TS EAMCET · Maths · Properties of Triangles
In a triangle \(A B C\), if \(c=9, s=10\) and \(\Delta=10 \sqrt{2}\) then \(b\left[1+\sqrt{2} \tan \left(\frac{A-B}{2}\right)\right]=\)
- A \(a\left[1-\sqrt{2} \tan \left(\frac{A-B}{2}\right)\right]\)
- B \(c\left[1-\sqrt{2} \tan \left(\frac{A-B}{2}\right)\right]\)
- C \(a\left[\sqrt{2} \tan \left(\frac{A-B}{2}\right)-1\right]\)
- D \(c\left[\sqrt{2} \tan \left(\frac{A-B}{2}\right)-1\right]\)
Answer & Solution
Correct Answer
(A) \(a\left[1-\sqrt{2} \tan \left(\frac{A-B}{2}\right)\right]\)
Step-by-step Solution
Detailed explanation
For a triangle \(A B C\), it is given that \(c=9, s=10\) and \(\Delta=10 \sqrt{2}\) \(\therefore \quad \tan \frac{A-B}{2}=\frac{a-b}{a+b} \cot \frac{c}{2}=\frac{a-b}{a+b} \frac{s(s-c)}{\Delta}\)…
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