Question

Given that A=27, B=43 and c=17, solve triangle ABC to the nearest hundredth

Answer 1

When we are given 3 sides, we try to solve the angles first by using the
law of cosines
cos (A) = [b^2 + c^2 - a^2] / (2 * b * c)
cos (A) = [43^2 + 17^2 -27^2] / (2 * 43 * 17)
cos (A) = [1,849 + 289 -729] / 1,462
cos (A) = 1,409 / 1,462
cos (A) = 0.96374829001368
Angle A = 15.475

Now that we have one angle, we next can use the
Law of Sines
sin(B) / side b = sin(A) / side a
sin(B) = sin(A) * sideb / sidea
sin(B) = sin(15.475) * 43 / 27
sin(B) = 0.26682 * 43 / 27
sin (B) = 0.424935555555
Angle B = 25.147 Degrees
Remember the arc sine (0.424935555555) also equals 154.85

Finally, calculating the third angle is quite easy
Angle C =  180 - Angle (A) - Angle(B)
Angle C =  180 - 15.475 - 154.85
Angle C =  9.675

Source:
http://www.1728.org/trigtut2.htm