HELP WITH MY TRIG HW: Side-Angle-Side Triangles *Best Answer Awarded*?

a = 13, b = 18, A = 27, all angles in degrees

a^2 = b^2 + c^2 - 2bc*cos(A)

13^2 = 18^2 + c^2 - 2*18*c*cos(27)

169 = 324 + c^2 - 36c*cos(27)

324 + c^2 - 36c*cos(27) - 169 = 0

c^2 - 36c*cos(27) + 155 = 0

c = (-(-36*cos(27)) +/- sqrt((-36*cos(27))^2 - 4*1*155)) / (2*1)

c = (36*cos(27) +/- sqrt(1296*cos^2(27) - 620)) / 2

c = 18*cos(27) +/- sqrt(324*cos^2(27) - 155)

c =~ 26.14856801246089603623175823901 or 5.92766685832034700871778633187

sin(B)/b = sin(A)/a

sin(B)/18 = sin(27)/13

sin(B) = sin(27)*18/13

B =~ 38.9471 or 141.053

Answers are to the nearest tenth are: B = 38.9 degrees, C = 114.1 degrees and side c = 26.2

// Use the law of sines to find B

a.............b

------ = -------- ⇒

sin A....sin B

13............18

-------- = -------- ⇒

sin 27°....sin B

..................18 sin 27°

.....sin B = -------------- ⇒

......................13

.....sin B = 0.6286

.....B = sin‾¹ 0.6286

.....B = 38.95°...................ANS

// Now find C: We know that the sum of

// all angles in a triangle is 180°

A+B+C = 180°

.....C = 180° - A - B

.....C = 180° - 27° - 38.95°

.....C = 114.05°.....................ANS

// And finally, use the law of sines again

// to find the length of side c

c.............a

------ = -------- ⇒

sin C....sin A

c......................13

-------------- = -------- ⇒

sin 114.05°....sin 27°

............13 sin 114.05°

.....c = -------------------- ⇒

..................sin 27°

.....c = 26.15....................ANS

// So the answers are:

B = 38.95° ; C = 114.05° and c = 26.15..................ANS

Use cosine law to solve for cos(A) or any other angle, then use cos^-1 to get the angle.

Once you have one angle, you can use sine law to get the others.

Nothing tricky about the problem--straight application of standard formulas.

That is not side-angle-side, which is when the angle is between the two sides.