For $\triangle A B C$



$$
\begin{aligned}
a & =6 \mathrm{~cm}, b=10 \mathrm{~cm} \text { and } c=14 \mathrm{~cm} \\
\cos A & =\frac{b^2+c^2-a^2}{2 b c} \\
& =\frac{10^2+14^2-6^2}{2 \times 10 \times 14}=\frac{100+196-36}{2 \times 10 \times 14} \\
& =\frac{296-36}{2 \times 10 \times 14}=\frac{260}{2 \times 10 \times 14} \\
\cos A & =\frac{13}{14}
\end{aligned}
$$
$\therefore A$ is acute angle.
$$
\begin{aligned}
\cos B & =\frac{a^2+c^2-b^2}{2 a c} \\
& =\frac{6^2+14^2-10^2}{2 \times 6 \times 14}=\frac{36+196-100}{2 \times 6 \times 14} \\
& =\frac{132}{2 \times 6 \times 14} \\
\cos B & =\frac{11}{14}
\end{aligned}
$$
$B$ is acute angle.
$$
\begin{aligned}
& \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{6^2+10^2-14^2}{2 \times 6 \times 10}=\frac{136-196}{2 \times 6 \times 10} \\
& =\frac{-60}{2 \times 6 \times 10}=-\frac{1}{2} \\
& \cos C=-\frac{1}{2} \\
& C=120^{\circ} \\
& \therefore \text { Sum of acute angle }=A+B=180-C \\
& =180-120=60^{\circ} \\
&
\end{aligned}
$$