In - ABC- if 1-b-c-1-c-a-3-a-b-c- then C is equal toA- 90-B- 45-C- 60-D- 30-

The correct option is C 60° Given that, 1/b+c+1/c+a=3/a+b+c ⇒a/b+c+b/c+a=1 ⇒ a(c+a)+b(b+c)=(b+c)(c+a) ⇒ a^2+b^2 c^2=ab ∴ cosC=a^2+b^2 c^2/2abab/2ab1/2 ⇒ C=60^0 ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

In Δ ABC, if ...

Question

In $Δ$ ABC, if $\frac{1}{b + c} + \frac{1}{c + a} = \frac{3}{a + b + c},$ then C is equal to

90°

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60°

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45°

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30°

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Solution

The correct option is B
60°

Given that,

$\frac{1}{b + c} + \frac{1}{c + a} = \frac{3}{a + b + c} \Rightarrow \frac{a}{b + c} + \frac{b}{c + a} = 1 \Rightarrow a (c + a) + b (b + c) = (b + c) (c + a) \Rightarrow a^{2} + b^{2} - c^{2} = a b ∴ c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \frac{a b}{2 a b} \frac{1}{2} \Rightarrow C = 60^{0}$

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In Δ ABC, if 1b+c+1c+a=3a+b+c, then C is equal to

The correct option is B 60° Given that, 1b+c+1c+a=3a+b+c⇒ab+c+bc+a=1⇒a(c+a)+b(b+c)=(b+c)(c+a)⇒a2+b2−c2=ab∴cosC=a2+b2−c22abab2ab12⇒C=600

In $Δ$ ABC, if $\frac{1}{b + c} + \frac{1}{c + a} = \frac{3}{a + b + c},$ then C is equal to