Question

In the diagram, $PW$ is parallel to $QX$, $S$ and $T$ lie on $QX$, and $U$ and $V$ are the points of intersection of $PW$ with $SR$ and $TR$, respectively. If $\angle SUV = 120^\circ$ and $\angle VTX = 112^\circ$, what is the measure of $\angle URV$?[asy]

size(6cm); defaultpen(linewidth(0.75) + fontsize(11));
// Useful function
path extend(pair a, pair b, int n=10) { return a--(a + n * (b - a));}

// Points
real ust = 60; real vtx = 112;
pair q = (0, 0); pair p = (0, 0.5); pair w = (2, p.y); pair x = (2, 0);
pair s = (0.3, 0); pair t = s + (1, 0);
pair u = shift(s) * (p.y / Tan(ust), p.y); pair v = shift(t) * (p.y / Tan(vtx), p.y);
pair r = IP(extend(s, u), extend(t, v));

// Draw lines
draw(s--r--t);
draw(p--w, Arrow(position=0.8, size=5));
draw(q--x, Arrow(position=0.8, size=5));

// Labels
label("$P$", p, W); label("$Q$", q, W); label("$W$", w, E); label("$X$", x, E);
label("$R$", r, N); label("$U$", u, NW); label("$V$", v, NE); label("$S$", s, NW);

label("$T$", t, NE);

[/asy]

Answer 1

Since $PW\parallel QX$, we have $\angle UPW = \angle RSQ$ and $\angle PVW = \angle RTQ$ (corresponding angles). Therefore, ∠URV=∠URP+∠PRV=∠UPW+∠PVW.∠URV=∠URP+∠PRV=∠UPW+∠PVW.Also, $\angle SUV + \angle VTX + \angle RSQ + \angle RTQ = 360^\circ$ (angles around a point). Substituting $\angle SUV = 120^\circ$ and $\angle VTX = 112^\circ$ and simplifying, we get $\angle RSQ = 64^\circ$ and $\angle RTQ = 64^\circ$. Thus, ∠URV=∠UPW+∠PVW=180∘−∠PWV.∠URV=∠UPW+∠PVW=180∘−∠PWV.To find $\angle PWV$, we consider $\triangle UVS$ and $\triangle TVV$. Since $UV = UV$ (common side), $VS = VT$ (both are segments of $QX$), and $\angle VTV = \angle USV = 180^\circ - \angle SUV - \angle VTX = 28^\circ$, the two triangles are congruent. Therefore, $PW = VW$, so $\angle PWV = \angle PVW = \angle VTX = 112^\circ$. Hence, ∠URV=180∘−∠PWV=68∘.∠URV=180∘−∠PWV=68∘.Therefore, $\angle URV = \boxed{68^\circ}$.