Question

In a triangle ABC, $A={45}^0$ and $c_1,c_2$ are the two values of side c in the ambiguous case, show that $cos{B}_{1}C{B}_2=\frac{2c_1{c}_2}{c_1^2+c_2^2}.$

Answer 1

Let the two triangles formed be $A{B}_{1}C$ and $A{B}_{2}C$. Draw CN $\perp$ to $A{B}_1$. Then since $△B_{1}C{B}_2$ is isosecles, we have $\angle B_{1}CN=\angle B_{2}CN=\theta$, say. Since $\angle A={45}^0$.
we have $CN=bsin{45}^0=b/sqrt2$.
Now $a^2=b^2+c^2-2bccosA$
= $b^2+c^2-2bccos{45}^0=b^2+c^2-\sqrt{2}bc$
or $c^2-\sqrt{2}bc+b^2-a^2=0$
If $c_1,c_2$ are the two values of c then
$c_1+c_2=\sqrt{2}b$ and $c_1{c}_2=b^2-a^2.$ .....(1)
$\therefore c_1^2+c_2^2=(c_1+c_2{)}^2-2c_1{c}_2$
=$2b^2-2(b^2-a^2)=2a^2$ ........(2)
Now $cos{B}_{1}C{B}_2=cos2\theta =2co{s}^2\theta -1.$
=$2.(\frac{CN}{a}{)}^2-1=2.\frac{b^2}{2a^2}-1$ ( $\because CN=b/\sqrt{2}$)
$\frac{b^2-a^2}{a^2}=\frac{c_1{c}_2}{\frac{1}{2}(c_1^2+c_2^2)}=\frac{2c_1{c}_2}{c_1^2+c_2^2}$ by (1) and (2)