The sides of $△ A B C$ satisfy the equation $2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c$ . Then

The triangle is isosceles

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The triangle is obtuse

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B = {cos}^{- 1} (\frac{7}{8})

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A = {cos}^{- 1} (\frac{1}{4})

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Solution

The correct option is D $A = {cos}^{- 1} (\frac{1}{4})$
$(a^{2} - 2 a c + c^{2}) + (a^{2} - 4 a b + 4 b^{2}) = 0$
Or, $(a - c)^{2} + (a - 2 b)^{2} = 0$
$\Rightarrow a = c$ and $a = 2 b$
Therefore, the triangle is isosceles.
Also, $cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{7 b^{2}}{8 b^{2}} = \frac{7}{8}$
$cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{1}{4}$

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Similar questions

Δ A B C

2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c .

△ A B C

2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c,

∠ A

∠ B

△ A B C

2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c

A B C

2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c

cos B

A B C, 2 a^{2} + 4 b^{2} + c^{2} = 4 a b + 2 a c

cos B

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