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

1 Answer
Feb 6, 2017

Answer:

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]