Question

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

Answer 1

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

Explanation:

Law of sines for triangle is that `a/sinA=b/sinB=c/sinC`

As `A= 45^o,a=60.5 and b=90`, we have

`(60.5/sin45^o)=(90/sinB)`, hence `sinB=(90xxsin45^o)/60.5=90/(60.5xxsqrt2)=1.052`

As `sinB` is out of range `[-1,1]`,

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