Question

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

Answer 1

Explained below

Explanation:

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