How do you solve the triangle of $A = 58^{\circ}, a = 4.5, b = 12.8$ $A=58^circ, a=4.5, b=12.8$ ?

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Answer 1 · 2017-05-27T07:33:02

This question requires use of the 'Sine Rule':

$\frac{a}{sin A} = \frac{b}{sin B}$ $a/(sin A) = b/(sin B)$

Rearranging:

$sin B = \frac{b sin A}{a} = \frac{12.8 \times sin 58}{4.5} = 2.41$ $sin B = (b sin A)/a = (12.8xxsin58)/4.5 = 2.41$

Taking the inverse sin, angle $B =$ $B=$ Error

There is no angle whose sin is > 1. There may be an error in the question.

How do you solve the triangle of A=58∘,a=4.5,b=12.8A=58^circ, a=4.5, b=12.8?