TS EAMCET · Maths · Properties of Triangles
In a triangle \(\mathrm{ABC}, a=5, b=4\) and \(\tan \frac{\mathrm{C}}{2}=\sqrt{\frac{7}{9}}\), then its inradius \(r=\)
- A \(\frac{\sqrt{7}}{2}\)
- B \(2 \sqrt{7}\)
- C \(\frac {9}{\sqrt{7}}\)
- D \(\frac{4}{\sqrt{7}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{7}}{2}\)
Step-by-step Solution
Detailed explanation
\(\cos C = \frac{1-\tan^2 \frac{C}{2}}{1+\tan^2 \frac{C}{2}} = \frac{1-\frac{7}{9}}{1+\frac{7}{9}} = \frac{\frac{2}{9}}{\frac{16}{9}} = \frac{1}{8}\) \(c^2 = a^2+b^2-2ab \cos C = 5^2+4^2-2(5)(4)(\frac{1}{8}) = 25+16-5 = 36 \Rightarrow c=6\)…
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