Towers A and B are located 5 miles apart. A ranger spots a fire at a 42-degree angle from tower A. Another fire ranger spots the
same fire at a 64-degree angle from tower B. To the nearest tenth of a mile, how far from tower B is the fire?
1 answer:
Answer: The fire is 3.5 miles from tower B
Step-by-step explanation: Please refer to the attached diagram. The triangle in the attached diagram illustrates the clues given in the question. Both rangers are standing at points A and B respectively with a distance of 5 miles between them, which is line AB. Also, one ranger spots a fire from a tower at an angle of 42 degrees, which is point A. Another ranger spots the same fire from another tower, but from an angle of 64 degrees, which is point B. The fire is at point C on the triangle. Now we have a triangle with only one side known (5 miles) and three angles known (the third angle is computed as 180 - {64+42} which equals 74) which are 64 degrees, 42 degrees and 74 degrees.
The distance from the fire to tower B is calculated using the law of sines. (Note that this is not a right angled triangle, hence we cannot use trigonometric ratios). The law of sines is expressed as follows;
a/SinA = b/SinB or
a/SinA = c/SinC
Depending on the sides and angles we are given and the ones we are to calculate.
The distance from the fire to tower B is line BC, labeled as a in our diagram. Using the law of sines
a/SinA = c/SinC
(Note also that a is directly facing angle A, c is directly facing angle C, and so on)
a/SinA = c/SinC
a/Sin 42 = 5/Sin 74
By cross multiplication we now have
a (Sin 74) = 5 (Sin 42)
Divide both sides of the equation by Sin 74 and we now arrive at
a = 5 (Sin 42)/Sin 74
a = 5 (0.6691)/0.9613
a = 3.3455/0.9613
a = 3.4802
{rounded to the nearest tenth of a mile, a equals 3.5}
Therefore the distance from tower B to the fire is approximately 3.5 miles
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