AP EAMCET · Maths · Properties of Triangles
In a triangle \(\mathrm{ABC}\), if \(\tan \left(\frac{A-B}{2}\right)=\frac{1}{3} \tan \left(\frac{A+B}{2}\right)\) then \(a: b=\)
- A \(2: 1\)
- B \(3: 1\)
- C \(4: 1\)
- D \(1: 3\)
Answer & Solution
Correct Answer
(A) \(2: 1\)
Step-by-step Solution
Detailed explanation
\(\frac{a-b}{a+b} = \frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)}\) \(\frac{a-b}{a+b} = \frac{1}{3}\) \(3(a-b) = a+b\) \(3a-3b = a+b\) \(2a = 4b\) \(a:b = 2:1\)
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