How do you use law of sines to solve the triangle given A = 71° , a = 9.3, b = 8.5?

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Answer 1 · 2018-03-11T03:00:00

$a = 9.3,quad b= 8.5,quad c = 4.4$
$A = 71^circ, B = 60^circ, C = 49^circ$

$a/ sin A = b/sin B = c/sin C$

Given $A=71^circ, a = 9.3, b = 8.5$ , we have

$sin B = sin A times b/a = sin 71^circ times 8.5/9.3 = 0.8642$

Thus $B = sin^-1 0.8642 = 60^circ$

Since $A+B+C=180^circ$ , we have

$C = 180^circ-71^circ-60^circ=49^circ$

and, by the sine law again

$c = a/sin A times sin C = 9.3/{sin 71^circ} times sin 49^circ~~7.4$