In a triangle ABC- if r1-r2-r3 are the exradii- then bcr1-car2-abr3-k-sa-3- where k is equal to

We have, bcr_1+car_2+abr_3 =bcs a+cas b+abs c =bc(s a)+ca(s b)+ab(s c) =1[bcs bca+cas cab+abs abc] =abc[sa+sb+sc 3] =abc[∑sa 3] On ...

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Relation between AM,GM,HM for 2 Numbers

In a triangle...

Question

In a triangle $A B C,$ if $r_{1}, r_{2}, r_{3}$ are the exradii, then $\frac{b c}{r_{1}} + \frac{c a}{r_{2}} + \frac{a b}{r_{3}} = k [\sum \frac{s}{a} - 3],$ where $k$ is equal to

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A B C

\frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b}

r_{1}, r_{2}

r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}

r_{1}, r_{2}

r_{3}

2 s

△ A B C, \frac{r_{1}}{b c} + \frac{r_{2}}{c a} + \frac{r_{3}}{a b}

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

A B C

r

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

△ A B C

a < b < c

r

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

A, B, C

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

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

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