In triangle ABC, a = 9, c = 5, and angle B = 120°. Find the measure of angle A to the nearest degree. Show your work.
hurr-durr
huh? lol @PeterPan
Just what I said. hurr-durr. Now this appears to be nothing more than a law-of-cosines problem :)
Yes, I did it but i got stuck and i did it again then i got stuck again. I used c^=a^2+b^2-2(a*b)cos(x)
@PeterPan
You should find b first. \[\large b^2 = a^2 + c^2 -2ac \ \cos(B)\]
okay one sec @PeterPan
i did b^ = 9^ + 5^ - 2(9*5)cos(120) and i got 151...now what? @PeterPan
well that's b^2 isn't it? get its square root ^^
yes and now i got 12.2 @PeterPan
I don't know.... maybe include three decimal places to be safe?
okay now what? @PeterPan
Well, you have angle B and side b, and you have side a. Use the Law of Sines now ^^
\[\Large \frac{\sin(A)}{a}=\frac{\sin(B)}{b}\]
2.01=.023?? oh gosh i suck @PeterPan
Best way to fix is to show your work. Then we'll see what went wrong ^^
sin9/9 = sin12.288/12.288 i crossed multiplied @PeterPan
a is not the same as A silly :P A is your missing angle a is the side length (9) b is also a side length (the one you just found, 12.288) B is the angle (120 degrees) Let's try again, shall we?
okay sin(120)*9=7.79 but what to do for A since i dont know it? im sorry @PeterPan
that's what we're trying to find, silly :P
Ohh okay . so.. 12.288sin(A)=2.133 soooooooooooooo 7.79=2.133 @PeterPan
it's really confusing with you skipping steps, so could you please show your work, step-by-step? :)
i need to know where you're getting all this.
okay lol 12.288sin(A)= 2.133 AND SIN(120)*9=7.79 @PeterPan
where does 2.133 come from?
NOOOOOOOOOOOO I GOT IT I GOT IT ! OKAY SO 12.288SINA=7.79 THE I DIVIDED 12.288 ON BOTH SIDES WHICH GAVE ME SINA=.634 THEN I DID SIN-1(.634 WHICH GAVE ME 39.3 @PeterPan
That's more like it ^^
Lol thanks for helping me. Even though i was so hard to help I bet ! @PeterPan
no worries ^^
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