 28 = (x tan 40) / (x + 50)

Cross multiply

x tan 28 + 50 tan 28 = x tan 40

Subtract x tan 28 to get both x-terms on one side

50 tan 28 = x tan 40 - x tan 28

Factor out x

50 tan 28 = x(tan 40 - tan 28)

Divide by (tan 40 - tan 28)

(50 tan 28) / (tan 40 - tan 28) = x

x is approximately 86.488 ft.

Substitute into either equation and solve for y.

86.488 tan 40 = y

the height of the pole is approximately 72.572 ft.

Answer 2

This gives you two right angle triangles that share the same "opposite" side (the pole), the same 90 degree angle (at the base of the pole) and the adjacent sides are overlapping (i.e., from the pole to point A, the segment belongs to both triangles).

Let h be the height of the pole.
Let x be the distance from the pole to point A

Then Tan(40) = h/x

In the longer triangle, the height is still h but the bottom side is now x+50

Tan(28) = h/(x+50)

h is common to both, so try substitution:

in first one, h = x Tan(40)
substitute in second:
Tan(28) = x Tan(40) / (x+50)
x Tan(28) + 20 Tan (28) = x Tan(40)
20 Tan(28) = x Tan(40) - x Tan(28)
20 Tan(28) = x (Tan(40) - Tan(28))
[20 Tan(28)] / [Tan(40) - Tan(28)] = x

Once you have x, you can find h.


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Or, going directly for h:

Tan(40) = h/x
then x = h/Tan(40)
Tan(28) = h/(x+50)
then x+50 = h/Tan(28), and
x = -50 + [h/Tan(28)]

subtract equation 1 from equation 2:

x - x = -50 + [h/Tan(28)] - [h/Tan(40)]

50 = h [ Cot(28) - Cot(40)]
where Cot(28) is the cotangent of 28, meaning 1/Tan(28)

solve for h.

Answer 3

Call the base of the pole o and the top of the pole C.
Call the point 50 feet past point A point Z.

angle azc = 28 and angle zac is supplement of angle
oac so angle zac is 140 degrees. Thus angle zca = 12 degrees.
usiing law of sines we get 50/ sin 12 = AC /sin 28
So AC = 50 sin28/sin12 = 112.9

OC/AC = sin 40 so OC = 112.9sin40 = 72.6 feet = pole height

Answer 4

the idea is that you make a right angled triangle with yourself, the ground and the top and bottom of the pole.

let the height of the pole be, h. and the inital distance (from A to pole), a. final distance, b.

tan40 = h/a

tan28 = h/(a+50)

dividing the 2 equations,

(a+50)/a = tan40/tan28

a[1-(tan40/tan28)] = -50

so, a = -50/([1-(tan40/tan28)]

using the previous equation, tan40 = h/a, we conclude

h= a*tan40 = -50*tan40/([1-(tan40/tan28)] which is measured in feet.

i dont have a calulator but that should be correct!!

Answer 5

Draw two right trianlges. Both share one leg (the pole = p). The smaller one will be a distance,d, away --so its leg will be d
the other one move 50 feet further so it will be d+50
So you have one trianlge with angle 40deg and adjacent d and opposite p. The other trianlge has angle 28 deg and adjacent (d+50) and opposite p.
Now you can set up two equations using SOHCAHTOA
using opposite and adjacent, we want tangent

tan 40 = p/d
tan 28 = p/(d+50) ===> two eqns with two unknowns.

p=d tan40
p=(d+50) tan28 ===> set them equal to get d
d tan40 = (d+50)tan28 ===> solve (make sure calc in degree mode
d tan40 = d tan28 + 50 tan28
d(tan40-tan28) = 50 tan28
d = (50tan28) / (tan40 - tan28)
d approx= 86.5


now plug in to get p(pole)
p = d tan40 = (86.5 )tan40 = 72.6

{I need to get my calvulator.. be back}

Answer 6

call the top of the pole t, the bottom b, the ast point a & the 2nd point c. we have 2 right triangles, abt & cbt. we also have triangle act. we know that ac is 50', angle c=28º, angle cat=180º-40º=140º therefore angle atc=180º-140º-28º=12º
using thre law of sines
50/sin 12º=ct/sin 140º so
ct=50*sin 140º /sin 12º
ct=154.58'
now looking at triangle cbt, ct is the hypotenuse=154.58' & bt, the length we want is the side opposite the 28º angle so
sin 28º=bt/154.58'
bt=154.58' *sin 28º
bt=72.57' or about 73'