In acute angled triangle ABC- r- r2-r3-r1 and - B- -3 then exhaustive range of a-c-b is

R= r_2+r_3 r_1 ⇒Δ/s= Δ/s b+Δ/s c Δ/s a ⇒1/s 1/s b= 1/s c 1/s a ⇒ b/s(s b)= s a s+c/(s c)(s a)= c a/(s c)(s a) ⇒(s a)(s c)/s(s b)= a c/b ⇒ tan^2B/2= a c/b Since ...

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Byju's Answer

Standard XI

Mathematics

Area Using Semiperimeter

In acute angl...

Question

In acute angled triangle $A B C, r = r_{2} + r_{3} - r_{1}$ and $∠ B > \frac{π}{3}$ then exhaustive range of $\frac{a - c}{b}$ is

$(\frac{1}{2}, 1)$

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$(\frac{1}{4}, \frac{1}{2})$

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$(\frac{1}{4}, 1)$

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None of these

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Solution

The correct option is D
None of these

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

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In acute angled triangle $A B C, r = r_{2} + r_{3} - r_{1}$ and $∠ B > \frac{π}{3}$ then exhaustive range of $\frac{a - c}{b}$ is

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In acute angled triangle ABC,r=r2+r3−r1 and ∠B>π3 then exhaustive range of a−cb is

The correct option is D None of these r=r2+r3−r1⇒Δs=Δs−b+Δs−c−Δs−a⇒1s−1s−b=1s−c−1s−a⇒−bs(s−b)=s−a−s+c(s−c)(s−a)=c−a(s−c)(s−a)⇒(s−a)(s−c)s(s−b)=a−cb⇒tan2B2=a−cbSinceBϵ(π3,π2)⇒B2ϵ(π6,π4)⇒a−cbϵ(13,1)

In acute angled triangle $A B C, r = r_{2} + r_{3} - r_{1}$ and $∠ B > \frac{π}{3}$ then exhaustive range of $\frac{a - c}{b}$ is