Question

Compute each angle of the given triangle. Where necessary, use a calculator and round to one decimal place. a = 31, b = 9, c = 37

∠A = ? °
∠B = ? °
∠C = ? °

Answer 1

\[c ^{2}=a ^{2}+b ^{2}-2abcos(C)\]

Answer 2

This formula I gave you is the Law of Cosines, which lets you to solve for each angle of a triangle if your given (3 side lengths) or (2 side lengths and 1 angle between them)

Answer 3

I worked it out, but idk how to get the top number

Answer 4

I know the bottom number is 260

Answer 5

Angle A?

Answer 6

Yup the denominator is 260

Answer 7

Idk how to get the numerator

Answer 8

So if you rewrite the law of cosines to find the angle your formula would then be
\[\frac{ c ^{2}-a ^{2}-b ^{2} }{-2ab}=\cos(C)\] taking the inverse cosine would then give you
\[C=\cos^{-1} (\frac{ c ^{2}-a ^{2}-b ^{2} }{-2ab})\]

Answer 9

\[\cos C=\frac{ a^2+b^2-c^2 }{ 2ab }\]

Answer 10

@surjithayer, yes thats correct with the -1 from the denominator distributed into the numerator and reorganized.

Answer 11

If C is obtuse then cos C is negative.

Answer 12

so, its 233/260 for A

Answer 13

?

Answer 14

\[\cos A=\frac{ b^2+c^2-a^2 }{ 2bc }\]

Answer 15

How about Cos B?

Answer 16

that would be cos(B)=(a^2+c^2-b^2)/2ac