TS EAMCET · Maths · Properties of Triangles
If \(a=3, b=5, c=7\) are the sides of a triangle ABC, then \(\cot \mathrm{A}+\cot \mathrm{B}+\cot \mathrm{C}=\)
- A \(\frac{15 \sqrt{3}}{4}\)
- B \(\frac{7}{\sqrt{3}}\)
- C \(\frac{83}{15 \sqrt{3}}\)
- D \(\frac{83 \sqrt{3}}{15}\)
Answer & Solution
Correct Answer
(C) \(\frac{83}{15 \sqrt{3}}\)
Step-by-step Solution
Detailed explanation
\(s = \frac{a+b+c}{2} = \frac{3+5+7}{2} = \frac{15}{2}\) \(K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{15}{2}(\frac{15}{2}-3)(\frac{15}{2}-5)(\frac{15}{2}-7)} = \sqrt{\frac{15}{2} \cdot \frac{9}{2} \cdot \frac{5}{2} \cdot \frac{1}{2}} = \frac{15\sqrt{3}}{4}\)…
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