Given ΔABC, m∠A = 60°, b = 9 and c = 6. Find m∠B. Round to the nearest whole degree
1 answer:
The problem above uses a combination of sine and cosine law of triangle to solve for the m∠B.
Given:
<span>m∠A = 60°, b = 9 and c = 6
Cosine Law:
a^2=b^2+c^2-2bccosA
a^2=(9)^2+(6)^2-2(9)(6)cos(60)
a^2=81+36-54
a^2= 63
a=</span>
Sine Law
sin B= 9/ sin 60
sin B= 0.98198
B= sin ^-1 (0.98198) = 79.10
Therefore, m∠B is equal to 79.10
Given:
<span>m∠A = 60°, b = 9 and c = 6
Cosine Law:
a^2=b^2+c^2-2bccosA
a^2=(9)^2+(6)^2-2(9)(6)cos(60)
a^2=81+36-54
a^2= 63
a=</span>
Sine Law
sin B= 9/
sin B= 0.98198
B= sin ^-1 (0.98198) = 79.10
Therefore, m∠B is equal to 79.10
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