SOLUTION: find the area of the triangle whose longest and shortest sides are 62 cm and 31 cm respectively and whose biggest angle is 140

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Question 776252: find the area of the triangle whose longest and shortest sides are 62 cm and 31 cm respectively and whose biggest angle is 140
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Two shorter sides are 31 and d in length and longest side is 62, opposite of the given angle. The given angle is between the 31 and d lengths.

Law of Cosines:
d%5E2%2B31%5E2-2%2A3%2Ad%2Acos%28140%29=62%5E2
d%5E2%2B31%5E2%2B2%2A3%2Ad%2Acos%2840%29=62%5E2
d%5E2%2B6%2Acos%2840%29%2Ad%2B31%5E2-62%5E2=0 ---------- a quadratic equation in d.

Finding value for d just requires computation through the general solution to the quadratic formula. After than, you can get the height of the triangle according to d%2Acos%2840%29. Area of the triangle would be %281%2F2%2931%2Ad%2Acos%2840%29. That just uses area as %281%2F2%29%2Abase%2Aheight.