Question

How do you solve the triangle given `B=125^circ40', a=32, c=32`?

Answer 1

Solution: `a= 32 ; b= 56.94(2dp) , c= 32 , /_A= 27^0 10' , /_B =125^0 40', /_C=27^0 10'`

Explanation:

`/_B= 125^0 40' ; a =32; c=32`. `/_A and /_C` are equal since they are opposite of equal sides `/_A=/_C= (180- 125^0 40')/2= 27^0 10'`
we can find `b` by using sine law `a/sinA = b/sin B :. b = a* sin B/sin A or b= 32* (sin 125^0 40')/ (sin27^0 10') = 56.94 (2dp)`

Triangle sides : `a= 32 ; b= 56.94(2dp) , c= 32`
Triangle angles : `/_A= 27^0 10' , /_B =125^0 40', /_C=27^0 10'` [Ans]