SOLUTION: Hi, I am not sure about this question. Any help is appreciated. Thank you! Triangle with sides a,b, and, c with corresponding angles A, B, and, C. If a is 2 units longer than b,

Question 847510: Hi, I am not sure about this question. Any help is appreciated. Thank you!
Triangle with sides a,b, and, c with corresponding angles A, B, and, C. If a is 2 units longer than b, angle C is 30 degrees, and the area is 12cm squared, find the length of a,b,and, c.
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
Notice that we are given two sides with its included angle.
Thus the area = 1/2 *a*b * sin(C)
12 = 1/2 * a*b*sin(30 degrees)
b = a -2
12 = 1/2 * a*(a-2)*sin(30 degrees)
12 = 1/2 * a*(a-2) * 1/2
48 = a*(a-2)
48 = a^2 - 2a
a^2 -2a - 48 = 0
(a+8)(a-6)= 0
a = 6
b = 4
So we have that a =6, b=4, we need to find c.
Law of cosines to finish this off:
c^2 = a^2 + b^2 - 2ab*cos(C)
c^2 = 6^2 + 4^2 - 2*6*4*cos(30 degrees)
c^2 = 36 + 16 - 48*(sqrt(3)/2)
c^2 = 10.43
c = 3.23
a = 6, b =4 , c = 3.23

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