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Andre45 [30]
3 years ago
14

A hot air balloon is flying above Groveburg. To the left side of the balloon, the balloonist measure the angle of depression to

the Groveburg soccer fields to be 20° 15'. To the right side of the balloon, the balloonist measures the angle of depression to the high school football field to be 62° 30'. The distance between the two athletic complexes is 4 miles.
What is the distance from the balloon to the football field?

Mathematics
1 answer:
const2013 [10]3 years ago
4 0

Answer:

The answer is 1.4 miles ⇒ answer (d)

Step-by-step explanation:

* Let the balloon is at the vertex A, and the soccer fields

  at vertex B and the football field at vertex C

∴ m∠B = 20° 15' = 20.25°

∴ m∠C = 62° 30' = 62.5°

∴ m∠A = 180 - 20.25 - 62.5 = 97.25°

∵ BC = 4 miles

* By using the sin Rule:

∵ AC/sin(B) = BC/sin(A)

∴ AC = (BC)(sinB)/sin(A)

∴ AC = [4 × sin(20.25)] ÷ sin(97.25) = 1.395 ≅ 1.4 miles

∴ The distance from the balloon to the football filed = 1.4 miles

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To solve this problem, we need to know 2 relationships:

<h2>1. AC = AB + BC</h2>

The distance of AC is the sum of AB and BC.

AC = AB + BC

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We know this since B is in the middle of AC, so the distance from B to A (BA) is the same as the distance from B to C (BC).

You can see the attached image (at the bottom) for a visualization of this.

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Let's put in the values of AB and BC:

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Before we put in x = 8, we can simplify this:

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We group x and 3x and add those together. Then we subtract 7 from 9.

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<h2>Answer</h2>

x = 8

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