Trig.

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I have answers for these problems, but I wanted to check if I had them right because I wasn't sure on some of them....Thanks.
Solve the triangle:
1. a=4, b=8, alpha=30 deg.
answer- beta=90 deg., gamma=60 deg., c=7
2. a=5, b=7, alpha=30 deg.
answer- beta=44 deg., gamma=106 deg., c=10
3. a=3, b=8, alpha=30 deg.
and this one I am doing something completely wrong because I cant find out the first answer...so I do not have an answer for it yet
4. alpha=53 deg., gamma=105 deg., b=42
answer- beta=22 deg., a=108, c=90
5. alpha= 65 deg., b=103, c=53
answer- a=155, gamma=25 deg., beta=90 deg.
6. a=43, b=48, c=53
answer- gamma=71 deg., alpha=50 deg., beta=59 deg.


this isn't a homework problem but something he told us about and I would like to know...how does the law of cosines relate to the Pythagorean theorem? if anyone knows that.

1. Is not quite right. It is true than beta = 90 degrees (from the law of sines), and that makes gamma = 180-30-90 = 60 degrees. However, the next step is
c/sin 60 = b/sin beta = 8
c = 8 sin 60 = 6.928..
You have the formulas right, but don't round off the side lengths to whole numbers.

As for your last question, the law of cosines reduces to the Pythagorean theorem when c is the hypotenuse and gamma equals 90 degrees

c^2 = a^2 + b^2 - 2 ab cos 90
= a^2 + b^2

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