Question

In a $ \Delta ABC $ , $ 2{a^2} + 4{b^2} + {c^2} = 4ab + 2ac $ , then $ \cos B $ is
1) $ 0 $
2) $ \dfrac{1}{8} $
3) $ \dfrac{3}{8} $
4) $ \dfrac{7}{8} $

Answer 1

Hint: -First we have to use the given condition to form a relation between the sides of the triangle. On obtaining the relation, we can use the cosine rule for triangles, that is,
 $ \cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} $
We will substitute the relations obtained in the cosine rule for triangles in order to get the required result.

Complete step-by-step answer:
The given condition is, $ 2{a^2} + 4{b^2} + {c^2} = 4ab + 2ac $ .
We are to find, $ \cos B $ .
We can write the given condition as,
 $ \Rightarrow {a^2} + {a^2} + 4{b^2} + {c^2} = 4ab + 2ac $
Now, taking all the terms on the left hand side, we get,
 $ \Rightarrow {a^2} + {a^2} + 4{b^2} + {c^2} - 4ab - 2ac = 0 $
 $ \Rightarrow ({a^2} - 4ab + 4{b^2}) + ({a^2} - 2ac + {c^2}) = 0 $
\[ \Rightarrow \{ {(a)^2} - 4ab + {(2b)^2}\} + \{ {(a)^2} - 2ac + {(c)^2}\} = 0\]
Now, using the formula, $ {x^2} - 2xy + {y^2} = {\left( {x - y} \right)^2} $ , we get,
 $ \Rightarrow {\left( {a - 2b} \right)^2} + {\left( {a - c} \right)^2} = 0 $
Therefore, we can say,
 $ {\left( {a - 2b} \right)^2} = 0 $ and $ {\left( {a - c} \right)^2} = 0 $
 $ \Rightarrow a - 2b = 0 $ and $ \Rightarrow a - c = 0 $
 $ \Rightarrow a = 2b $ and $ \Rightarrow a = c $
 $ \Rightarrow b = \dfrac{a}{2} $ and $ \Rightarrow a = c $
Therefore, we get the relations between the sides of the triangle, that is,
 $ b = \dfrac{a}{2} $ and $ a = c $
Now, we know, according to the cosine rule of triangle, we have,
 $ \cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} $
Now, substituting the relations obtained between the sides of the triangle, in the cosine rule, we get,
 $ \Rightarrow \cos B = \dfrac{{{a^2} + {{\left( a \right)}^2} - {{\left( {\dfrac{a}{2}} \right)}^2}}}{{2a\left( a \right)}} $
Computing the squares, we get,
 $ \Rightarrow \cos B = \dfrac{{{a^2} + {a^2} - {{\dfrac{a}{4}}^2}}}{{2{a^2}}} $
Now, simplifying the equation, we get,
 $ \Rightarrow \cos B = \dfrac{{2{a^2} - {{\dfrac{a}{4}}^2}}}{{2{a^2}}} $
Now, taking LCM in the numerator, we get,
 $ \Rightarrow \cos B = \dfrac{{\dfrac{{8{a^2} - {a^2}}}{4}}}{{2{a^2}}} $
 $ \Rightarrow \cos B = \dfrac{{\dfrac{{7{a^2}}}{4}}}{{2{a^2}}} $
Now, simplifying the equation further, we get,
 $ \Rightarrow \cos B = \dfrac{{7{a^2}}}{{8{a^2}}} $
Now, cancelling the term $ {a^2} $ in the numerator and denominator, we get,
 $ \Rightarrow \cos B = \dfrac{7}{8} $
Therefore, the value of $ \cos B $ is $ \dfrac{7}{8} $ , the correct option is 4.
So, the correct answer is “Option 4”.

Note: The cosine rule is useful to find all the three angles of the triangle. Those are,
 $ \cos A = \dfrac{{{b^2} + {c^2} - {c^2}}}{{2bc}} $
 $ \cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} $
 $ \cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} $
Also, like the cosine rule, there is also the sine rule, that is,
 $ \dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} $
These rules are used to find the angles and measure of sides of the triangle. This gives us the idea how the various angles and measure of sides are interrelated to each other. Calculations must be done with accuracy to get to the correct answer.