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Special Integrals -4

In the ambigu...

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
(i) $c_{1} + c_{2} = 2 b cos A$ and $c_{1} c_{2} = b^{2} - a^{2}$ .
(ii) $c_{1}^{2} - 2 c_{1} c_{2} cos 2 A + c_{2}^{2} = 4 a^{2} {cos}^{2} A .$
(iii) $(c_{1} - c_{2})^{2} + (c_{1} + c_{2})^{2} {tan}^{2} A = 4 a^{2}$ .

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Solution

the quadratic given the two values $c_{1} . c_{2}$ of $c^{2} - 2 b c c o s A + (b^{2} - a^{2}) = 0$
Then we have
(i) $c_{1} + c_{2} = 2 b c o s a$ ........(1)
and $c_{1} c_{2} = b^{2} - a^{2}$ ........(2)
(ii) From (1) and (2),
$(c_{1} + c_{2})^{2} = 4 b^{2} c o s^{2} A = 4 (c_{1} c_{2} + a^{2}) c o s^{2} A$
or $c_{1}^{2} + 2 c_{1} c_{2} + c_{z}^{2} - 4 c_{1} c_{2} c o s^{2} A = 4 a^{2} c o s^{2} A$
or $c_{1}^{2} - 2 c_{1} c_{2} (2 c o s^{2} A - 1) + c_{2}^{2} = 4 a^{2} c o s^{2} A$
or $c_{1}^{2} - 2 c_{1} c_{2} c o s 2 A + c_{2}^{2} = 4 a^{2} c o s^{2} A$

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