Construct a triangle with b, c and $|\angle B - \angle C|$ How can we construct a triangle with given b, c and $|\angle B - \angle C|$?

 A: A synthetic solution:
Assume WLOG that $b > c$ then $\angle B > \angle C$. Then $|\angle B -\angle C|=\angle B -\angle C$.
Now, draw a line segment of length $b$.
Now, construct angle $\angle DAC = \angle B -\angle C$, such that $DA=c$.
Join $D$ and $C$.
Now, with $A$ as the center, draw a circle of radius $c$ and let it cut $DC$ (extended maybe) at $B$.
You have your triangle.
A: Let's call $\alpha$ the angle in $A$, $\beta$ the angle in $B$ and $\gamma$ the angle in $C$. I will discuss the case $\beta\gt\gamma$ and call $\theta=\beta-\gamma$, so that $\beta=\theta+\gamma$.
Place $A$ in the origin of a cartesian plane, and let $B$ lie on the $x$ axis, you'll have $B\equiv(b,0)$, $C\equiv(c\cos\alpha,c\sin\alpha)$ so:$$
\tan\gamma=\frac{c\sin\alpha}{b-c\cos\alpha}
$$
but we know that $\beta=\pi-\alpha-\gamma$ so $\theta+\gamma=\pi-\alpha-\gamma$ and $\alpha=\pi-\theta-2\gamma$.
Substitution for $\alpha$ gives the following equation in $\gamma$: $$
\tan\gamma=\frac{c\sin(\theta+2\gamma)}{b+c\cos(\theta+2\gamma)}
$$
Use this to find $\gamma$ so you can find the other two angles and the coordinates of $C$.
A: Let $B-C=x$ we know $B+C=180-A$ so we get a relation between B and A its $B=90-A/2+x/2$ so $C=90-A/2-x/2$ so now you said you know the value of $x$ ie $B-C$ so you know $A+B+C=180$ now you have all three angles in terms of A so get it. Then find the remaining side if you want use Sine rule ie $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$ get a ie side opposite to angle A and you can construct a triangle.
