How do you use law of sines to solve the triangle given A = 45° , a = 60.5 , b = 90?

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How do you use law of sines to solve the triangle given A = 45° , a = 60.5 , b = 90?

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Answer 1 · 2016-03-14T07:17:08

Such a triangle is not possible (i.e. with $A = 45^{o}, a = 60.5 and b = 90$ $A= 45^o,a=60.5 and b=90$ ) and solution does not arise

Explanation:

Law of sines for triangle is that $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a/sinA=b/sinB=c/sinC$

As $A = 45^{o}, a = 60.5 and b = 90$ $A= 45^o,a=60.5 and b=90$ , we have

$(\frac{60.5}{{sin 45}^{o}}) = (\frac{90}{sin B})$ $(60.5/sin45^o)=(90/sinB)$ , hence $sin B = \frac{90 \times {sin 45}^{o}}{60.5} = \frac{90}{60.5 \times \sqrt{2}} = 1.052$ $sinB=(90xxsin45^o)/60.5=90/(60.5xxsqrt2)=1.052$

As $sin B$ $sinB$ is out of range $[- 1, 1]$ $[-1,1]$ ,

Such a triangle is not possible (i.e. with $A = 45^{o}, a = 60.5 and b = 90$ $A= 45^o,a=60.5 and b=90$ ) and solution does not arise

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How do you use law of sines to solve the triangle given A = 45° , a = 60.5 , b = 90?

1 Answer

Explanation:

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