How do you solve the triangle given $A=110^circ15', a=48, b=16$ ?

Question

2920 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-19T19:14:17

Explained below

Apply sin e rule, $(sin A)/a = (sin B)/b$ to get angle B.

$sin 110^o 15^' /48 = sin(B)/16$

$sin B= 1/3 Sin 110^o 15^'$ =0.3127......... ( $110^o 15'= 110.25^o$ )

$B =18.22^o$ . Therefore $C= 180- (110.25+18.22)= 51.53^o$

Then use $SinA /a= SinC/c$ that is $sin 110.25/48= sin51.53^o/c$

$c=48sin 51.53^o/ sin 110.25^o$ = 40.05

How do you solve the triangle given A=110^circ15', a=48, b=16A=110^circ15', a=48, b=16?