How do you solve the triangle given #A=24.3^circ, C=54.6^circ, c=2.68#?

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Answer 1 · 2018-01-29T12:15:57

Solution of triangle is #a ~~ 1.35, b ~~ 3.23, c=2.68 #
and #/_A=24.3 , /_B=101.1 , /_C=54.6 #

Angle between Sides #a and b# is # /_C= 54.6^0#

Angle between Sides # b and c# is # /_A=24.3^0 :.#

Angle between Sides #c and a# is

# /_B=180-(54.6+24.3)=101.1^0# The sine rule states if

#a, b and c# are the lengths of the sides

and opposite angles are #A, B and C# in a triangle, then:

#a/sinA = b/sinB=c/sinC ; c=2.68 ; b/sinB=c/sinC# or

#b/sin101.1=2.68/sin54.6 or b= 2.68* (sin101.1/sin54.6) # or

#b ~~ 3.23 (2dp) # Similarly #a/sinA=c/sinC # or

#a/sin24.3=2.68/sin54.6 or a= 2.68*(sin24.3/sin54.6) # or

#a ~~ 1.35 (2dp) :. # Solution:#a ~~ 1.35, b ~~ 3.23, c=2.68 #

and #/_A=24.3 , /_B=101.1 , /_C=54.6 # [Ans]