ACCORDING TO QUES,:
Sides of a triangle are
\[a\text{ }=\text{ }18,\text{ }b\text{ }=\text{ }24\text{ }and\text{ }c\text{ }=\text{ }30\]
Hence, using the formulas:
\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]
\[Cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2ac\]
\[Cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2ab\]
Now, substituting the values of\[a,\text{ }b\text{ }and\text{ }c,\]in the equation:
\[Cos\text{ }A\text{ }=\text{ }({{b}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{a}^{2}})/2bc\]
\[=\text{ }({{24}^{2}}~+\text{ }{{30}^{2}}~\text{ }{{18}^{2}})/2\times 24\times 30\]
Or,
\[=\text{ }1152/1440\]
\[=\text{ }4/5\]
Now,
\[Cos\text{ }B\text{ }=\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~\text{ }{{b}^{2}})/2ac\]
\[=\text{ }({{18}^{2}}~+\text{ }{{30}^{2}}~\text{ }{{24}^{2}})/2\times 18\times 30\]
Or,
\[=\text{ }648/1080\]
\[=\text{ }3/5\]
Now,
\[Cos\text{ }C\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}}~\text{ }{{c}^{2}})/2ab\]
\[=\text{ }({{18}^{2}}~+\text{ }{{24}^{2}}~\text{ }{{30}^{2}})/2\times 18\times 24\]
Or,
\[=\text{ }0/864\]
\[=\text{ }0\]
\[\therefore cos\text{ }A\text{ }=\text{ }4/5,\text{ }cos\text{ }B\text{ }=\text{ }3/5,\text{ }cos\text{ }C\text{ }=\text{ }0\]