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solve the acute triangle
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|dw:1358738191351:dw|
Draw an altitude of the triangle, PS with S on QR. Then \[PS=11sinx=12\sin61\]if x is the angle at Q. The equation will give us a solution for x. The angle at P is 180-61-x. QR is 11cosx+12cos61.
im confused do i have to use the sine law?
Actually, I think the construction gives the same information that you would get from the sin rule. Let me look for a second....
yes, use the sin law
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11cm/sin(61) = 12cm / sin(angle opposite 12cm side)
I did it by construction, but the sin rule will give exactly the same answer.
then solve for the next angle. use that angle to solve the sin law for the next side length. if you need more help just let me know.
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