At Grade 2, every angle question draws on just five facts:

• Angles on a straight line sum to $180°$.
• Angles around a point sum to $360°$.
• Vertically opposite angles (formed where two lines cross) are equal.
• Angles in a triangle sum to $180°$.
• Angles in a quadrilateral sum to $360°$.

More advanced facts — angles in parallel lines, angles in polygons, circle theorems — belong to higher grades and are not required here.

Show the calculation on one line and state the matching geometric reason on the next. For a missing angle on a straight line next to a given $55°$:

Reason: angles on a straight line add up to $180°$.

Diagrams in these questions are deliberately not drawn to scale — an angle that looks like $90°$ might really be $88°$. Never measure; only use the numbers given in the question together with the five angle facts above. Treat the diagram as a schematic, not a scale drawing.

Calculate and label every angle you can from the given information, even ones the question doesn't explicitly ask for — intermediate angles are often the key that unlocks the final one.

$ABCD$ is a quadrilateral. Angle $A = 95°$, angle $B = 110°$, and angle $C$ is twice the size of angle $D$. Find angle $D$.

1. Angles in a quadrilateral sum to $360°$: $A + B + C + D = 360°$.
2. Known angles: $95° + 110° = 205°$.
3. Remaining: $360° - 205° = 155°$, shared between $C$ and $D$ where $C = 2D$.
4. So $2D + D = 155°$, giving $3D = 155°$, so $D = 51.\dot{6}°$.

Answer: angle $D \approx 51.7°$ (1 d.p.), using the reason that angles in a quadrilateral sum to $360°$.

Substitute your found values back into the original fact you used. Here, $155° \div 3 = 51.\overline{6}°$, so $D \approx 51.7°$ and $C = 2D \approx 103.3°$; adding everything — $95+110+103.3+51.7$ — should land back on $360°$. This single check catches almost every arithmetic slip in a multi-angle problem.