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klio [65]
4 years ago
9

Andrew wants to measure the height of a traffic light. He walks exactly 20 feet from the base of the traffic light and looks up.

The angle from his eyes to the top of the traffic light is 40 degrees. Andrews eyes are at a height of 5 feet when hes looking up . How tall is the traffic light?
The traffic light is approximatly _ feet tall

Mathematics
2 answers:
slava [35]4 years ago
4 0

<u>Answer-</u>

<em>The height of the traffic light is 21.8 feet.</em>

<u>Solution-</u>

As shown in figure, we can see the right triangle formed from the situation,

From the question,

BC = the distance walked by Andrew from the traffic light = 20 feet

Height of Andrew's eye = 5 feet

Height of the traffic light = AB + Height of Andrew's eye

From the properties of triangle,

\Rightarrow \tan \theta =\frac{Height}{Base}

\Rightarrow \tan C =\frac{AB}{BC}

\Rightarrow AB=\tan C\times BC

\Rightarrow AB=\tan 40\times 20

\Rightarrow AB=16.8

Therefore, height of the traffic light = AB + Height of Andrew's eye = 16.8+5 = 21.8 feet


Alexus [3.1K]4 years ago
4 0

Answer:

The other person is right, the correct answer is 21.8.

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<h2>~AnonymousHelper1807</h2>
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A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depressio
guapka [62]
Answer: 4.5 miles

Explanation:

When you draw the situation you find two triangles.

1) Triangle to the east of the helicopter

a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°

b) hypotensue = distance between the high school and the helicopter

c) opposite-leg to angle 20° = heigth of the helicopter

d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x

2) triangle to the west of the helicopter

a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°

b)  distance between the helicopter and the elementary school = hypotenuse

c) opposite-leg to angle 62° = height of the helicopter

d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x

3) tangent ratios

a) triangle with the helicpoter and the high school

tan 20° = Height / x ⇒ height = x tan 20°

b) triangle with the helicopter and the elementary school

tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°

c) equal the height from both triangles:

x tan 20° = (5 - x) tan 62°

x tan 20° = 5 tan 62° - x tan 62°

x tan 20° + x tan 62° = 5 tan 62°

x  (tan 20° + tan 62°) = 5 tan 62°

⇒ x = 5 tant 62° / ( tan 20° + tan 62°)

⇒ x = 4,19 miles

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Rounded to the nearest tenth = 4.5 miles

That is the distance between the helicopter and the high school.
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