Question

How do you use law of sines to solve the triangle given C = 37°, a = 19, c = 8?

Answer 1

`A=115.4^o`
`B=26.6^o`
`b=6.16`

Explanation:

`(sinA)/a=(sinB)/b=(sinC)/c`

`(sinA)/19=(sinC)/8`
Multiply both sides by 19 and `sinA=19(sin37)/8` which is `sinA=1.429`
The next step is to take the `sin^-1` but you can not do an inverse of sin > 1 . We know that `sin^-1(1)=90^o` `sin^-1(0.429)=25.4` `90+25.4=115.4^o` That's your angle A.

Angle B is `180-115.4-37=27.6^o`

`b/(sin27.6)=8/(sin37` so `b=sin27.6(8/(sin37))`
`b=6.16`