Expert: Sherman D. - 6/12/2005

Question
find the measure of the largest angle in triangle ABC.

1) b=10, c=5, angle B=20 degrees
2) a=12, b=10, c=15

find area of triangle ABC
3) a=10, b=17, c=15

Answer
I'm not 100% sure if these are correct.

1.)
b/(sinB) = c/(sinC)
10/(sin(20)) = 5/sin(C)
5sin(20) = 10sin(C)
(1/2)sin(20) = sin(C)
C = 9.84655194
C = about 10

A = 180 - (10 + 20)
A = 180 - 30
A = 150

ANS : 150°

If you were to find the other side, it would be

10/sin20 = a/sin150
10sin(150)/sin20 = a
a = about 14.55

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2.)
a = 12
b = 10
c = 15

a^2 = b^2 + c^2 - 2 b c cos(A)
b^2 = c^2 + a^2 - 2 c a cos(B)
c^2 = a^2 + b^2 - 2 a b cos(C)

12^2 = 10^2 + 15^2 - 2(10)(15)cos(A)
144 = 100 + 225 - 300cos(A)
144 = 325 - 300cos(A)
-181 = -300cos(A)
(181/300) = cos(A)
A = 52.89099505

100 = 225 + 144 - 2(15)(12)cos(B)
100 = 369 - 360cos(B)
-269 = -360cos(B)
cos(B) = (269/360)
B =  41.64967227

225 = 144 + 100 - 2(12)(10)cos(C)
225 = 244 - 240cos(C)
-19 = -240cos(C)
cos(C) = (19/240)
C = 85.45933267

ANS : C = 85.5°

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3.)

To make this short.

When using the formulas in #2, you get
A = 35.73129198
B = 83.10789742
C = 61.1608106

Now using the Area Formula.
a*c*sin(B)/2 = b*c*sin(A)/2 = a*b*sin(C)/2
150sin(83.10789742) = 255sin(35.73129198) = 170sin(61.1608106)

You will get that they work out.

the answer to this problem is area = 148.9160838
or

ANS : area = 149

Info found at www.ping.be/~ping1339/gonio.htm#Sine-rule