In $\triangle A B C, a=6 \mathrm{~cm}, b=10 \mathrm{~cm}$ and $c=14 \mathrm{~cm}$. Then, the sum of the…

1. $180^{\circ}$
2. $120^{\circ}$
3. $90^{\circ}$
4. $60^{\circ}$

Solution

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} $

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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