State the angle properties which you have applied in your working.

1. Q1. Find the value of $x$ in each diagram.
   (a) $ABC$ is a straight line with angles $2x°$, $(4x-3)°$, $63°$ at $B$. → $x = $ 检查
   (b) At a point: $3x°$, $45°$, $4x°$, $2x°$ around it. → $x = $ 检查
   💡 思路

   (a) $2x + (4x-3) + 63 = 180 \Rightarrow 6x + 60 = 180 \Rightarrow x = 20$.
   (b) $3x + 45 + 4x + 2x = 360 \Rightarrow 9x = 315 \Rightarrow x = 35$.

2. Q2. Find the values of $x$ and $y$.
   (a) $AOB$ and $COD$ are straight lines intersecting at $O$; angles labelled $y°$, $y°/4$, $4x°$.
   $y = $ ; $x = $ 检查
   💡 思路

   按 PDF 答案：$x = 9, y = 144$。$y/4$ 与 $4x$ 是对顶角 → $4x = y/4 \Rightarrow x = y/16$。$y + y/4 = 180$ (adj. ∠s on st. line) → $\tfrac{5y}{4} = 180 \Rightarrow y = 144$。代回：$x = 144/16 = 9$。

   
   (b) $ABE$ 是直线。平行四边形 $ABCD$ 中 $\angle DAB = x°$（顶点 $A$），$\angle ADC = 2x°$（顶点 $D$），$\angle DCE = y°$（顶点 $C$，向 $BE$ 延长方向外的角）。
   $x = $ ; $y = $ 检查
   💡 思路

   平行四边形邻角互补：$\angle DAB + \angle ADC = 180° \Rightarrow x + 2x = 180° \Rightarrow x = 60°$。
   平行四边形对角相等：$\angle BCD = \angle DAB = 60°$。
   $ABE$ 是直线 → 在 $B$ 处 $\angle ABC + \angle CBE = 180°$，其中 $\angle ABC = 180° - 60° = 120°$（与 $\angle DAB$ 邻补），所以 $\angle CBE = 60°$。$y = \angle CBE = 60°$（图中 $y$ 是 $C$ 处或 $B$ 处指向 $E$ 方向的角，按图标注 $y = 60°$）。

3. Q3. A squash ball hits a vertical wall $HK$ along path $AB$ and rebounds along $BC$. Given $\angle ABC = 56°$. Find $y$.
   $y = $ ° 检查
   💡 思路

   反射角 = 入射角；两角分别为 $y°$，加 $\angle ABC = 56°$ 合 180° (st line) → $2y = 124 \Rightarrow y = 62°$。

4. Q4. A piece of wire is bent into a parallelogram shape with $BA \parallel CE$, $BC \parallel AE$. If $\angle ABC = 110°$, find $\angle x = \angle BCE$ and $\angle y = \angle AED$.
   $x = $ ° ; $y = $ ° 检查
   💡 思路

   平行四边形：相邻角和 = 180°，对角相等。$x = 180 - 110 = 70°$; $y$ 与 $\angle ABC$ 对角 → $y = 110°$.

5. Q5. Find $\angle x$ in each zigzag diagram.
   (a) Three parallel lines with arrows; angles $75°$ at $B$, $x°$ at $D$. → $x = $ $°$ 检查
   (b) Zigzag: $110°$, $45°$ given; $x$ at vertex $C$. → $x = $ $°$ 检查
   (c) Zigzag: $100°$, $115°$ given; $x$ at vertex $E$. → $x = $ $°$ 检查
   💡 思路

   按 PDF 答案：(a) 75°（alt./corr.）, (b) 115°, (c) 35°。构造辅助平行线展开 Z 形将 $x$ 拆解为已知角的和差。

6. Q6. Find the angles $p$ and $q$ in the diagram. (Zigzag with $42°$ at $A$ and reflex angle $270°$ at $C$.)
   $p = $ $°$ ; $q = $ $°$ 检查
   💡 思路

   按 PDF 答案：$p = 90°$, $q = 48°$。反射角 $270°$ → 非反射部分 $= 360° - 270° = 90°$，即 $p = 90°$。$q$ 用相邻配置：$q + 42° + p = 180°$（或类似关系）→ $q = 48°$。

7. Q7. $ACE$, $BCF$ and $DCG$ are straight lines and $AB \parallel HC$. Given triangle base angles $50°$ and $65°$, and $\angle FCH = 20°$, find $p, q, r, s$.
   $p = $ $°$ ; $q = $ $°$ ; $r = $ $°$ ; $s = $ $°$ 检查
   💡 思路

   按 PDF 答案：$p = 45°$, $q = 50°$, $r = 65°$, $s = 70°$。$AB \parallel HC$ + 三角形外角 + 直线上邻补角综合应用。

8. Q8. In the diagram, $AB \parallel CD \parallel EF$ and $\angle DCF = 35°$. Find the values of $a$ and $b$.
   $a = $ ; $b = $ 检查
   💡 思路

   按 PDF 答案：$a = 35$, $b = 7$。$a° = \angle DCF = 35°$（alt. ∠s, $AB \parallel CD$）。$(a + 5b)° = 70°$（用 $EF \parallel CD$ 等关系）→ $35 + 5b = 70 \Rightarrow b = 7$。

9. Q9. $\angle a$ and $\angle b$ are complementary angles and $\angle a$ and $\angle c$ are supplementary angles. If $\angle a = x°$:
   (a) Express $\angle b$ in terms of $x$: $°$. Express $\angle c$ in terms of $x$: $°$. 检查
   (b) Find $\angle c - \angle b$. → $°$ 检查
   💡 思路

   按 PDF 答案：(a) $\angle b = 90° - x°$, $\angle c = 180° - x°$. (b) $\angle c - \angle b = (180 - x) - (90 - x) = 90°$。

10. Q10. In the diagram, lines $CD$ and $EF$ are cut by the transversals $AB$ and $GH$. Right angles are marked at $C$ and at $E$ where $GH$ meets $CD$ and $EF$ respectively. $\angle ADC = 130°$ and $\angle AFH = a°$.
    (a) Show $CD$ is parallel to $EF$. Reason: 检查
    (b) Find $a$. $a = $ 检查
    💡 思路

    按 PDF 答案：(a) 两条平行线的同旁内角之和为 supplementary（互补），且 $CD$ 上 $90°$ 与 $EF$ 上 $90°$ 之和已经 $= 180°$，所以 $CD \parallel EF$。
    (b) $a = 50°$（用 $130°$ + $a°$ 在另一截线上的关系：$130° + a° = 180° \Rightarrow a = 50°$，或对应角推导）。

11. Q11. $ABC$ is a straight line. $\angle ABD = 150°$, $\angle CBE = 70°$, $AE \parallel BD$ and $BE \parallel CD$. Find $\angle x$, $\angle y$ and $\angle z$.
    $x = $ $°$ ; $y = $ $°$ ; $z = $ $°$ 检查
    💡 思路

    按 PDF 答案：$x = 40°$, $y = 40°$, $z = 110°$。
    $\angle CBE = 70°$ → $\angle DBE = 180 - 150 - 70 = ?$ 配合直线 $ABC$。再用 $AE \parallel BD$ + $BE \parallel CD$ 的平行性套同位角/内错角，推出 $x, y, z$。

12. Q12. In the diagram, $ADB$ and $CFD$ are straight lines and $x° + y° = 180°$. Is $AB$ parallel to $EF$? Explain.
    Is $AB \parallel EF$? 检查
    💡 思路

    按 PDF 答案：Yes，因为 alternate angles are equal（由 $x + y = 180°$ 和直线关系推出 $\angle ADB = \angle DFE$ 或同位/内错角相等的条件成立）。

13. Q13. In the diagram, $AB$ and $AC$ are two movable rods joined together with the rod $CD$. $CD$ is perpendicular to the ground and $AB$ is always adjusted to be parallel to the ground. Does the sum of $\angle BAC$ and $\angle ACD$ change when the rods $AB$ and $AC$ move? If yes, explain. If not, find $\angle BAC + \angle ACD$.
    Sum constant? 检查; $\angle BAC + \angle ACD = $ $°$ 检查
    💡 思路

    按 PDF 答案：No, sum does not change. $\angle BAC + \angle ACD = 270°$。
    $AB$ 与地面平行；$CD$ 垂直于地面 → $CD \perp AB$（垂直关系传递）→ 在 $C$ 处 $AC$ 与 $CD$ 之间的角 + 在 $A$ 处 $AB$ 与 $AC$ 之间的角，加上 $CD$ 与 $AB$ 之间的 $90°$ 关系，总和恒为 $360° - 90° = 270°$（用同位角性质 + 直线关系推导）。

14. Q14. In the diagram, $PM$ and $NM$ represent plane mirrors. Incident ray $AB$ reflects at $B$, then at $C$, emerging as $CD$. Given that $a° = b° = c° = d°$, $AB \parallel NM$ and $CD \parallel MP$, find $x$ (angle of the V-shape at $M$).
    $x = $ $°$ 检查
    💡 思路

    按 PDF 答案：$x = 60°$。
    $a = b = c = d$ → 反射对称。$AB \parallel NM$ 和 $CD \parallel MP$ 给出两组同位角。三角形 $BCM$ 内角和 $= 180°$；$\angle MBC = 2a$, $\angle MCB = 2c = 2a$, $\angle BMC = x$ → $4a + x = 180°$。结合反射规律（入射角 = 反射角）解得 $x = 60°$, $a = 30°$。

15. Q15. The diagram is not drawn to scale. Hexagon $ABCDEF$ with arrows indicating $AB \parallel DC$, $BC \parallel ED$, $FA \parallel CB$（or similar parallels per PDF）. $\angle FAB = x°$, $\angle CDE = y°$.
    (a) Show $\angle FAB = \angle CDE$. Construct: 检查
    (b) Given also $\angle EFA = \angle BCD$, show $FE \parallel BC$. Construct: 检查
    💡 思路

    按 PDF 答案：(a) Connect $AD$。然后用 $AD$ 作为截线分两次应用平行线性质，得到 $\angle FAB = \angle CDE$。
    (b) Connect $FC$。同样把 $FC$ 当截线，配合 $\angle EFA = \angle BCD$ 推出 $FE \parallel BC$。