if three sides of a triangle are 18.5,13.5,7 then find the angles between them?
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a = 18.5 b = 13.5 c = 7
cos A = (b² + c² - a² )/ 2 bc = 125.965 deg
cos B = (a² + c² - b²) / 2 ac = 36.2 deg
COS C = (a² + b² - c²) / 2 ab or 180 - A - B = 17.834 deg
cos A = (b² + c² - a² )/ 2 bc = 125.965 deg
cos B = (a² + c² - b²) / 2 ac = 36.2 deg
COS C = (a² + b² - c²) / 2 ab or 180 - A - B = 17.834 deg
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