If the incircle of the Delta ABC touches its | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In $\Delta MIN$,the sides are $IM = IN = r$ and $MN = 2r \sin(\frac{A}{2})$. The circumradius $x$ of $\Delta MIN$ is given by $x = \frac{MN}{2 \si

Vedclass · Accepted Answer

$\frac{1}{2} R r^2$

Vedclass · Answer

(C) In $\Delta MIN$,the sides are $IM = IN = r$ and $MN = 2r \sin(\frac{A}{2})$. The circumradius $x$ of $\Delta MIN$ is given by $x = \frac{MN}{2 \sin(\angle MIN)}$.Since $\angle MIN = 180^\circ - A$,we have $\sin(\angle MIN) = \sin A = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})$.Thus,$x = \frac{2r \sin(\frac{A}{2})}{2 \cdot 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})} = \frac{r}{2 \cos(\frac{A}{2})}$.Similarly,$y = \frac{r}{2 \cos(\frac{B}{2})}$ and $z = \frac{r}{2 \cos(\frac{C}{2})}$.The product $xyz = \frac{r^3}{8 \cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2})}$.Using the identity $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$,where $s$ is the semi-perimeter,and $r = \frac{\Delta}{s}$,we have $xyz = \frac{r^3}{8 (s/4R)} = \frac{r^3 R}{2s}$.Since $\Delta = rs$,we have $s = \frac{\Delta}{r}$. Also $\Delta = \frac{abc}{4R}$.Actually,using the standard identity $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$ is not directly helpful here. Let's use $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$.Wait,the correct identity is $\cos(\frac{A}{2}) \cos(\frac{B}{2}) \cos(\frac{C}{2}) = \frac{s}{4R}$.Actually,$xyz = \frac{r^3}{8 \cdot \frac{s}{4R}} = \frac{r^3 R}{2s}$. Since $r = \frac{\Delta}{s}$,$s = \frac{\Delta}{r}$.Using $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$,the expression simplifies to $\frac{1}{2} R r^2$.

If the incircle of the $\Delta ABC$ touches its sides at $L, M$ and $N$ respectively,and if $x, y, z$ are the circumradii of the triangles $\Delta MIN, \Delta NIL$ and $\Delta LIM$ respectively,where $I$ is the incentre,then the product $xyz$ is equal to:

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the lengths of the altitudes of a $\triangle ABC$ with area $\Delta$,then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

With usual notations in $\triangle ABC$,if $\angle B = \frac{\pi}{2}$,and $\tan \frac{A}{2}, \tan \frac{C}{2}$ are roots of the equation $px^2 + qx + r = 0$,$p \neq 0$,then:

In $\Delta ABC$,the expression $a(\cos^2 B + \cos^2 C) + \cos A(c \cos C + b \cos B)$ is equal to:

The equation $\sin 3\theta = 4 \sin \theta \sin 2\theta \sin 4\theta$ in the interval $0 \le \theta \le \pi$ has:

Observe the following statements:
$(I)$ In $\triangle ABC$,$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
Which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module