Question

How do you solve the triangle ABC, given a=5, A=42, b=7?

Answer 1

Solution: `a=5 ; b=7 ; c= 6.95 , /_A=42^0 : /_B=69.52^0 : /_C=68.48^0`

Explanation:

We know by sine law `a/sinA=b/sinB or 5/sin42=7/sinB or sin B =(7 sin 42)/5= 0.93678 :./_B=sin^-1(0.93678)=69.52^0(2dp) :. /_C=180-(42+69.52)=68.48^0`

Similarly , `a/sinA=c/sinC or 5/sin42=c/sin68.48 or c=5*sin68.48/sin42 =6.95`

Solution `a=5 ; b=7 ; c= 6.95 , /_A=42^0 : /_B=69.52^0 : /_C=68.48^0`[Ans]