How do you solve the triangle given $B=28^circ, C=104^circ, a=3 5/8$ ?

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Answer 1 · 2018-02-18T14:24:10

$a = 3(5/8), b = 2.29, c = 4.733$

$hatA = 48^@, hatB = 28^@, hatC = 104^@$

$hatB = 28, hatC = 104, a = 29/8$

To find $hatA, b, c$

$hatA = 180 - 28 - 104 = 48^@$

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

$b = (a * sin B) / sin A = ((29/8)*sin 28) / sin 48 = 2.29$

$c = (a * sin C) / sin A = ((29/8)*sin 104) / sin 48 = 4.733$

How do you solve the triangle given B=28^circ, C=104^circ, a=3 5/8B=28^circ, C=104^circ, a=3 5/8?