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.

2 answers

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 ID
   524973

   Created
   April 4, 2011 12:31am UTC

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   URL
   https://questions.llc/answers/524973

2. Thank you soooo much!

   Answer ID
   507165

   Created
   April 4, 2011 2:21am UTC

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   URL
   https://questions.llc/answers/507165