IB Syllabus Alignment

Standard Level Topic 3.5: Definition of $\cos\theta$ and $\sin\theta$ in terms of the unit circle; definition of $\tan\theta = \frac{\sin\theta}{\cos\theta}$. Exact values of trigonometric ratios of $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$ and their multiples across all four quadrants. The sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$, including the ambiguous case (SSA condition) and testing supplementary angles.

1. The Unit Circle & ASTC Quadrant Rules

In right-angled triangle trigonometry, angles are restricted between $0^\circ$ and $90^\circ$. The unit circle (a circle of radius $r = 1$ centered at the origin $x^2 + y^2 = 1$) extends trigonometric functions to all real angles $\theta \in \mathbb{R}$.

  • Coordinate Definitions: For any point $P(x, y)$ on the unit circle rotated by counterclockwise angle $\theta$ from the positive $x$-axis: $$x = \cos\theta, \qquad y = \sin\theta, \qquad \tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta} \quad (x \neq 0)$$
  • The Pythagorean Identity: Because $x^2 + y^2 = 1$, substituting the coordinate definitions immediately proves: $$\cos^2\theta + \sin^2\theta = 1$$
  • The ASTC Quadrant Rule: The signs of trigonometric ratios in the four quadrants are dictated by the signs of the $x$ and $y$ coordinates:
    • Quadrant I ($0 < \theta < \frac{\pi}{2}$): $x > 0, y > 0 \implies$ All ratios are positive ($\sin, \cos, \tan > 0$).
    • Quadrant II ($\frac{\pi}{2} < \theta < \pi$): $x < 0, y > 0 \implies$ Only Sine is positive ($\sin > 0; \cos, \tan < 0$).
    • Quadrant III ($\pi < \theta < \frac{3\pi}{2}$): $x < 0, y < 0 \implies$ Only Tangent is positive ($\tan > 0; \sin, \cos < 0$).
    • Quadrant IV ($\frac{3\pi}{2} < \theta < 2\pi$): $x > 0, y < 0 \implies$ Only Cosine is positive ($\cos > 0; \sin, \tan < 0$).
  • Reference Angles ($\alpha$): The acute angle made with the horizontal $x$-axis: $$\text{QI: } \alpha = \theta, \quad \text{QII: } \alpha = \pi - \theta, \quad \text{QIII: } \alpha = \theta - \pi, \quad \text{QIV: } \alpha = 2\pi - \theta$$
Worked Example 1.1 — Finding Exact Ratios Across Quadrants

Given that $\sin\theta = -\frac{3}{5}$ and $\pi < \theta < \frac{3\pi}{2}$, find the exact values of $\cos\theta$ and $\tan\theta$ without using a calculator.

  1. Step 1: Identify quadrant: The interval $\pi < \theta < \frac{3\pi}{2}$ is Quadrant III. In Quadrant III, $\cos\theta < 0$ and $\tan\theta > 0$.
  2. Step 2: Apply the Pythagorean identity: $$\cos^2\theta = 1 - \sin^2\theta = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$$ Since $\theta$ is in Quadrant III, $\cos\theta$ is strictly negative: $$\cos\theta = -\sqrt{\frac{16}{25}} = -\frac{4}{5}$$
  3. Step 3: Calculate $\tan\theta$: $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-3/5}{-4/5} = \frac{3}{4}$$
Answers: $\cos\theta = -\frac{4}{5}$, $\tan\theta = \frac{3}{4}$.

2. Exact Special Values Table

In Paper 1 (Non-Calculator), you are required to know exact radical values for standard angles derived from the $45^\circ$-$45^\circ$-$90^\circ$ and $30^\circ$-$60^\circ$-$90^\circ$ special triangles.

Angle $\theta$ (rad) Angle $\theta$ (deg) $\sin\theta$ $\cos\theta$ $\tan\theta$
$0$$0^\circ$$0$$1$$0$
$\frac{\pi}{6}$$30^\circ$$\frac{1}{2}$$\frac{\sqrt{3}}{2}$$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
$\frac{\pi}{4}$$45^\circ$$\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$1$
$\frac{\pi}{3}$$60^\circ$$\frac{\sqrt{3}}{2}$$\frac{1}{2}$$\sqrt{3}$
$\frac{\pi}{2}$$90^\circ$$1$$0$$\text{Undefined}$
$\pi$$180^\circ$$0$$-1$$0$
$\frac{3\pi}{2}$$270^\circ$$-1$$0$$\text{Undefined}$
$2\pi$$360^\circ$$0$$1$$0$

Symmetry Relations: $$\sin(\pi - \theta) = \sin\theta, \qquad \cos(\pi - \theta) = -\cos\theta, \qquad \tan(\pi - \theta) = -\tan\theta$$ $$\sin(-\theta) = -\sin\theta, \qquad \cos(-\theta) = \cos\theta, \qquad \tan(-\theta) = -\tan\theta$$

Worked Example 1.2 — Non-Calculator Evaluation in Higher Quadrants

Find the exact value of each expression without a calculator:
(a) $\cos\left(\frac{5\pi}{6}\right)$
(b) $\sin\left(\frac{4\pi}{3}\right)$
(c) $\tan\left(\frac{7\pi}{4}\right)$

  1. Step 1: Evaluate $\cos\left(\frac{5\pi}{6}\right)$ in Quadrant II: Reference angle $\alpha = \pi - \frac{5\pi}{6} = \frac{\pi}{6}$.
    In QII, cosine is negative: $$\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$
  2. Step 2: Evaluate $\sin\left(\frac{4\pi}{3}\right)$ in Quadrant III: Reference angle $\alpha = \frac{4\pi}{3} - \pi = \frac{\pi}{3}$.
    In QIII, sine is negative: $$\sin\left(\frac{4\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$
  3. Step 3: Evaluate $\tan\left(\frac{7\pi}{4}\right)$ in Quadrant IV: Reference angle $\alpha = 2\pi - \frac{7\pi}{4} = \frac{\pi}{4}$.
    In QIV, tangent is negative: $$\tan\left(\frac{7\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1$$
Answers: (a) $-\frac{\sqrt{3}}{2}$; (b) $-\frac{\sqrt{3}}{2}$; (c) $-1$.

The Ambiguous Case of the Sine Rule (SSA)

When given two sides and a non-included acute angle (Side-Side-Angle, SSA), the information does not always specify a unique triangle. Depending on the length of side $a$ compared to the altitude $h = b\sin A$, zero, one, or two distinct triangles can be formed.

Baseline Side b h = b sin A Side a Side a A C B₂ (Obtuse) B₁ (Acute) A B₁ B₂=180°-B₁ Condition for 2 Triangles h < a < b (b sin A < a < b) Supplementary Angles sin(B₁) = sin(180° - B₁)
Figure 3.2 — The Ambiguous Case of the Sine Rule: Side $a$ swings from vertex $C$ and intersects the baseline twice at $B_1$ (acute triangle $\triangle AB_1C$) and $B_2$ (obtuse triangle $\triangle AB_2C$), with $B_2 = 180^\circ - B_1$.

3. Resolving the Ambiguous Case Step-by-Step

In the Sine Rule $\frac{\sin B}{b} = \frac{\sin A}{a}$, solving for $\sin B$ yields $\sin B = \frac{b\sin A}{a}$. Because $\sin\theta = \sin(180^\circ - \theta)$, there are two potential candidates for angle $B$ in the range $(0^\circ, 180^\circ)$.

  1. Calculate the altitude: $h = b\sin A$.
  2. Examine the ratio:
    • If $a < h$: $\sin B > 1$. No triangle exists ($0$ solutions). The side is too short to reach the baseline.
    • If $a = h$: $\sin B = 1 \implies B = 90^\circ$. Exactly 1 right-angled triangle.
    • If $a \ge b$: Exactly 1 triangle. Because $a \ge b$, angle $A \ge B$. Since $A$ is already an angle in the triangle, $B$ cannot be obtuse ($B$ must be acute).
    • If $h < a < b$: Exactly 2 distinct triangles exist!
      • Triangle 1 (Acute): $B_1 = \arcsin\left(\frac{b\sin A}{a}\right)$.
      • Triangle 2 (Obtuse): $B_2 = 180^\circ - B_1$ (or $\pi - B_1$).
      • Validation Check: Check that $A + B_2 < 180^\circ$, ensuring the third angle $C_2 = 180^\circ - (A + B_2) > 0$.
Worked Example 1.3 — Complete Solution of an Ambiguous Triangle

In $\triangle ABC$, angle $A = 30^\circ$, side $b = 10\text{ cm}$, and side $a = 7\text{ cm}$.
(a) Show that two distinct triangles can be formed.
(b) Find the two possible values of angle $B$, giving answers to 1 decimal place.
(c) For each triangle, find the corresponding length of side $c$.

  1. Step 1: Check conditions for ambiguity: Altitude $h = b\sin A = 10\sin(30^\circ) = 10 \times 0.5 = 5\text{ cm}$.
    Since $h < a < b$ ($5 < 7 < 10$), exactly two triangles exist.
  2. Step 2: Solve for angle $B$: $$\frac{\sin B}{b} = \frac{\sin A}{a} \implies \frac{\sin B}{10} = \frac{\sin 30^\circ}{7} \implies \sin B = \frac{10 \times 0.5}{7} = \frac{5}{7} \approx 0.7143$$ Acute candidate $B_1$: $$B_1 = \arcsin\left(\frac{5}{7}\right) \approx 45.58^\circ \approx 45.6^\circ$$ Obtuse candidate $B_2$: $$B_2 = 180^\circ - 45.58^\circ \approx 134.42^\circ \approx 134.4^\circ$$ Check validity: $A + B_2 = 30^\circ + 134.4^\circ = 164.4^\circ < 180^\circ$. Valid!
  3. Step 3: Calculate side $c$ for both cases: Case 1 ($B_1 = 45.58^\circ$): $$C_1 = 180^\circ - (30^\circ + 45.58^\circ) = 104.42^\circ$$ $$\frac{c_1}{\sin 104.42^\circ} = \frac{7}{\sin 30^\circ} \implies c_1 = 14\sin(104.42^\circ) \approx 13.56 \approx 13.6\text{ cm}$$ Case 2 ($B_2 = 134.42^\circ$): $$C_2 = 180^\circ - (30^\circ + 134.42^\circ) = 15.58^\circ$$ $$\frac{c_2}{\sin 15.58^\circ} = \frac{7}{\sin 30^\circ} \implies c_2 = 14\sin(15.58^\circ) \approx 3.76 \approx 3.76\text{ cm}$$
Answers: (a) $h = 5 < 7 < 10$, confirming two triangles.
(b) $B = 45.6^\circ$ or $B = 134.4^\circ$.
(c) $c_1 \approx 13.6\text{ cm}$, $c_2 \approx 3.76\text{ cm}$.

4. Ambiguous Case & Exact Value Applications

Worked Example 1.4 — Solving the Ambiguous Case (Two Triangles)

In $\triangle ABC$, side $a = 8\text{ cm}$, side $b = 10\text{ cm}$, and angle $A = 35^\circ$.
(a) Show that two distinct triangles can be formed with these dimensions. [3 marks]
(b) Find the two possible values of angle $B$, giving answers to 1 decimal place. [3 marks]
(c) Calculate the two possible areas of $\triangle ABC$. [4 marks]

  1. Part (a): Altitude comparison test: $$\text{Altitude } h = b\sin A = 10\sin(35^\circ) \approx 5.736\text{ cm} \quad [M1]$$

    Since $h < a < b$ ($5.736 < 8 < 10$), the side $a$ can pivot to meet the base line at two separate points, forming exactly two non-congruent triangles [R1][A1].

  2. Part (b): Compute the two values of angle $B$: $$\frac{\sin B}{b} = \frac{\sin A}{a} \implies \sin B = \frac{10\sin(35^\circ)}{8} \approx 0.71697 \quad [M1]$$ $$\text{Acute solution: } B_1 = \arcsin(0.71697) \approx 45.8^\circ \quad [A1]$$ $$\text{Obtuse solution: } B_2 = 180^\circ - 45.8^\circ = 134.2^\circ \quad [A1]$$ Check: $A + B_2 = 35^\circ + 134.2^\circ = 169.2^\circ < 180^\circ$, so $B_2$ is geometrically valid.
  3. Part (c): Compute third angles and areas:

    Case 1 ($B_1 = 45.8^\circ$):

    $$C_1 = 180^\circ - (35^\circ + 45.8^\circ) = 99.2^\circ \quad [M1]$$ $$\text{Area}_1 = \frac{1}{2}ab\sin C_1 = \frac{1}{2}(8)(10)\sin(99.2^\circ) = 40\sin(99.2^\circ) \approx 39.5\text{ cm}^2 \quad [A1]$$

    Case 2 ($B_2 = 134.2^\circ$):

    $$C_2 = 180^\circ - (35^\circ + 134.2^\circ) = 10.8^\circ \quad [M1]$$ $$\text{Area}_2 = \frac{1}{2}ab\sin C_2 = 40\sin(10.8^\circ) \approx 7.49\text{ cm}^2 \quad [A1]$$
Answers: (a) $h \approx 5.74 < a < b$; (b) $B = 45.8^\circ$ or $134.2^\circ$; (c) $\text{Area} = 39.5\text{ cm}^2$ or $7.49\text{ cm}^2$.
Worked Example 1.5 — Exact Values in the Second Quadrant

An angle $\theta$ satisfies $\sin\theta = \frac{5}{13}$, where $\frac{\pi}{2} < \theta < \pi$.
Without using a calculator, find the exact value of:
(a) $\cos\theta$ [2 marks]
(b) $\tan\theta$ [1 mark]
(c) $\sin(2\theta)$ [3 marks]

  1. Part (a): Exact cosine in Quadrant II: $$\cos^2\theta = 1 - \sin^2\theta = 1 - \left(\frac{5}{13} \right)^2 = 1 - \frac{25}{169} = \frac{144}{169} \quad [M1]$$ Since $\frac{\pi}{2} < \theta < \pi$ (Quadrant II), $\cos\theta$ must be negative [R1]: $$\cos\theta = -\sqrt{\frac{144}{169}} = -\frac{12}{13} \quad [A1]$$
  2. Part (b): Exact tangent: $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{5/13}{-12/13} = -\frac{5}{12} \quad [A1]$$
  3. Part (c): Double angle $\sin(2\theta)$: $$\sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{5}{13} \right)\left(-\frac{12}{13} \right) = -\frac{120}{169} \quad [M1][A1]$$
Answers: (a) $\cos\theta = -\frac{12}{13}$; (b) $\tan\theta = -\frac{5}{12}$; (c) $\sin(2\theta) = -\frac{120}{169}$.

10 Tiered Practice Exercises

Progress from fundamental non-calculator exact evaluations in Tier 1 to full ambiguous case analyses in Tier 2, concluding with algebraic radical proofs in Tier 3.

Tier 1: Core Fluency (Problems 1–4)

Exercise 1 Non-Calculator Exact Values

Find the exact value of each of the following without using a calculator:
(a) $\sin\left(\frac{2\pi}{3}\right)$
(b) $\cos\left(\frac{7\pi}{4}\right)$
(c) $\tan\left(\frac{5\pi}{6}\right)$

View Full Solution

(a) $\sin\left(\frac{2\pi}{3}\right)$:
In Quadrant II, reference angle is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$.
Sine is positive in QII: $\sin\left(\frac{2\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

(b) $\cos\left(\frac{7\pi}{4}\right)$:
In Quadrant IV, reference angle is $2\pi - \frac{7\pi}{4} = \frac{\pi}{4}$.
Cosine is positive in QIV: $\cos\left(\frac{7\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ (or $\frac{1}{\sqrt{2}}$).

(c) $\tan\left(\frac{5\pi}{6}\right)$:
In Quadrant II, reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.
Tangent is negative in QII: $\tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{3}$ (or $-\frac{1}{\sqrt{3}}$).

Final Answers: (a) $\frac{\sqrt{3}}{2}$; (b) $\frac{\sqrt{2}}{2}$; (c) $-\frac{\sqrt{3}}{3}$.

Exercise 2 Solving Elementary Exact Equations

Find all exact solutions to the equation $\sin\theta = -\frac{1}{2}$ in the interval $0 \le \theta \le 2\pi$.

View Full Solution

Step 1: Identify reference angle:
$\sin\alpha = \frac{1}{2} \implies \alpha = \frac{\pi}{6}$.

Step 2: Locate quadrants:
$\sin\theta < 0$ in Quadrant III and Quadrant IV.

Step 3: Compute angles:
Quadrant III: $\theta_1 = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
Quadrant IV: $\theta_2 = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$.

Final Answer: $\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$.

Exercise 3 Testing Triangle Existence (SSA)

Determine whether $0$, $1$, or $2$ triangles can be formed with each given set of measurements:
(a) $A = 30^\circ$, $b = 12$, $a = 5$
(b) $A = 30^\circ$, $b = 12$, $a = 6$
(c) $A = 30^\circ$, $b = 12$, $a = 15$

View Full Solution

Calculate altitude $h = b\sin A = 12\sin 30^\circ = 12(0.5) = 6$.

(a) $a = 5$: Since $a < h$ ($5 < 6$), side $a$ cannot reach the base. 0 triangles.

(b) $a = 6$: Since $a = h = 6$, exactly one right-angled triangle is formed. 1 triangle.

(c) $a = 15$: Since $a \ge b$ ($15 \ge 12$), only one triangle can be formed because $A \ge B$. 1 triangle.

Final Answers: (a) $0$ triangles; (b) $1$ triangle; (c) $1$ triangle.

Exercise 4 Pythagorean Unit Circle Ratios

Let $\cos\theta = -\frac{5}{13}$ where $\frac{\pi}{2} < \theta < \pi$.
(a) State the exact value of $\sin\theta$.
(b) State the exact value of $\tan\theta$.
(c) Evaluate $\sin(2\pi - \theta)$.

View Full Solution

(a) $\sin\theta$:
$\theta$ is in Quadrant II, so $\sin\theta > 0$. $$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$

(b) $\tan\theta$:
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{12/13}{-5/13} = -\frac{12}{5}$$

(c) $\sin(2\pi - \theta)$:
By periodicity and symmetry, $\sin(2\pi - \theta) = \sin(-\theta) = -\sin\theta = -\frac{12}{13}$.

Final Answers: (a) $\frac{12}{13}$; (b) $-\frac{12}{5}$; (c) $-\frac{12}{13}$.

Tier 2: Exam Modeling (Problems 5–8)

Exercise 5 Two-Triangle Sine Rule Calculation

In triangle $PQR$, angle $P = 40^\circ$, side $p = 9\text{ cm}$, and side $q = 11\text{ cm}$.
(a) Find the two possible values of angle $Q$, correct to the nearest $0.1^\circ$.
(b) For each possible value of $Q$, calculate the third angle $R$.

View Full Solution

(a) Find possible values of $Q$:
$$\frac{\sin Q}{q} = \frac{\sin P}{p} \implies \frac{\sin Q}{11} = \frac{\sin 40^\circ}{9} \implies \sin Q = \frac{11\sin 40^\circ}{9}$$ $$\sin Q \approx \frac{11(0.642788)}{9} \approx 0.78563$$ Acute candidate: $$Q_1 = \arcsin(0.78563) \approx 51.78^\circ \approx 51.8^\circ$$ Obtuse candidate: $$Q_2 = 180^\circ - 51.78^\circ \approx 128.22^\circ \approx 128.2^\circ$$ Validation: $P + Q_2 = 40^\circ + 128.2^\circ = 168.2^\circ < 180^\circ$. Both are valid.

(b) Calculate angle $R$:
Case 1: $R_1 = 180^\circ - (40^\circ + 51.78^\circ) = 88.22^\circ \approx 88.2^\circ$.
Case 2: $R_2 = 180^\circ - (40^\circ + 128.22^\circ) = 11.78^\circ \approx 11.8^\circ$.

Final Answers: (a) $Q = 51.8^\circ$ or $Q = 128.2^\circ$; (b) $R = 88.2^\circ$ or $R = 11.8^\circ$.

Exercise 6 Area Difference in Ambiguous Triangles

Referring to triangle $PQR$ in Exercise 5 ($P = 40^\circ, p = 9\text{ cm}, q = 11\text{ cm}$):
(a) Calculate the area of the acute triangle ($\triangle P Q_1 R_1$).
(b) Calculate the area of the obtuse triangle ($\triangle P Q_2 R_2$).
(c) Find the difference between the two areas, correct to 3 significant figures.

View Full Solution

(a) Area of acute triangle:
$$\text{Area}_1 = \frac{1}{2} p q \sin R_1 = \frac{1}{2}(9)(11)\sin(88.22^\circ) \approx 49.5(0.9995) \approx 49.48\text{ cm}^2$$

(b) Area of obtuse triangle:
$$\text{Area}_2 = \frac{1}{2} p q \sin R_2 = \frac{1}{2}(9)(11)\sin(11.78^\circ) \approx 49.5(0.2041) \approx 10.10\text{ cm}^2$$

(c) Area difference:
$$\Delta\text{Area} = 49.48 - 10.10 = 39.38 \approx 39.4\text{ cm}^2$$

Final Answers: (a) $49.5\text{ cm}^2$; (b) $10.1\text{ cm}^2$; (c) $39.4\text{ cm}^2$.

Exercise 7 Rejection of Ambiguous Angle

In triangle $XYZ$, angle $X = 75^\circ$, side $x = 12\text{ cm}$, and side $y = 10\text{ cm}$.
(a) Use the sine rule to show that $\sin Y \approx 0.805$.
(b) Explain why there is only ONE valid triangle in this scenario, despite $\sin Y$ having two supplementary solutions.

View Full Solution

(a) Calculate $\sin Y$:
$$\frac{\sin Y}{10} = \frac{\sin 75^\circ}{12} \implies \sin Y = \frac{10\sin 75^\circ}{12} \approx \frac{10(0.9659)}{12} \approx 0.8049 \approx 0.805$$

(b) Why only one triangle is valid:
The acute candidate is $Y_1 = \arcsin(0.8049) \approx 53.6^\circ$.
The obtuse candidate is $Y_2 = 180^\circ - 53.6^\circ = 126.4^\circ$.
Testing the angle sum for $Y_2$: $$X + Y_2 = 75^\circ + 126.4^\circ = 201.4^\circ > 180^\circ$$ Since the sum of angles in a flat triangle cannot exceed $180^\circ$, $Y_2$ is geometrically impossible.
Geometric reason: Because $x = 12 > y = 10$, angle $X$ must be strictly greater than angle $Y$. Since $X = 75^\circ$, $Y$ must be smaller than $75^\circ$, ruling out any obtuse angle.

Final Answers: (a) $\sin Y \approx 0.805$; (b) $X + Y_2 = 201.4^\circ > 180^\circ$, so the obtuse angle is rejected.

Exercise 8 Surveying Bearing Ambiguity

A lighthouse $L$ is located due North of port $A$. A ship $S$ is at a distance of $8\text{ km}$ from port $A$ on a bearing of $035^\circ$. The distance from the ship $S$ to the lighthouse $L$ is $6\text{ km}$.
(a) Set up a sine rule equation to find angle $A\hat{L}S$.
(b) Determine the two possible distances from port $A$ to the lighthouse $L$.

View Full Solution

(a) Sine rule equation:
In $\triangle ALS$: angle at $A$ between North and $035^\circ$ is $35^\circ$. Opposite side $LS = 6\text{ km}$. Side $AS = 8\text{ km}$. $$\frac{\sin(\angle ALS)}{8} = \frac{\sin 35^\circ}{6} \implies \sin(\angle ALS) = \frac{8\sin 35^\circ}{6} \approx 0.7648$$

(b) Two possible distances $AL$:
$$\angle ALS_1 = \arcsin(0.7648) \approx 49.89^\circ \approx 49.9^\circ$$ $$\angle ALS_2 = 180^\circ - 49.89^\circ = 130.11^\circ \approx 130.1^\circ$$ Third angle $\angle ASL$:
Case 1: $180^\circ - (35^\circ + 49.89^\circ) = 95.11^\circ$.
$$AL_1 = \frac{6\sin(95.11^\circ)}{\sin 35^\circ} \approx \frac{6(0.9960)}{0.5736} \approx 10.42 \approx 10.4\text{ km}$$ Case 2: $180^\circ - (35^\circ + 130.11^\circ) = 14.89^\circ$.
$$AL_2 = \frac{6\sin(14.89^\circ)}{\sin 35^\circ} \approx \frac{6(0.2570)}{0.5736} \approx 2.688 \approx 2.69\text{ km}$$

Final Answers: (a) $\sin(\angle ALS) = \frac{4\sin 35^\circ}{3}$; (b) $AL \approx 10.4\text{ km}$ or $AL \approx 2.69\text{ km}$.

Tier 3: Synthesis & Problem-Solving (Problems 9–10)

Exercise 9 Exact Radical Sine Rule Proof

In triangle $ABC$, angle $A = 45^\circ$, angle $B = 30^\circ$, and side $a = 6\sqrt{2}$.
(a) Find the exact value of side $b$.
(b) Use the sine sum formula $\sin(75^\circ) = \sin(45^\circ + 30^\circ)$ to show that $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$.
(c) Hence find the exact length of side $c$ in the form $p + q\sqrt{3}$, where $p, q \in \mathbb{Z}$.

View Full Solution

(a) Find side $b$:
$$\frac{b}{\sin 30^\circ} = \frac{a}{\sin 45^\circ} \implies \frac{b}{1/2} = \frac{6\sqrt{2}}{\sqrt{2}/2} \implies 2b = 12 \implies b = 6$$

(b) Show $\sin 75^\circ$:
Angle $C = 180^\circ - (45^\circ + 30^\circ) = 105^\circ$. Note $\sin 105^\circ = \sin(180^\circ - 75^\circ) = \sin 75^\circ$. $$\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ$$ $$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$$

(c) Find exact length of side $c$:
$$\frac{c}{\sin 105^\circ} = \frac{a}{\sin 45^\circ} = \frac{6\sqrt{2}}{\sqrt{2}/2} = 12$$ $$c = 12 \sin 105^\circ = 12 \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = 3(\sqrt{6} + \sqrt{2})$$ Let's write this in exact radical form: $$c = 3\sqrt{2}(\sqrt{3} + 1) = 3\sqrt{2} + 3\sqrt{6}$$ If normalized with standard base: $c = 3\sqrt{2} + 3\sqrt{6}$.

Final Answers: (a) $b = 6$; (b) $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$; (c) $c = 3\sqrt{6} + 3\sqrt{2}$.

Exercise 10 Parametric Ambiguity Interval

In $\triangle ABC$, angle $A = \frac{\pi}{6}$ and side $b = 16\text{ cm}$.
(a) Find the exact range of values of side length $a$ for which exactly TWO distinct triangles can be formed.
(b) State the exact geometric shape of $\triangle ABC$ when $a = 8\text{ cm}$.
(c) If $a = 10\text{ cm}$, find the exact ratio of the areas of the two triangles formed.

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(a) Range of $a$ for 2 triangles:
Altitude $h = b\sin A = 16\sin\left(\frac{\pi}{6}\right) = 16\left(\frac{1}{2}\right) = 8\text{ cm}$.
For two triangles, we require $h < a < b$: $$8 < a < 16$$

(b) Geometric shape when $a = 8\text{ cm}$:
When $a = h = 8\text{ cm}$, $\sin B = 1 \implies B = \frac{\pi}{2}$ ($90^\circ$).
$\triangle ABC$ is a right-angled triangle (a $30^\circ$-$60^\circ$-$90^\circ$ special triangle).

(c) Area ratio when $a = 10\text{ cm}$:
$$\sin B = \frac{16\sin(\pi/6)}{10} = \frac{8}{10} = 0.8$$ $$B_1 = \arcsin(0.8) \approx 0.9273\text{ rad} \approx 53.13^\circ$$ $$B_2 = \pi - 0.9273 \approx 2.2143\text{ rad} \approx 126.87^\circ$$ Third angles:
$C_1 = 180^\circ - (30^\circ + 53.13^\circ) = 96.87^\circ$.
$C_2 = 180^\circ - (30^\circ + 126.87^\circ) = 23.13^\circ$.
Ratio of areas equals ratio of $\sin C$: $$\frac{\text{Area}_1}{\text{Area}_2} = \frac{\sin C_1}{\sin C_2} = \frac{\sin(96.87^\circ)}{\sin(23.13^\circ)} \approx \frac{0.9928}{0.3928} \approx 2.53$$

Final Answers: (a) $8 < a < 16$; (b) Right-angled triangle ($B = 90^\circ$); (c) Area ratio is $\approx 2.53 : 1$.

Exact Values & Sine Rule Reference Sheet

In the IB Formula Booklet (Topic 3.5 & 3.6)

The official IB Mathematics AA SL formula booklet provides the following references:

The Sine Rule $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
The Cosine Rule $$c^2 = a^2 + b^2 - 2ab\cos C$$
Area of a Triangle $$A = \frac{1}{2}ab\sin C$$
Fundamental Identities $$\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cos^2\theta + \sin^2\theta = 1$$
Must Memorize & Ambiguity Decision Tree
  • Exact Values Table: $$\sin 30^\circ = \cos 60^\circ = \frac{1}{2}, \quad \sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}$$
  • The SSA Decision Protocol (Given $a, b, A$ with $A < 90^\circ$):
    1. Compute $h = b\sin A$.
    2. If $a < h$: $0$ triangles ($\sin B > 1$).
    3. If $a = h$: $1$ triangle ($B = 90^\circ$).
    4. If $a \ge b$: $1$ triangle ($B$ is acute).
    5. If $h < a < b$: $2$ triangles ($B_1$ acute, $B_2 = 180^\circ - B_1$ obtuse).
  • Angle Sum Verification: Always verify that $A + B_{\text{obtuse}} < 180^\circ$. If this condition fails, discard the obtuse candidate immediately.
Critical IB Examination Pitfalls
  • Pitfall 1: Forgetting the Second Angle on Paper 2: Calculators automatically output only the principal value $\arcsin(x) \in [-90^\circ, 90^\circ]$. If you don't actively calculate $180^\circ - B$, you will forfeit half the marks on any ambiguous case question.
  • Pitfall 2: Premature Rounding: Rounding $\sin B$ to $0.78$ instead of storing the full calculator precision before applying inverse sine leads to inaccurate angle and side values that fail IB $3\text{ s.f.}$ checks.
  • Pitfall 3: Quadrant Signs: Forgetting that cosine is negative in Quadrant II: $\cos(120^\circ) = -\frac{1}{2}$, NOT $+\frac{1}{2}$.
  • Pitfall 4: Misidentifying Opposite Pairs: Pairing angle $A$ with side $b$ in the sine rule. The ratio must strictly compare a side to the sine of its opposite angle: $\frac{a}{\sin A} = \frac{b}{\sin B}$.

Section A: Diagnostic Multiple Choice

Answer the questions to check your understanding and score.

Score: 0 / 4

Question 1: What is the exact value of $\sin\left(\frac{5\pi}{6}\right)$?

Question 2: In $\triangle ABC$, $A = 30^\circ$, $b = 14\text{ cm}$, and $a = 6\text{ cm}$. How many distinct triangles can be formed?

Question 3: In $\triangle ABC$, $A = 70^\circ$, $a = 15$, and $\sin B = \frac{\sqrt{2}}{2}$. What are the geometrically possible values of angle $B$?

Question 4: What is the exact value of $\tan\left(\frac{3\pi}{4}\right)$?

Section B: Exam-Style Free-Response Questions

Authentic IB style questions with detailed markschemes showing method ($M$), accuracy ($A$), and reasoning ($R$) marks.

Question 5 (Paper 1 Style) 6 Marks

The following diagram shows $\triangle ABC$ with $A = 30^\circ$, $b = 8\text{ cm}$, and $a = 4\sqrt{2}\text{ cm}$.
(a) Use the sine rule to show that $\sin B = \frac{\sqrt{2}}{2}$. [2]
(b) State the two possible exact values of angle $B$ in degrees. [2]
(c) For the obtuse angle candidate, find the exact area of the triangle. [2]

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(a) Show $\sin B = \frac{\sqrt{2}}{2}$: [2 marks]
$$\frac{\sin B}{b} = \frac{\sin A}{a} \implies \frac{\sin B}{8} = \frac{\sin 30^\circ}{4\sqrt{2}} \quad (M1)$$ $$\sin B = \frac{8 \times \frac{1}{2}}{4\sqrt{2}} = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \quad (A1)$$

(b) Two possible values of $B$: [2 marks]
Acute: $B_1 = 45^\circ$ (A1)
Obtuse: $B_2 = 180^\circ - 45^\circ = 135^\circ$ (A1)
Check: $30^\circ + 135^\circ = 165^\circ < 180^\circ$, so both are valid.

(c) Exact area for $B = 135^\circ$: [2 marks]
Third angle: $C = 180^\circ - (30^\circ + 135^\circ) = 15^\circ$.
Alternatively, use side $c$: $$\text{Area} = \frac{1}{2} a b \sin C = \frac{1}{2}(4\sqrt{2})(8)\sin(15^\circ) = 16\sqrt{2}\sin(15^\circ) \quad (M1)$$ Since $\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$: $$\text{Area} = 16\sqrt{2}\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) = 4\sqrt{2}(\sqrt{6} - \sqrt{2}) = 4\sqrt{12} - 8 = 8\sqrt{3} - 8\text{ cm}^2 \quad (A1)$$

Question 6 (Paper 2 Style) 8 Marks

A park ranger at watchtower $A$ spots smoke at point $C$ on a bearing of $040^\circ$. A second ranger at station $B$, located $12\text{ km}$ due East of tower $A$, spots the same smoke at a distance of $10\text{ km}$.
(a) Find the measure of angle $C\hat{A}B$. [1]
(b) Use the sine rule to determine the two possible values of angle $A\hat{C}B$, correct to the nearest $0.1^\circ$. [3]
(c) Hence determine the two possible distances from watchtower $A$ to the fire at $C$. [3]
(d) If ranger $A$ confirms that the distance to the fire is greater than $10\text{ km}$, state which location corresponds to the fire. [1]

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(a) Angle $C\hat{A}B$: [1 mark]
Due East is bearing $090^\circ$. Smoke is on bearing $040^\circ$. $$\angle CAB = 90^\circ - 40^\circ = 50^\circ \quad (A1)$$

(b) Two possible values of $\angle ACB$: [3 marks]
In $\triangle ABC$: $AB = 12\text{ km}$, $BC = 10\text{ km}$, $\angle CAB = 50^\circ$. $\frac{\sin(\angle ACB)}{AB} = \frac{\sin(\angle CAB)}{BC} \implies \frac{\sin(\angle ACB)}{12} = \frac{\sin 50^\circ}{10} \quad (M1)$ $\sin(\angle ACB) = \frac{12\sin 50^\circ}{10} \approx 0.9193 \quad (A1)$ $\angle ACB_1 = \arcsin(0.9193) \approx 66.8^\circ \quad (A1)$ $\angle ACB_2 = 180^\circ - 66.8^\circ = 113.2^\circ \quad (A1)$ Validation: $50^\circ + 113.2^\circ = 163.2^\circ < 180^\circ$. Both values are geometrically valid.

(c) Two possible distances $AC$: [3 marks]
Case 1 ($\angle ACB = 66.8^\circ$): $$\angle B_1 = 180^\circ - (50^\circ + 66.8^\circ) = 63.2^\circ \quad (M1)$$ $$\frac{AC_1}{\sin 63.2^\circ} = \frac{10}{\sin 50^\circ} \implies AC_1 = \frac{10\sin 63.2^\circ}{\sin 50^\circ} \approx 11.65 \approx 11.7\text{ km} \quad (A1)$$ Case 2 ($\angle ACB = 113.2^\circ$): $$\angle B_2 = 180^\circ - (50^\circ + 113.2^\circ) = 16.8^\circ$$ $$AC_2 = \frac{10\sin 16.8^\circ}{\sin 50^\circ} \approx \frac{10(0.2890)}{0.7660} \approx 3.77\text{ km} \quad (A1)$$

(d) Selection: [1 mark]
Since $AC > 10\text{ km}$, the true distance is $AC_1 \approx 11.7\text{ km}$ (corresponding to acute angle $\angle ACB \approx 66.8^\circ$). (R1)