Question

How do you solve the triangle if C=60 degrees, A= 12, c=15?

Answer 1

`/_A ~= 43.9^@`
`/_B ~= 76.1^@`
`b ~= 16.8`

Explanation:

The Law of Sines tells us
`color(white)("XXX")sin(A)/a=sin(B)/b=sin(C)/c`

#sin(C)/c = sin(60^@)/15 = (sqrt(3)/2)/15 = sqrt(3)/30 ~= 0.057735

`sin(A) = sin(C)/c*a = 0.057735*12 = 0.69282`
`/_A = arcsin(sin(A)) = arcsin(0.69282) ~= 43.85378^@`

`/_B = 180^@ - (/_A + /_C) = 180^@ - (43.85378^@+60^@) = 76.14622^@`

`sin(B)/b =sin(C)/c=0.057735`
`rarr b = sin(B)/0.057735 = sin(76.14622^@)/0.057735 = 16.81655`