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

1 Answer
Binayaka C.
Feb 6, 2017

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]

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