1. **Problem statement:** We have two triangles PQR and RST sharing vertex R on the same baseline. Given angles $\angle Q = 76^\circ$ and $\angle S = 102^\circ$, sides QP = QR and RS = ST (isosceles triangles), and bases PR = RT. We need to find the angle $x = \angle R$ between the two triangles. 2. **Key properties:** - In isosceles triangle PQR, sides QP = QR imply $\angle P = \angle R$. - In isosceles triangle RST, sides RS = ST imply $\angle R = \angle T$. - Since PR = RT, the segments PR and RT are equal. 3. **Find angles in triangle PQR:** - Sum of angles in triangle PQR is $180^\circ$. - Given $\angle Q = 76^\circ$, and $\angle P = \angle R = y$ (say). - So, $y + y + 76 = 180 \Rightarrow 2y = 104 \Rightarrow y = 52^\circ$. - Therefore, $\angle P = 52^\circ$ and $\angle R = 52^\circ$. 4. **Find angles in triangle RST:** - Sum of angles in triangle RST is $180^\circ$. - Given $\angle S = 102^\circ$, and $\angle R = \angle T = z$ (say). - So, $z + z + 102 = 180 \Rightarrow 2z = 78 \Rightarrow z = 39^\circ$. - Therefore, $\angle R = 39^\circ$ and $\angle T = 39^\circ$. 5. **Analyze angle $x = \angle R$ between the two triangles:** - At vertex R, the angle is split between the two triangles. - From triangle PQR, $\angle R = 52^\circ$. - From triangle RST, $\angle R = 39^\circ$. - The angle between the two triangles at R is the sum of these two angles minus the straight line angle (since PR = RT and they form a straight line). 6. **Calculate $x$:** - Since PR and RT are equal and marked with double parallel lines, they form a straight line at R. - The total angle around point R on the baseline is $180^\circ$. - The angle between the two triangles is the external angle formed by the difference between $52^\circ$ and $39^\circ$. - So, $x = 180^\circ - (52^\circ + 39^\circ) = 180^\circ - 91^\circ = 89^\circ$. **Final answer:** $$x = 89^\circ$$