In a - ABC- a-b-cb-c-a- - bc- then

The correct option is C 0 lt; λ lt;4 In ΔABC, #160; #160; #160; #160; (a+b+c)(b+c a)=λbc #160; #160; #160; #160; 2 ...

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Byju's Answer

Standard IX

Mathematics

Exponential Representation for Irrational Numbers

In a Δ ABC,...

Question

In a $Δ A B C, (a + b + c) (b + c - a) = λ b c,$ then

λ < - 6

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λ > 6

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0 < λ < 4

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λ > 4

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Solution

The correct option is C $0 < λ < 4$
In $Δ A B C,$
$(a + b + c) (b + c - a) = λ b c$
$2 s (2 s - 2 a) = λ b c$
$\frac{s (s - a)}{b c} = \frac{λ}{4}$
${cos}^{2} \frac{A}{2} = \frac{λ}{4}$

$0 < {cos}^{2} \frac{A}{2} < 1$
$0 < \frac{λ}{4} < 1$
$\Rightarrow 0 < λ < 4$ is the right answer.

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In a ΔABC,(a+b+c)(b+c−a)=λbc, then

The correct option is C 0<λ<4In ΔABC, (a+b+c)(b+c−a)=λbc 2s(2s−2a)=λbc s(s−a)bc=λ4 cos2A2=λ40<cos2A2<10<λ4<1⇒0<λ<4 is the right answer.

In a $Δ A B C, (a + b + c) (b + c - a) = λ b c,$ then

The correct option is C $0 < λ < 4$
In $Δ A B C,$
$(a + b + c) (b + c - a) = λ b c$
$2 s (2 s - 2 a) = λ b c$
$\frac{s (s - a)}{b c} = \frac{λ}{4}$
${cos}^{2} \frac{A}{2} = \frac{λ}{4}$

$0 < {cos}^{2} \frac{A}{2} < 1$
$0 < \frac{λ}{4} < 1$
$\Rightarrow 0 < λ < 4$ is the right answer.