The distance of the incentre of the triangle $ABC$ from the point $A$ is?
- $4 R \sin \dfrac{A}{2}$
- $4 R \sin \dfrac{B}{2} \sin \dfrac{C}{2}$
- $4R \cos \dfrac{A}{2}$
- $4 R \cos \dfrac{B}{2} \cos \dfrac{C}{2}$
Reveal answer
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B
Correct answer
Explanation
The distance from a vertex A to the incentre I is given by r / sin(A/2), where r is the inradius. Using r = 4R sin(A/2) sin(B/2) sin(C/2), the distance AI = 4R sin(B/2) sin(C/2).
AI explanation
Using the triangle angle bisector and the extended sine rule, the distance from vertex A to the incenter is given by the formula IA equals r multiplied by the cosecant of A divided by 2. By substituting r equals 4R times the sine of A divided by 2 times the sine of B divided by 2 times the sine of C divided by 2, the r and sine of A divided by 2 terms cancel out. The expression simplifies perfectly to 4R times the sine of B divided by 2 times the sine of C divided by 2.