Question & Answer

Find the distance of the incentre of the triangle ABC from the point A.

A. $4R\sin \dfrac{A}{2}$
B. $4R\sin \dfrac{B}{2}\sin \dfrac{C}{2}$
C. $4R\cos \dfrac{A}{2}$
D. $4R\cos \dfrac{B}{2}\cos \dfrac{C}{2}$

ANSWER Verified Verified
Hint: Incentre is the point in the triangle whose distances from this point to sides are equal. We have to use basic trigonometric angle formulas to solve this problem.

Complete step-by-step answer:
From the figure: ‘I’ is the incentre of triangle ABC.
Let a, b, c be the sides AB, BC and CA respectively.
In $\vartriangle AIC,\;\angle AIC = \pi - \left( {\dfrac{{A + C}}{2}} \right)$
Sine rule states that:
 $\dfrac{a}{{\operatorname{Sin} a}} = \dfrac{b}{{\operatorname{Sin} b}} = \dfrac{c}{{\operatorname{Sin} c}}$
So, from sine rule we can write:
$\dfrac{{AI}}{{\sin \dfrac{C}{2}}} = \dfrac{b}{{\sin \angle AIC}}$
$ \Rightarrow AI = \dfrac{{b\sin \dfrac{C}{2}}}{{\sin \left( {\pi - \dfrac{{A + C}}{2}} \right)}}$
$ \Rightarrow AI = \dfrac{{2R\sin Bsin\dfrac{C}{2}}}{{\sin \dfrac{B}{2}}} = \dfrac{{2R\left( {2\sin \dfrac{B}{2}\cos \dfrac{B}{2}} \right)\sin \dfrac{C}{2}}}{{\sin \dfrac{B}{2}}}$
$ \Rightarrow AI = 4R\sin \dfrac{B}{2}\sin \dfrac{C}{2}$
$\therefore $ Option B is the correct answer.

Note: The three angle bisectors of any triangle are always concurrent and meet in the triangles interior. The incenter is the center of the incircle. It is always present inside of the triangle regardless of the type of the triangle.