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That is a bit complicated. You need to know sines and cosines and more geometry.

Given area K and 2 sides a and b ,the relation is K = (1/2) (absinC); solve for angle C; then the third side c, from law of cosines is c =sqrt (a squared +b squared -2abcosC); then sin A = 2K/bc and

sin B = 2K/ac. solve for angles B and C. In your case, approximately, C = 56 degrees, A =52 degrees and B = 72 degrees

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Q: How do you find an angle in a triangle using the area and two sides Area 12.5cm2 Side a 6cm Side b 5cm?
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