1. **State the problem:** We have a diamond-shaped quadrilateral ABCD with diagonals AC and BD intersecting at right angles (90°). Given: $AC=10$, $DB=17$, and angle near vertex A is $18^\circ$. We need to find angle $A$. 2. **Recall properties:** In a rhombus (diamond), diagonals intersect at right angles and bisect each other. So, the diagonals split into halves: $AO=OC=\frac{AC}{2}=5$ and $BO=OD=\frac{DB}{2}=8.5$ where $O$ is the intersection point. 3. **Analyze triangle AOB:** Triangle $AOB$ is right-angled at $O$ because diagonals intersect at $90^\circ$. We know $AO=5$, $BO=8.5$, and angle $A$ is the angle at vertex $A$ in triangle $AOB$. 4. **Find angle $A$ using trigonometry:** In right triangle $AOB$, angle $A$ is adjacent to side $AO$ and opposite to side $BO$. So, $$\tan(A) = \frac{BO}{AO} = \frac{8.5}{5} = 1.7$$ 5. **Calculate angle $A$:** $$A = \arctan(1.7) \approx 59.04^\circ$$ 6. **Conclusion:** The angle $A$ in the diamond-shaped quadrilateral is approximately $59.04^\circ$.