Find side AB in triangle ABC if side AC = 300 and angleBAC = 62 and angleBCA = 53.

Question

Find side AB in triangle ABC if side AC=300 and angleBAC=62 degrees and angleBCA=53 degrees.

Hint: law of sines. 

Answer 1

angle ABC = 180 - (62 + 53) degrees = 65 degrees

 

AB / sin(53) = AC / sin(angle ABC) = 300 / sin(65)

 

AB = [sin(53) / sin(65)] (300) = 264.36