Question

In \( \triangle ABC \), if \[ \angle ABC = \delta, \delta = \cos^{-1} \left( \sqrt{ \frac{r_2}{r_3 r_1} } \right), \] then the expression \[ \angle ABC = \delta = \cos^{-1} \left( \sqrt{ \frac{r_2}{r_3 r_1} } \right) \]

• A. \( (r_3 - r_2)(r_1 - r_2) \)
• B. \( r_3 + r_1 \)
• C. \( \dfrac{b}{r_3 - r_1} \)
• D. \( \dfrac{b}{r_3 + r_1} \)

Hint:

When you see expressions involving \( r_1, r_2, r_3 \), recall that these are inradii corresponding to sub-triangles in \( \triangle ABC \), and relations can often be converted using identities involving angles and triangle side properties.

Answer 1

The Correct Option is D

In triangle \( \triangle ABC \), let:
- \( r_1, r_2, r_3 \) be the inradii of the triangles formed by the angles \( A, B, C \), respectively.
- The given expression inside the cosine inverse suggests a geometric identity involving inradii and the angle \( B \).
Given:
\[ \cos \angle B = \sqrt{ \frac{r_2}{r_3 r_1} } \] Recall the identity involving the inradii of the sub-triangles:
\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \] However, based on known triangle identities involving inradii, there's a relation:
\[ \cos B = \frac{r_2}{\sqrt{r_1 r_3}} \Rightarrow \cos B = \sqrt{ \frac{r_2}{r_3 r_1} } \] Now, manipulating this relation in terms of triangle sides and corresponding radii results in the final simplification:
\[ \cos \angle B = \sqrt{ \frac{r_2}{r_3 r_1} } \Rightarrow \angle B = \cos^{-1} \left( \sqrt{ \frac{r_2}{r_3 r_1} } \right) \Rightarrow \delta = \cos^{-1} \left( \sqrt{ \frac{r_2}{r_3 r_1} } \right) \] Now relating this to side \( b \) using a known geometric identity:
\[ \tan\left( \frac{B}{2} \right) = \frac{r_2}{s - b} \\ \text{and} \quad \tan\left( \frac{B}{2} \right) = \frac{r}{s - b} \\ \Rightarrow b \propto (r_3 + r_1) \\ \Rightarrow \frac{b}{r_3 + r_1} \] Thus,
\[ \delta = \cos^{-1} \left( \sqrt{ \frac{r_2}{r_3 r_1} } \right) = \frac{b}{r_3 + r_1} \]