In a - ABC - 2 a 2-4 b 2- c 2-4 ab -2 ac- then cos B is equal toA- 0B- 1 - 8C- 3 - 8D- 7 - 8

The correct option is D 7/8 2a^2+4b^2+c^2=4ab+ac ⇒ a^2+(2b)^2 4ab+a^2+c^2 2ac=0 ⇒ (a 2b)^2+(a c)^2=0 ⇒ a=2b=c cosB=a^2+c^2 b^2/2ac =c^2+c^2 ( c/2 )^2/2 ...

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Byju's Answer

Standard XI

Mathematics

Properties of Conjugate of a Complex Number

In a Δ ABC , ...

Question

In a $Δ$ ABC, $2 a ^{2} + 4 b^{2}$ + $c^{2} = 4 a b + 2 a c,$ then cos B is equal to

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1/8

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3/8

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7/8

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Solution

The correct option is D
7/8

$2 a^{2} + 4 b^{2} + c^{2} = 4 a b + a c \Rightarrow a^{2} + (2 b)^{2} - 4 a b + a^{2} + c^{2} - 2 a c = 0 \Rightarrow (a - 2 b)^{2} + (a - c)^{2} = 0 \Rightarrow a = 2 b = c c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{c^{2} + c^{2} - {(\frac{c}{2})}^{2}}{2 \times c \times c} = \frac{2 c^{2} - \frac{c^{2}}{4}}{2 c^{2}} \Rightarrow c o s B = \frac{7}{8}$

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In a Δ ABC, 2a­2+4b2 + c2=4ab+2ac,then cos B is equal to

The correct option is D 7/8 2a2+4b2+c2=4ab+ac⇒a2+(2b)2−4ab+a2+c2−2ac=0⇒ (a−2b)2+(a−c)2=0⇒a=2b=ccosB=a2+c2−b22ac=c2+c2−(c2)22×c×c=2c2−c242c2⇒cosB=78

In a $Δ$ ABC, $2 a ^{2} + 4 b^{2}$ + $c^{2} = 4 a b + 2 a c,$ then cos B is equal to