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marshall27 [118]
3 years ago
6

A hiker is lost in the forest, but has his cell phone with a weak signal. Cell phones with GPS can give an approximate location

through triangulation, which works by giving distances from two known points. Suppose the hiker is within distance of two cell phone towers that are 22.5 miles apart along a straight highway (running east to west, double-dashed line). Based on the signal delay, it can be determined that the signal from the hiker's phone is 14.2 miles from Tower A and 10.9 miles from Tower B. Assume the hiker is traveling a straight path south reach the highway quickly. How far must the hiker travel to reach the highway
Mathematics
1 answer:
lianna [129]3 years ago
5 0

Answer:

The distance the hiker must travel is approximately 5.5 miles

Step-by-step explanation:

The distance between the two cell phone towers = 22.5 miles

The distance between the hiker's phone and Tower A = 14.2 miles

The distance between the hiker's phone and Tower B = 10.9 miles

The direction of the highway along which the towers are located = East to west

The direction in which the hiker is travelling to reach the highway quickly = South

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Let 'a', 'b', and 'c', represent the sides of the triangle formed by the imaginary line between the two towers, the hiker's phone and Tower A, and the hiker's hone and tower B respectively, we have;

a = 22.5 miles

b = 14.2 miles

c = 10.9 miles

Therefore, we have;

22.5² = 14.2² + 10.9² - 2 × 14.2 × 10.9 × cos(A)

cos(A) = (22.5² - (14.2² + 10.9²))/( - 2 × 14.2 × 10.9) ≈ -0.6

∠A = arccos(-0.6) ≈ 126.9°

By sine rule, we have;

a/(sin(A)) = b/(sin(B)) = c/(sin(C))

∴ sin(B) = b × sin(A)/a

∴ sin(B) = 14.2×(sin(126.9°))/22.5

∠B = arcsine(14.2×(sin(126.9°))/22.5) ≈ 30.31°

∠C = 180° - (126.9° - 30.31°) = 22.79° See No Evil

The distance the hiker must travel, d = c × sin(B)

∴ d = 10.9 × sin(30.31°) ≈ 5.5

Therefore, the distance the hiker must travel, d ≈ 5.5 miles.

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