Question

If \[a,b,\]A are given in a triangle and \[{c_1}\]and \[{c_2}\]are two possible values of third side such that \[c_1^2 + {c_1}{c_2} + c_2^2 = {a^2},\]then A is equal to
A. \[{30^o}\]
B. \[{60^o}\]
C. \[{90^o}\]
D. $ {120^o} $

Answer 1

Hint: Firstly, we will convert into the quadratic equation and we will find the sum of zeroes, product of zeroes. Further we will put the value of zeroes in the given equation and then we will find the angle A.

Complete step-by-step answer:
As we know that \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
Now, cross multiplying the number
 $ 2bc\,\cos A = {b^2} + {c^2} - {a^2} $
 $ \Rightarrow {b^2} + {c^2} - {a^2} - 2bc\,\cos A = 0 $
In the above equation, we can do in the form of quadratic equation and C is the variable in which $ {c_1},{c_2} $ are zeros of equation:
  $ {c^2} - 2bc\cos A + {b^2} - {a^2} = 0 $
Here, $ A = 1,\,B = - 2b\cos A,C = {b^2} - {a^2} $
Then, Sum of the zeroes $ \left( {{c_1} + {c_2}} \right) $ $ = \dfrac{{ - b}}{a} $
Sum of zeroes $ \left( {{c_1} + {c_2}} \right) $ $ = \dfrac{{ - ( - 2b\cos A)}}{1} $
Sum of zeroes $ \left( {{c_1} + {c_2}} \right) $ $ = 2b\cos A $
And, product of zeroes $ ({c_1} \times {c_2}) = \dfrac{C}{a} $
product of zeroes $ {\left( {{c_1} \times c} \right)_2} = \dfrac{{{b^2} - {a^2}}}{1} $
product of zeroes $ \left( {{c_1} \times {c_2}} \right) = {b^2} - {a^2} $
It is given that: $ c_1^2 + {c_1}{c_2} + c_2^2 = {a^2} $ ……(i)
Now,
 $ c_1^2 + {c_1}{c_2} + c_2^2 = {a^2} $ ……(ii)
We will add $ {c_1}{c_2} $ and subtracted the equation (ii) , we have
\[c_1^2 + {c_1}{c_2} + c_2^2 + {c_1}{c_2} - {c_1}{c_2} = {a^2}\]
\[c_1^2 + 2{c_1}{c_2} + c_2^2 - c_2^2 - {c_2}{c_2} = {a_2}\]
As, we know that \[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
Then $ {({c_1} + {c_2})^2} - {c_1}{c_2} = {a^2} $ ……(iii)
Then, we will substitute the sum of zeroes $ ({c_1} + {c_2}) $ and product of zeroes $ ({c_1}{c_2}) $ in equation (iii), we have
\[{(2b\cos A)^2} - ({b^2} - {a^2}) = {a^2}\]
 $ 4{b^2}{\cos ^2}A - {b^2} + {a^2} = {a^2} $
\[4{b^2}{\cos ^2}A - {b^2} + {a^2} - {a^2} = 0\]
 $ \Rightarrow 4{b^2}{\cos ^2}A - {b^2} = 0 $
 $ \Rightarrow 4{b^2}{\cos ^2}A = {b^2} $
 $ \Rightarrow {\cos ^2}A = \dfrac{{{b^2}}}{{4{b^2}}} $
 $ \Rightarrow {\cos ^2}A = \dfrac{1}{4} $
 $ \Rightarrow {\cos ^2}A = {\left( {\dfrac{1}{2}} \right)^2} $
 $ \Rightarrow \cos A = \pm \dfrac{1}{2} $
 $ \therefore \cos A = 60 $
 $ A = {60^o} $
So, the correct answer is “Option B”.

Note: When students convert the equation into quadratic equation then take c as a variable. Then put the exact value of zeroes in the equation to get the answer. We can also find the roots by sridharas method.