In acute angled triangle ABC- r-r1-r2-r3 and - B- -3 then

R r_2=r_3 r_1 ⇒Δ/s Δ/s b= Δ/s c Δ/s a ⇒ b/s(s b)=c a/(s a)(s c) ⇒(s a)(s c)/s(s b)= a c/b⇒ tan^2 B/2= a c/b But B/2∈(π/6,π/4 ) ⇒ tan^2B/2∈(1/3, 1 ) ⇒1/3 lt; ...

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Byju's Answer

Standard XI

Mathematics

Area Using Semiperimeter

In acute angl...

Question

In acute angled triangle $A B C, r + r_{1} = r_{2} + r_{3}$ and $∠ B > \frac{π}{3}$ then

$b + 2 c < 2 a < 2 b + 2 c$

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$b + 4 c < 4 a < 2 b + 4 c$

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$b + 4 c < 4 a < 4 b + 4 c$

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$b + 3 c < 3 a < 3 b + 3 c$

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Solution

The correct option is D
$b + 3 c < 3 a < 3 b + 3 c$

$r - r_{2} = r_{3} - r_{1} \Rightarrow \frac{Δ}{s} - \frac{Δ}{s - b} = \frac{Δ}{s - c} - \frac{Δ}{s - a} \Rightarrow \frac{- b}{s (s - b)} = \frac{c - a}{(s - a) (s - c)} \Rightarrow \frac{(s - a) (s - c)}{s (s - b)} = \frac{a - c}{b} \Rightarrow t a n^{2} \frac{B}{2} = \frac{a - c}{b} But \frac{B}{2} \in (\frac{π}{6}, \frac{π}{4}) \Rightarrow t a n^{2} \frac{B}{2} \in (\frac{1}{3}, 1) \Rightarrow \frac{1}{3} < \frac{a - c}{b} < 1 \Rightarrow b < 3 a - 3 c < 3 b \Rightarrow b + 3 c < 3 a < 3 b + 3 c$

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In acute angled triangle ABC,r+r1=r2+r3 and ∠B>π3 then

The correct option is D b+3c<3a<3b+3c r−r2=r3−r1⇒Δs−Δs−b=Δs−c−Δs−a⇒−bs(s−b)=c−a(s−a)(s−c)⇒(s−a)(s−c)s(s−b)=a−cb⇒tan2B2=a−cbBut B2∈(π6,π4)⇒tan2B2∈(13,1)⇒13<a−cb<1⇒b<3a−3c<3b⇒b+3c<3a<3b+3c

In acute angled triangle $A B C, r + r_{1} = r_{2} + r_{3}$ and $∠ B > \frac{π}{3}$ then