(B) Let $I$ be the incentre of $\Delta ABC$. The distance from the incentre to the vertices are given by $IA = \frac{r}{\sin(A/2)}$,$IB = \frac{r}{\si

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(B) Let $I$ be the incentre of $\Delta ABC$. The distance from the incentre to the vertices are given by $IA = \frac{r}{\sin(A/2)}$,$IB = \frac{r}{\sin(B/2)}$,and $IC = \frac{r}{\sin(C/2)}$.The product of these distances is $IA \cdot IB \cdot IC = \frac{r^3}{\sin(A/2) \sin(B/2) \sin(C/2)}$.We know that $\sin(A/2) \sin(B/2) \sin(C/2) = \frac{r}{4R}$.Therefore,$IA \cdot IB \cdot IC = \frac{r^3}{r/(4R)} = 4 R r^2$.

Class 11 Mathematics Trigonometrical Equations Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

The product of the distances of the incentre $I$ from the angular points $A, B, C$ of a $\Delta ABC$ is:

A
$4 \,R^2\, r$
B
$4\, Rr^2$
C
$\frac{(abc)R}{s}$
D
$\frac{(abc)s}{R}$

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Mix Examples-Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance

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The product of the distances of the incentre $I$ from the angular points $A, B, C$ of a $\Delta ABC$ is:

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If in a triangle $ABC$,$AB=5$ units,$\angle B=\cos ^{-1}\left(\frac{3}{5}\right)$ and the radius of the circumcircle of $\triangle ABC$ is $5$ units,then the area (in sq. units) of $\triangle ABC$ is:

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