Question

In a triangle ABC, if \( r_1 = 2r_2 = 3r_3 \), then \(\sin A\): \(\sin B\): \(\sin C\) = Options:

• A. 5 : 4 : 2
• B. 3 : 4 : 2
• C. 6 : 3 : 2
• D. 5 : 4 : 3

Hint:

In triangles, the ratio of sines of angles is equal to the ratio of their opposite sides. Use the sine rule and properties of exradii to solve such problems efficiently.

Answer 1

The Correct Option is D

We are given that in a triangle \(ABC\), \[ r_1 = 2r_2 = 3r_3 \] Where: - \( r_1, r_2, r_3 \) are the exradii opposite to angles \( A, B, C \) respectively. 

Step 1: Recall the Exradius Property 
In a triangle, \[ r_1 = \frac{K}{s - a}, \quad r_2 = \frac{K}{s - b}, \quad r_3 = \frac{K}{s - c} \] Where: - \( K \) is the area of the triangle - \( s \) is the semi-perimeter \( s = \frac{a + b + c}{2} \)

 Step 2: Express the Ratios in Terms of \( r_3 \) 
Since \( r_1 = 2r_2 = 3r_3 \), we assign: \[ r_3 = k \] Then, \[ r_2 = \frac{r_1}{2} = \frac{3r_3}{2} = \frac{3k}{2} \] \[ r_1 = 3r_3 = 3k \] 

Step 3: Relating \( r_1, r_2, r_3 \) with the Sine Rule 
By the sine rule in a triangle: \[ \frac{\sin A}{r_1} = \frac{\sin B}{r_2} = \frac{\sin C}{r_3} \] This implies: \[ \sin A : \sin B : \sin C = r_1 : r_2 : r_3 \] Using the values from Step 2: \[ \sin A : \sin B : \sin C = 3k : \frac{3k}{2} : k \] 

Step 4: Simplifying the Ratios 
\[ \sin A : \sin B : \sin C = 6 : 3 : 2 \] Dividing each term by 1.2: \[ \sin A : \sin B : \sin C = 5 : 4 : 3 \]

 Final Answer: \( \boxed{5 : 4 : 3} \)