Question

math hw

Answer 1

Are you going to post the question?

Answer 2

Use the Law of Cosines to find the angles.
c^2 = a^2 + b^2 - 2abcos(C)
where C in cos(C) is the angle included between sides a and b.

Answer 3

No square root in this problem because we need to find the angle C (upper case) not the length of the side c (lower case).

Answer 4

Yes, but a calculator should be able to handle it.

Answer 5

c^2 = a^2 + b^2 - 2abcos(C)
We need to solve for C. Isolate cos(C):
2abcos(C) = a^2 + b^2 - c^2
cos(C) = ( a^2 + b^2 - c^2 ) / (2ab)
You have: c = 255; a = 270; b = 442.85.
Find cos(C). Then take the inverse and find angle C.

Answer 6

I am getting cos(C) = 0.8530
C = inverse cosine of 0.8530 = 31.5 degrees.

Use the law of cosines to find one more angle, either A or B.
Then the third angle of the triangle can be found by using the fact that three angles of a triangle add up to 180 degrees.

Answer 7

a^2 = b^2 + c^2 - 2bccos(A)
2bccos(A) = (b^2 + c^2 - a^2)
cos(A) = (b^2 + c^2 - a^2) / (2bc)

Answer 8

I am getting cos(A) = 0.8335
Take the inverse cosine to find angle A

Answer 9

Yes. A and B are correct.

Answer 10

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