In the ambiguous case, if the remaining angles of the triangles formed with a, b and A be $B_{1}, C_{1}$ and $B_{2}, C_{2}$ , then prove that $\frac{s i n C_{1}}{s i n B_{1}} + \frac{s i n C_{2}}{s i n B_{2}} = 2 c o s A .$

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Solution

$c_{1}, c_{2}$ are two values of c, we have
$c_{1} + c_{2} = 2 b c o s A .$ ...(1)
Also $B_{1}, B_{2}$ are supplementary angles, that is $B_{2} = 180^{0} - B_{1}$ so that $s i n B_{2} = s i n B_{1}$
Hence $\frac{s i n C_{1}}{s i n B_{1}} + \frac{s i n C_{2}}{s i n B_{2}} = \frac{s i n C_{1}}{s i n B_{1}} + \frac{s i n C_{2}}{s i n B_{1}}$
$= \frac{s i n C_{1} + s i n C_{2}}{s i n B_{1}}$
= $\frac{k c_{1} + k c_{2}}{k b} = \frac{c_{1} + c_{2}}{b} = 2 c o s A$ , from (1)

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