SOLUTION: Jarrod walked 3 miles due north before turning left and walking 3.5 miles due west. He turned again and returned to his starting point by walking a straight path. What was the angl
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-> SOLUTION: Jarrod walked 3 miles due north before turning left and walking 3.5 miles due west. He turned again and returned to his starting point by walking a straight path. What was the angl
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Question 1111148: Jarrod walked 3 miles due north before turning left and walking 3.5 miles due west. He turned again and returned to his starting point by walking a straight path. What was the angle of the last turn? Round to the nearest degree. Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website! 3 miles, turn a 90 degree angle? You don't say, I'll assume that's what you meant.
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Side a: 3
Side b: 3.5
Side c: sqrt(3^2+3.5^2) = 4.61
We know that angles
A+B+C = 180 (the angles of every triangle = 180)
And we know that angle C = 90. Let's find the other two, using SOH CAH TOA, in this case let's choose SOH (sin(o/h). But instead of using the "sin"key in your calculator we are going to use the "sin^-1" key. In the expression sin^-1, -1 is not an exponent, it is the inverse of the sine function.
So we are going to find sin^-1 instead of sin:
Angle A = sin^-1(3/4.61) = 40.6
Angle B = sin^-1(3.5/4.61) = 49.4
and of course, Angle C = 90
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Check: IN every triangle A+B+C = 180
40.6+49.4+90 = 180 Correct.
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P.S.:
Did you draw it? I did:
