Question

Triangle ABC has side a = 12, b = 16,
and c = 19.
Find the measure of angle C to the nearest whole number.

Answer 1

Use the cosine rule here.
If you are not familiar with this rule then this site might help explain it: http://www.y-maths.co.uk/trisincosrl.htm

Answer 2

Show Steps! Plz

Answer 3

have you read the details on the link I posted?

Answer 4

Yess

Answer 5

so how far did you get before you got stuck?

Answer 6

Im not sure how to get angel A.
I got 39 but it saids is not the answer
i dont know what i did wrong

Answer 7

please show your steps so that I can spot where you went wrong

Answer 8

\[Cos A= 12^{2}-16^{2}-19^{2} \div(-2(16)(19))\]
\[Cos A=(14-256-361)/-608\]
\[Cos A= 0.7780 =39\]

Answer 9

the cosine rule states, for angle A:\[a^2=b^2+c^2-2bc\cos(A)\]therefore:\[\cos(A)=\frac{b^2+c^2-a^2}{2bc}\]

Answer 10

however, you state you want angle C in your question, in which case it should be:\[\cos(C)=\frac{a^2+b^2-c^2}{2ab}\]

Answer 11

So I dont need to find angle A to get to C?

Answer 12

why find angle A when you can find angle C directly using the cosine rule?

Answer 13

I Though I Had To .
Thank You I Got The Answer Now (:

Answer 14

ok - glad you were able to work it out in the end :)