In any ∆ABC, prove: if a = 18, b = 24, c = 30, find cos A, cos B an

1.	In any ∆ABC, prove: if a = 18, b = 24, c = 30, find cos A, cos B and cos C
Answer» Given as Sides of a triangle are a = 18, b = 24 and c = 30 On using the formulas, Cos A = (b² + c² – a²)/2bc Cos B = (a² + c² – b²)/2ac Cos C = (a² + b² – c²)/2ab Therefore now let us substitute the values of a, b and c we get, Cos A = (b² + c² – a²)/2bc = (24² + 30² – 18²)/2 × 24 × 30 = 1152/1440 = 4/5 Cos B = (a² + c² – b²)/2ac = (18² + 30² – 24²)/2 × 18 × 30 = 648/1080 = 3/5 Cos C = (a² + b² – c²)/2ab = (18² + 24² – 30²)/2 × 18 × 24 = 0/864 = 0 ∴ cos A = 4/5, cos B = 3/5, cos C = 0

In any ∆ABC, prove: if a = 18, b = 24, c = 30, find cos A, cos B and cos C

Answer»

Given as

Sides of a triangle are a = 18, b = 24 and c = 30

On using the formulas,

Cos A = (b² + c² – a²)/2bc

Cos B = (a² + c² – b²)/2ac

Cos C = (a² + b² – c²)/2ab

Therefore now let us substitute the values of a, b and c we get,

Cos A = (b² + c² – a²)/2bc

= (24² + 30² – 18²)/2 × 24 × 30

= 1152/1440

= 4/5

Cos B = (a² + c² – b²)/2ac

= (18² + 30² – 24²)/2 × 18 × 30

= 648/1080

= 3/5

Cos C = (a² + b² – c²)/2ab

= (18² + 24² – 30²)/2 × 18 × 24

= 0/864

= 0

∴ cos A = 4/5, cos B = 3/5, cos C = 0