Given angle A and two sides, find Angle B.

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Homework Statement



In triangle ABC, A=65 degrees, b=9, and a=10. Find B.

Homework Equations


The Attempt at a Solution



I honestly don't know how to start this problem... I tried to use the law of cosines to find side C, and then again to find Angle B, but that answer was incorrect. I haven't been in Trig for two years, and can't think of how else to start it.

Thanks for any help!
 
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Triangle ABC with vertices A at left, B at right and C at top, altitude h drawn from C to base c. Labels b=9 on side AC, a=10 on side BC, and a 65° angle at A.


First use the sine to find h, and then use again the sine to find B, using arcsin(B).

[tex]\sin A = \frac{h}{b}\text{ and } \sin B = \frac{h}{a}.[/tex]

Regards.
 
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Дьявол said:
Triangle ABC with vertex C at the top and altitude h drawn from C to base c. Side b = 9 runs from A to C, side a = 10 from C to B, and the angle at A is 65 degrees.


First use the law of sine to find h, and then use again the law of sine to find B, using arcsin(B).

[tex]\sin A = \frac{h}{b}\text{ and } \sin B = \frac{h}{a}.[/tex]

Regards.
so that h= b sin A= a sin B and, as a result, [itex]\frac{sin A}{a}= \frac{sin B}{b}[/itex]. That is the "sine law" Bohrok is talking about.
 
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HallsofIvy said:
so that h= b sin A= a sin B and, as a result, [math]\frac{sin A}{a}= \frac{sin B}{b}[/math]. That is the "sine law" Bohrok is talking about.
Yes, you're right. I misspelled the words. I thought of "sine" and not "sine law".

Anyway, the things are same, and he would come up with the same result.
 
Yes, I wasn't criticizing, just pointing out to JacklinH that you were both giving the same advice!