Properties of Triangles

In triangle $\triangle ABC$, given $a=26, b=30$, and $\cos C=\;\frac{63}{65}$, we need to find the length of side $c$.We know the cosine rule for any triangle is given by:$\cos C=\;\frac{a^2+b^2-c^2}{2ab}$Substituting the known values:$\;\frac{a^2+b^2-c^2}{2 \times a \times b}=\;\frac{63}{65}$This implied expression becomes:$\;\frac{26^2+30^2-c^2}{2 \times 26 \times 30}=\;\frac{63}{65}$Now, solve this step-by-step:Calculate $26^2$ and $30^2$ :$26^2=676, \;\; 30^2=900$Substitute these into the equation:$\;\frac{676+900-c^2}{1560}=\;\frac{63}{65}$Simplify the numerator:$676+900=1576$Substitute back:$\;\frac{1576-c^2}{1560}=\;\frac{63}{65}$Cross-multiply to solve for $c^2$ :$65 (1576-c^2) =63 \times 1560$Compute $63 \times 1560$ :$63 \times 1560=98280$Now, solve for $c^2$ :$65 \times (1576-c^2)=98280$Simplify the expressions:$102440-65c^2=98280$Rearrange to find $c^2$ :$102440-98280=65c^2$Solve:$4160=65c^2$Divide by 65 to isolate $c^2$ :$c^2=\;\frac{4160}{65}=64$Take the square root:$c=\sqrt{64}=8$Thus, the value of $c$ is 8 .