Suppose that we have constructed the triangle $ABC$ First of all, it's obvious that $b>c \ \ \ \ (1)$. Indeed, the triangles $AMC,AMB$ have $AM=AM, MC=MB$ and $\widehat{AMC}>90^0>\widehat{AMB}$ Now, we bring a parallel from $C$ to $AM$, which intersects the line $AB$ at the point $B'$. $MC=MB \Rightarrow AB=AB'$ The triangle $BB'C$ have median $CA=b$, a side $BB'=2c$ and $\widehat{BCB'}=w$ Construction We construct the segment $BB'=2c$, and let $A$ its midpoint. Then, we construct the arc $h$ of the circle, which pass through $B,B'$ and $\forall P \in h,\ \widehat{BPB'}=w$. (There is also an arc $h'$ on the same circle of $h$, such that $h \cap h' = \{A,B\}$ and $h \cup h'$ is a circle. But $\forall P \in h', \ \widehat{BPB'}=180^0-w$. Also, since $w<90^0$, $h$ is greater than $h'$. That is, the ray $AO$ intersects the arc $h$ (at its midpoint), where $O$ is the center of the arc) For every $P \in h$, the triangle $BPB'$ has $\widehat{BPB'}=w$ and $BB'=2c$. Now we have to find a point $P$, such that $AP=b$. Let $K$ be the intersection point of $h$ with the perpendicular bisector of $BB'$ ($\equiv$ the line $AO$). Of course, the max length of $AP$ is when $P \equiv K$ (from the triangle $AOP$ we have $AP\leq OA+OP=OA+R$, so we have a max value for $AP$ when $A,O,P$ are collinear). There exists a solution, if the given $b$ is not greater than $AK$. If $b>AK$ then there are not solutions. Let's find the length of $AK$. $\triangle AKB : \widehat{AKB} = \frac{w}{2}$ $\tan{\frac{w}{2}}=\frac{AB}{AK} =\frac{c}{AK} \Rightarrow$ $AK \cdot \tan{\frac{w}{2}}=c$ $b\leq AK \Rightarrow$ $b \cdot \tan{\frac{w}{2}} \leq AK \cdot \tan{\frac{w}{2}} \Rightarrow$ $b \cdot \tan{\frac{w}{2}} \leq c \ \ \ \ \ (2)$ If both conditions $(1),(2)$ hold, then we construct a circle $C1=(A,b)$ and let $C$ an intersection point of $C1$ and $h$ (the other is symmetric to $C$ w.r.t. $AK$, so it makes a symmetric triangle) The triangle $BCB'$ has side $BB'=2c, \ \ \widehat{BCB'}=w$ and median $AC$. Let $M,M'$ the midpoints of $BC,BC'$ respectively. We have $\widehat{AMB}=\widehat{AM'B}=\widehat{BCB'}=w \ \ (CMAM'$ is a parellelogram $)$ So there are two triangles with the specified properties: The triangle $ABC$ has $AB=c, AC=b$, and $\widehat{AMB}=w$ The triangle $AB'C$ has $AB'=c, AC=b$, and $\widehat{AMB'}=w$ If it is $b \tan{\frac{w}{2}}=c$, then $b=AK$, so we have only one solution (the triangles $ABC,AB'C$ are equal) and $\triangle ABC$ is a right triangle $(A=90^0)$

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