Question

In the ambiguous case, if b and A are given and $c_1$, $c_2$ are the two values of third side, prove that-
${\text{(i)}}{\text{. }}{{\text{c}}_1} + {c_2} = 2b\cos A$ and ${c_1}{c_2} = {b^2} - {a^2}$.
${\text{(ii)}}{\text{. }}{{\text{c}}_1}^2 - 2{c_1}{c_2}\cos 2A + c_2^2 = 4{a^2}{\cos ^2}A$.
\[{\text{(iii)}}{\text{. (}}{{\text{c}}_1} - {c_2}{)^2} + {({c_1} + {c_2})^2}{\tan ^2}A = 4{a^2}\]

Answer 1

Hint: The quadratic given the two values ${c_1},c{}_2$ of ${c^2} - 2bc\cos A + ({b^2} - {a^2}) = 0$, using this prove the question asked.

Complete step-by-step answer:
We know that the quadratic given the two values ${c_1},c{}_2$ of ${c^2} - 2bc\cos A + ({b^2} - {a^2}) = 0$,
Then we have the following-
Part (i) ${{\text{c}}_1} + {c_2} = 2b\cos A - (1)$ {by the relation, sum of the roots of the equation is equal to -b/a}
Also, by the product of roots of the equation is equal to c/a, we can write-
${c_1}{c_2} = {b^2} - {a^2} - (2)$
Part (ii) From equation (1) and (2).
${{\text{(}}{{\text{c}}_1} + {c_2})^2} = {(2b\cos A)^2} = 4{b^2}{\cos ^2}A$
Solving further we get-
\[
  {{\text{(}}{{\text{c}}_1} + {c_2})^2} = 4{b^2}{\cos ^2}A \\
   \Rightarrow c_1^2 + c_2^2 + 2{c_1}{c_2} = 4({c_1}{c_2} + {a^2}){\cos ^2}A \\
  or, c_1^2 + c_2^2 + 2{c_1}{c_2} - 4{c_1}{c_2}{\cos ^2}A = 4{a^2}{\cos ^2}A \\
  or, c_1^2 - 2{c_1}{c_2}(2{\cos ^2}A - 1) + c_2^2 = 4{a^2}{\cos ^2}A \\
   \Rightarrow c_1^2 - 2{c_1}{c_2}\cos 2A + c_2^2 = 4{a^2}{\cos ^2}A \\
 \]
Part (iii) using the cosine formula, $\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$
Solving it further,
$
  2b\cos A.c = {c^2} + ({b^2} - {a^2}) \\
   \Rightarrow {c^2} - 2b\cos A.c + {b^2} - {a^2} = 0 \\
$
Which is a quadratic expression, whose roots are $c_1$, $c_2$.
So, sum of roots ${c_1} + {c_2} = 2b\cos A$ and product of roots ${c_1}{c_2} = {b^2} - {a^2}$
Now LHS of part 3-
\[{{\text{(}}{{\text{c}}_1} - {c_2})^2} + {({c_1} + {c_2})^2}{\tan ^2}A\]
\[{{\text{(}}{{\text{c}}_1} - {c_2})^2} = {({c_1} + {c_2})^2} - 4{c_1}{c_2}\] using this we get-
\[
  {{\text{(}}{{\text{c}}_1} + {c_2})^2} - 4{c_1}{c_2} + {({c_1} + {c_2})^2}{\tan ^2}A \\
   \Rightarrow {{\text{(}}{{\text{c}}_1} + {c_2})^2}(1 + {\tan ^2}A) - 4{c_1}{c_2} \\
 \]
Now we know \[(1 + {\tan ^2}A) = {\sec ^2}A\] and ${c_1} + {c_2} = 2b\cos A$ we get-
\[
   \Rightarrow {{\text{(2bcosA}})^2}({\sec ^2}A) - 4({b^2} - {a^2}) \\
   \Rightarrow 4{b^2}{\cos ^2}A.{\sec ^2}A - 4{b^2} + 4{a^2}\{ \because {\cos ^2}A = \dfrac{1}{{{{\sec }^2}A}}\} \\
   \Rightarrow 4{b^2} - 4{b^2} + 4{a^2} = 4{a^2} \\
\]
Hence, LHS = RHS [Hence Proved]

Note: Whenever such types of questions appear, then write down the things given in question and then by using the relation between the zeros and the coefficients of the polynomial, prove the following given parts.