In a triangle ABC- r1-bc - r2-ca- r3-ab is equal to Note - r1- r2 and r3 are ex-radii of the triangle

The correct option is D 1/r 1/2R nbsp; r _ 1 / bc + r _ 2 / ca + r _ 3 / ab = stan A / 2 nbsp; nbsp;/ bc + stan B / 2 nbsp; nbsp;/ ac ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

In a triangle...

Question

In a triangle $A B C$ , $\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b}$ is equal to
(Note : $r_{1}, r_{2}$ and $r_{3}$ are ex-radii of the triangle)

\frac{1}{2 R} - \frac{1}{r}

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2 R - r

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r - 2 R

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\frac{1}{r} - \frac{1}{2 R}

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Solution

The correct option is D $\frac{1}{r} - \frac{1}{2 R}$
$\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b} = \frac{s tan \frac{A}{2}}{b c} + \frac{s tan \frac{B}{2}}{a c} + \frac{s tan \frac{C}{2}}{a b} = \frac{s}{a b c} [a tan \frac{A}{2} + b tan \frac{B}{2} + c tan \frac{C}{2}]$
Now using $a = 2 R sin A = 2 R .2 sin \frac{A}{2} . cos \frac{A}{2} = 4 R sin \frac{A}{2} . cos \frac{A}{2}$ , and similarly using the formula for $b$ and $c$ , we get
$\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b} = \frac{s}{a b c} .4 R [{sin}^{2} \frac{A}{2} + {sin}^{2} \frac{B}{2} + {sin}^{2} \frac{C}{2}]$
Using $r = \frac{△}{s}$ and $R = \frac{a b c}{4 △}$ , we get $\frac{s}{a b c} .4 R = \frac{1}{r}$ .
.
$\Rightarrow \frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b} = \frac{1}{r} [\frac{1 - cos A}{2} + \frac{1 - cos B}{2} + \frac{1 - cos C}{2}]$
.
$= \frac{1}{r} [\frac{3}{2} - \frac{cos A + cos B + cos C}{2}]$
Now using $cos A + cos B + cos C = 1 + \frac{r}{R}$ , we get
$\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b} = \frac{1}{r} [1 - \frac{r}{2 R}]$
.
$\Rightarrow \frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b} = \frac{1}{r} - \frac{1}{2 R}$

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Similar questions

Q. Let

A B C

be a triangle with in centre

I

and in radius

r

. Let

D, E, F

be the feet of the perpendiculars from

I

to the sides

B C, C A

and

A B

respectively. If

r_{1}, r_{2} r_{3}

are the radii of circles inscribed in then prove

\frac{r_{1}}{r - r_{1}}_\frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}

Q. If

r_{1}, r_{2}, r_{3}

are the radii of the escribed circles of a triangle

A B C

and if

r

is the radius of its incircle,then

r_{1} r_{2} r_{3} - r (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1})

is equal to

Q. If

r_{1} < r_{2} < r_{3}

are the ex-radii of a right angled triangle and

r_{1} = 1, r_{2} = 2

, then

r_{3} = . . . .

Q. The radii of described circle of

△ A B C

are

r_{1}, r_{2}

and

r_{3}

respectively (opposite to vertices

A, B

and

C

). If

r_{2} + r_{3} = 2 R

and

r_{1} + r_{2} = 3 R

then

Q. Assertion :In any

△ A B C

maximum value of

r_{1} + r_{2} + r_{3} = \frac{9 R}{2}

because Reason: In any

△ A B C

R \geq 2 r

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In a triangle ABC, r1bc+r2ca+r3ab is equal to (Note : r1,r2 and r3 are ex-radii of the triangle)

In a triangle $A B C$ , $\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b}$ is equal to
(Note : $r_{1}, r_{2}$ and $r_{3}$ are ex-radii of the triangle)