In ∆ABC Prve that (b) partr1.r2 + r2.r3 + r3.r1 = s^2

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Arun · Answer

As we know r = /s r1 = /s-a r2 = /s-b r3= /s-c Put these value in LHS We get ^2[1/(s-b)*(s-a) +1/(s-b)(s-c) +1/ (s-c)(s-a)] Solve thia and you will get s 2

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In ∆ABC Prve that (b) partr1.r2 + r2.r3 + r3.r1 = s^2

In ∆ABC Prve that (b) partr1.r2 + r2.r3 + r3.r1 = s^2

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In ∆ABC Prve that (b) partr1.r2 + r2.r3 + r3.r1 = s^2

In ∆ABC Prve that (b) partr1.r2 + r2.r3 + r3.r1 = s^2

1 Answers

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Other Related Questions on Trigonometry

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