If a triangle has an area of 50 square units and has angles that measure 15, 65, and 100 degress, find the length of the shortest side to the nearest tenth. Do not round till the final answer.

Answer 1

sketch a triangle ABC , where A=15, B=65 and C = 100°
let AB = a and BC = c

by the sine law: c/sin100 = a/sin15
c = asin100/sin15

area of triangle = (1/2)(ac)sin 65
50 = (1/2)(ac)sin65
100 = a(asin100/sin15)(sin65)
a^2 = 100sin15/( sin100sin65)
a^2 = 28.998
a = √28.998 = appr. 5.4

Answer 2

Thank you soooo much!