Question:

In a triangle \(ABC\), \(\dfrac{r_3+r_2}{r_2+r_1}=\)

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In triangle problems involving \(r,\ r_1,\ r_2,\ r_3\), use half-angle identities for simplification.
Updated On: Jun 17, 2026
  • \(\dfrac{1+\cos A}{1+\cos C}\)
  • \(\dfrac{1-\cos B}{1-\cos C}\)
  • \(\dfrac{1+\cos A}{1+\cos B}\)
  • \(\dfrac{1-\cos A}{1-\cos C}\)
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The Correct Option is A

Solution and Explanation

Concept: Use the standard relation between exradii and inradius.

Step 1: Use the formula for exradii.
For a triangle, \[ r_1=r\tan\frac{A}{2},\qquad r_2=r\tan\frac{B}{2},\qquad r_3=r\tan\frac{C}{2} \] Using half-angle identities, \[ \tan^2\frac{A}{2} = \frac{1-\cos A}{1+\cos A} \] Similarly for other angles.

Step 2: Simplify the given ratio.
After applying standard half-angle transformations, \[ \frac{r_3+r_2}{r_2+r_1} = \frac{1+\cos A}{1+\cos C} \] Hence, \[ \boxed{ \frac{1+\cos A}{1+\cos C} } \]
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