Question

In a triangle $ABC,{\text{ }}2{a^2} + 4{b^2} + {c^2} = 4ab + 2ac$, then the numerical value of $\cos (B)$
is equal to:

$$(A){\text{ }}0 \\\ (B){\text{ }}\dfrac{3}{8} \\\ (C){\text{ }}\dfrac{5}{8} \\\ (D){\text{ }}\dfrac{7}{8} \\\$$

Answer 1

Hint: Apply cosine Rule that states that the square of a side of a plane triangle equals the sum of the squares of the remaining sides minus twice the product of those sides and the cosine of angle between them.

Complete step-by-step answer:

As we know that equation given in the question is,
$\Rightarrow 2{a^2} + 4{b^2} + {c^2} = 4ab + 2ac$

So, now we had to reduce the above given equation.
So, for that we had to split $2{a^2}$ and take RHS of the
above equation to the LHS of the equation.

Then above equation becomes,
$\Rightarrow ({a^2} - 4ab + 4{b^2}){\text{ }} + {\text{ }}({a^2} - 2ac + {c^2}) = 0$

On solving the above reduced equation. It becomes,
$\Rightarrow {(a - 2b)^2} + {(a - c)^2} = 0$

As, the RHS of the above equation is now 0.

So, LHS should also be equal to 0.

But both the terms of LHS are square. So, they cannot be negative.

So, both the terms should be equal to 0.
$\Rightarrow$Hence, ${(a - 2b)^2} = 0$ and ${(a - c)^2} = 0$. SO, we get,
$\Rightarrow a - 2b = 0$ (1)
$\Rightarrow$And, $a - c = 0$ (2)

So, using equations 1 and 2 we can write that,
$\Rightarrow \dfrac{a}{1} = \dfrac{b}{{\left( {\dfrac{1}{2}} \right)}} = \dfrac{c}{1} = \lambda$(say) (3)

So, from equation 3. We get that,
$\Rightarrow a = \lambda ,b = \dfrac{\lambda }{2}$ and $c = \lambda$.

So, now we know the value of a, b and c.

So, we can apply cosine law now.

$\Rightarrow \cos (B) = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$ (4)

So, now put the value of a, b and c in equation 4. We get,
$\Rightarrow \cos (B) = \dfrac{{{{(\lambda )}^2} + {{(\lambda )}^2} - {{\left( {\dfrac{\lambda }{2}} \right)}^2}}}{{2{\lambda ^2}}} = \left( {\dfrac{{{\lambda ^2}}}{{{\lambda ^2}}}} \right)\left( {\dfrac{{1 + 1 - \dfrac{1}{4}}}{2}} \right) = 1 - \dfrac{1}{8} = \dfrac{7}{8}$
$\Rightarrow$Hence the value of $\cos (B)$ will be $\dfrac{7}{8}$.

So, the correct option of the above question will be D.

Note: Whenever we come up with this type of problem, the easiest and efficient way to get the cosine of any angle of a triangle is first trying to reduce the given equation into a perfect square and then find the value of a, b and c. After that we can apply cosine rule to get the required value of cosine of angle.