Asked by A

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.
Answered by Reiny
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
Answered by A
Thank you soooo much!
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