In triangle ABC, measure of angle A=33, a=12,b=15, what is the measure of angle b to the nearest degree?
@geoffb
Are you given a diagram? You should be able to use sin, cos, or tan to find angle B, since you know two sides (a and b).
|dw:1353915380080:dw|
Does it look like that?
If so, use a rule like \(\tan B = \frac{15}{12}\), though it might be different depending on your diagram.
my diagram looks exactly like yours
Perfect! You can solve for B using \(\tan^{-1} (\frac{15}{12})\) then.
i got a long decimal, although the choices im giveb are all whole numbers
Okay, wait... If the triangle was a right-angle triangle, you could have solved for B by simply subtracting 90-33. So I imagine that's not the case. It's not a right-angle triangle, I'm guessing.
wait, dont i use the law of sine?
Yes.
idk man do you know why i keep getting the decimal, all im doing is putting in the inverse of tan(15/12) into my calculator
No, forget I said that. That was for a right-angle triangle. Like you said, you need to use the law of sines. \(\Large\frac{a}{\sin A} = \frac{b}{\sin B}\) \(\Large\frac{12}{\sin 33} = \frac{15}{\sin B}\)
\[\sin B = \frac{15 \sin 33}{12}\]
Is 43 one of the possible answers?
yea, so i dont find the inverse?
Yes, you do. We got: \[\sin B = \frac{15 \sin 33}{12}\] so... \[B = \sin^{-1} (\frac{15 \sin 33}{12})\]
I see now, thanks out :D
No problem. Have a good night. :)
same for you :)
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