PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 21 marks不可使用计算器 · 简答题 · 21 分
Section A · Short ResponseA 节 · 简答题
Give exact values where possible (use surds, not decimal approximations). Quote the rule you are applying ("sine rule", "cosine rule", "SOHCAHTOA") before each substitution. No calculator permitted.能给精确值就给精确值（用根号，不用十进制近似）。每次代入前先注明所用定理（"正弦定理"、"余弦定理"、"SOHCAHTOA"）。不可使用计算器。
In right-angled triangle $PQR$ the right angle is at $R$. The hypotenuse $PQ = 13$ and the side $QR$ opposite angle $P$ has length $5$.直角三角形 $PQR$ 中直角在 $R$。斜边 $PQ = 13$，与角 $P$ 相对的边 $QR$ 长为 $5$。
(a)Find the length $PR$ in exact form.求 $PR$ 的精确长度。[2]
(b)Write down the exact values of $\sin P$, $\cos P$, and $\tan P$.写出 $\sin P$、$\cos P$、$\tan P$ 的精确值。[2]
Q2MEDIUMPaper 1AC2.2 Sine Rule (AAS)[5 marks]
In triangle $ABC$, $A = 40^{\circ}$, $B = 60^{\circ}$, and the side $a$ (opposite $A$) has length $8$.三角形 $ABC$ 中 $A = 40^{\circ}$、$B = 60^{\circ}$，对边 $a$（与 $A$ 相对）长为 $8$。
(a)State the size of angle $C$.写出 $C$ 的度数。[1]
(b)Use the sine rule to express $b$ exactly in the form $\dfrac{k \sin 60^{\circ}}{\sin 40^{\circ}}$ for an integer $k$.用正弦定理把 $b$ 写成精确形式 $\dfrac{k \sin 60^{\circ}}{\sin 40^{\circ}}$，其中 $k$ 为整数。[2]
(c)Show that the area of triangle $ABC$ may be written $\dfrac{32 \sin 60^{\circ} \sin 80^{\circ}}{\sin 40^{\circ}}$. (Do not evaluate.)证明三角形 $ABC$ 的面积可写成 $\dfrac{32 \sin 60^{\circ} \sin 80^{\circ}}{\sin 40^{\circ}}$。（无需求值。）[2]
Q3MEDIUMPaper 1AC2.3 Cosine Rule (SAS)[6 marks]
In triangle $ABC$, $b = 5$, $c = 7$, and the included angle $A = 60^{\circ}$.三角形 $ABC$ 中 $b = 5$、$c = 7$，夹角 $A = 60^{\circ}$。
(a)Use the cosine rule to find the exact length of $a$.用余弦定理求 $a$ 的精确长度。[3]
(b)Find the exact area of triangle $ABC$.求三角形 $ABC$ 的精确面积。[2]
(c)Hence write down the value of $bc \sin A$ in surd form.由此用根号形式写出 $bc \sin A$ 的值。[1]
(a)Use the sine rule to show that $\sin B = \dfrac{5}{8}$.用正弦定理证明 $\sin B = \dfrac{5}{8}$。[2]
(b)Find the two possible values of $B$, giving each to the nearest tenth of a degree by exact identification of the acute and obtuse candidates. (You may write $B_{1} = \arcsin\tfrac{5}{8}$ and $B_{2} = 180^{\circ} - B_{1}$.)求 $B$ 的两个可能值，分别精确给出锐角与钝角候选（可写 $B_{1} = \arcsin\tfrac{5}{8}$，$B_{2} = 180^{\circ} - B_{1}$）。[2]
(c)Verify that both values of $B$ give a valid triangle (that is, $A + B < 180^{\circ}$ in each case), and state the two corresponding values of $C$.验证两个 $B$ 值均给出有效三角形（即每种情况下 $A + B < 180^{\circ}$），并写出对应的两个 $C$ 值。[2]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分
Section B · Extended ResponseB 节 · 长答题
A single multi-part problem that strings together the area formula, the cosine rule, and the derived quantities (semiperimeter, inradius, circumradius). Quote each rule before applying it. Leave irrational answers in surd form.一道把面积公式、余弦定理与衍生量（半周长、内切圆半径、外接圆半径）串联的多步题。每用一条定理前先注明。无理数答案保留根号形式。
In triangle $ABC$, $a = 5$, $b = 8$, and the included angle $C = 60^{\circ}$.三角形 $ABC$ 中 $a = 5$、$b = 8$，夹角 $C = 60^{\circ}$。
(a)Find the exact area of triangle $ABC$ using the formula $\tfrac{1}{2} a b \sin C$.用公式 $\tfrac{1}{2} a b \sin C$ 求三角形 $ABC$ 的精确面积。[2]
(b)Use the cosine rule to find the exact length of $c$.用余弦定理求 $c$ 的精确长度。[3]
(c)State the perimeter of the triangle, and hence write down the semiperimeter $s$.写出三角形的周长，并由此写出半周长 $s$。[1]
(d)The inradius of a triangle satisfies $r = \dfrac{\text{Area}}{s}$. Find the exact inradius $r$.三角形内切圆半径满足 $r = \dfrac{\text{Area}}{s}$。求精确内切圆半径 $r$。[2]
(e)The circumradius satisfies $R = \dfrac{a b c}{4 \cdot \text{Area}}$. Find the exact circumradius $R$, rationalising any denominator.外接圆半径满足 $R = \dfrac{a b c}{4 \cdot \text{Area}}$。求精确外接圆半径 $R$，并对分母进行有理化。[3]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 18 marks可使用计算器 · 混合题型 · 18 分
Paper 2 · Calculator Permitted第二卷 · 允许使用计算器
A graphing calculator is required. Give angles to one decimal place, lengths to three significant figures unless an exact answer is requested. For bearings problems, draw a clearly labelled diagram with north arrows at each station before applying any rule.需要图形计算器（GDC）。角度精确到一位小数，长度精确到三位有效数字，除非要求精确答案。方位角题在套用任何定理前，先在每个测站画出标注清楚的"北"指向草图。
A ship leaves harbour $A$ and sails on a bearing of $060^{\circ}$ for $100$ km to reach buoy $B$. From $B$ it then sails on a bearing of $150^{\circ}$ for $80$ km to reach island $C$.一艘船从港口 $A$ 沿方位角 $060^{\circ}$ 航行 $100$ km 到达浮标 $B$；从 $B$ 沿方位角 $150^{\circ}$ 再航行 $80$ km 抵达岛屿 $C$。
(a)Show that the interior angle $\widehat{ABC}$ of triangle $ABC$ equals $90^{\circ}$.证明三角形 $ABC$ 的内角 $\widehat{ABC}$ 等于 $90^{\circ}$。[3]
(b)Find the distance $AC$ in exact form, and give a decimal value to three significant figures.求 $AC$ 的精确距离，并给出三位有效数字的十进制值。[2]
(c)Find the bearing of $C$ from $A$, giving the answer as a three-digit bearing to one decimal place.求 $C$ 相对 $A$ 的方位角，给出三位数字、一位小数。[3]
Q7HARDPaper 2C2.6 3D Trigonometry (Square Pyramid)[10 marks]
A right pyramid has a square base $PQRS$ with side length $6$ cm. The apex $T$ lies directly above the centre $O$ of the base. Each slant edge $TP$, $TQ$, $TR$, $TS$ has length $10$ cm.一座正四棱锥的底面 $PQRS$ 是边长 $6$ cm 的正方形。顶点 $T$ 在底面中心 $O$ 的正上方。各斜棱 $TP$、$TQ$、$TR$、$TS$ 长为 $10$ cm。
(a)Show that the half-diagonal $OP$ of the base equals $3\sqrt{2}$ cm, and hence find the exact height $TO$.证明底面对角线之半 $OP = 3\sqrt{2}$ cm，并由此求精确高 $TO$。[3]
(b)Find the angle that the slant edge $TP$ makes with the base, giving the answer to one decimal place.求斜棱 $TP$ 与底面所成角，答案精确到一位小数。[2]
(c)In triangle $TPQ$ (a face of the pyramid), use the cosine rule to find the apex angle $\widehat{PTQ}$ between two adjacent slant edges, to one decimal place.在三角形 $TPQ$（棱锥的一面）中，用余弦定理求两相邻斜棱之间的顶角 $\widehat{PTQ}$，精确到一位小数。[3]
(d)Let $M$ be the midpoint of $PQ$. Find the dihedral angle between the face $TPQ$ and the base $PQRS$ (that is, the angle $\widehat{TMO}$), to one decimal place.设 $M$ 为 $PQ$ 的中点。求面 $TPQ$ 与底面 $PQRS$ 之间的二面角（即角 $\widehat{TMO}$），精确到一位小数。[2]