SOLUTION: To the nearest tenth of a degree, calculate the angles of the triangle with vertices (0, 0), (6, 3), and (1, 8).

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Question 1154772: To the nearest tenth of a degree, calculate the angles of the triangle with vertices (0, 0), (6, 3), and (1, 8).
Answer by VFBundy(438) About Me  (Show Source):
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Use the Law of Cosines:

C² = A² + B² - 2AB(cos c)

sqrt%2865%29%5E2 = sqrt%2850%29%5E2+%2B+sqrt%2845%29%5E2+-+2sqrt%2850%29sqrt%2845%29%28cos%28c%29%29

65 = 50+%2B+45+-+2sqrt%282250%29%28cos%28c%29%29

65 = 95+-+2sqrt%282250%29%28cos%28c%29%29

-30 = -2sqrt%282250%29%28cos%28c%29%29

15 = sqrt%282250%29%28cos%28c%29%29

15%2Fsqrt%282250%29 = cos%28c%29

cos%28c%29 = 0.3162

c = arccos%280.3162%29

c = 1.2491

Now, use the Law of Sines:

b%2Fsqrt%2845%29 = c%2Fsqrt%2865%29

b%2Fsqrt%2845%29 = 1.2491%2Fsqrt%2865%29

Cross-multiply:

b%2Asqrt%2865%29 = 1.2491%2Asqrt%2845%29

b%2Asqrt%2865%29 = 8.3792

b = 8.3792%2Fsqrt%2865%29

b = 1.0393

So, we know b = 1.0393 radians. This is 59.6 degrees. We also know c = 1.2491 radians. This is 71.6 degrees.

Subtracting both these numbers from 180 degrees, this means a = 48.8 degrees.

In summary:

a = 48.8 degrees
b = 59.6 degrees
c = 71.6 degrees