If a=3, b=5, c=7 are the sides of a ABC, then A+ B+ C=

Correct Answer is : 83153{83}{15{3}}15383

Correct Answer is : 83153{83}{15{3}}15383

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TG EAMCET 4 May 2025 Shift 1 Paper

Question : 28 of 160

Marks: +1, -0

a=3, b=5, c=7

\triangle ABC

\cot A+\cot B+\cot C=

Solution:

\;=\;\frac{\cos A}{\sin A}\;+\;\frac{\cos B}{\sin B}\;+\;\frac{\cos C}{\sin C}

\;=\;\frac{2R\cos A}{a}\;+\;\frac{2R\cos B}{b}\;+\;\frac{2R\cos C}{c}

\;=2R\left(\;\frac{\cos A}{a}\;+\;\frac{\cos B}{b}\;+\;\frac{\cos C}{c}\right)

\;=2R\left(\;\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2abc}\right)

\;=\;\frac{R}{abc}(a^2+b^2+c^2)

\;=\;\frac{a^2+b^2+c^2}{4\Delta}\;\; (\Delta=\;\text{area of triangle}\;)

Now,

s=\;\frac{a+b+c}{2}=\;\frac{15}{2}

s-a\;=\;\frac{9}{2},\; s-b=\;\frac{5}{2}\;\text{ and }\; s-c=\;\frac{1}{2}

\Delta\;=\sqrt{s \times (s-a) \times (s-b) \times (s-c)}

\;=\sqrt{\;\frac{15}{2} \times \frac{9}{2} \times \frac{5}{2} \times \frac{1}{2}}

\;=\;\frac{15\sqrt{3}}{4}

\Rightarrow \cot A+\cot B+\cot C\;=\;\frac{3^2+5^2+7^2}{4 \times \frac{15}{4} \times \sqrt{3}}

\;=\;\frac{83}{15\sqrt{3}}

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