Question

How do you use law of sines to solve the triangle given c= 59ft, A= 65 degrees, B= 64 degrees?

Answer 1

Solution: `/_A=65^0 , /_B=64^0 ,/_C=51^0, a=68.81 ft, b= 68.24 ft,c=59ft`

Explanation:

`/_C=180-(65+64)=51^0 ;c= 59 ft`

Applying sine law `a/sinA=b/sinB=c/sinc` we get

`a=c*sinA/sinC=59*sin65/sin51=68.81(2dp)` and
`b=c*sinB/sinC=59*sin64/sin51=68.24(2dp)`

Solution:`/_A=65^0 , /_B=64^0 ,/_C=51^0, a=68.81 ft, b= 68.24 ft,c=59ft` [Ans]