Question -
Answer -
Given:
Sides of a triangle are a = 18, b = 24 and c = 30
By using the formulas,
Cos A = (b2 + c2 – a2)/2bc
Cos B = (a2 + c2 – b2)/2ac
Cos C = (a2 + b2 – c2)/2ab
So now let us substitute the values of a, b and c weget,
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