MYP Inquiry Framework
Key Concept Relationships & Space
Related Concepts Measurement, Generalization, Models
Global Context Orientation in space and time (Satellite geodetic surveying & marine navigation)
Statement of Inquiry Generalizing trigonometric relations to non-right triangles enables triangulation across curved planetary surfaces and navigational grids.
3.1

The Sine Rule & The Ambiguous Case (SSA)

  • The Sine Rule Formula: In any planar triangle $ABC$ with opposite sides $a, b, c$: $$\mathbf{\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}} \quad \text{or} \quad \mathbf{\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}}$$
  • When to Use the Sine Rule:
    • Given two angles and any side (AAS or ASA).
    • Given two sides and a non-included angle (SSA) — beware the ambiguous case!
  • The Ambiguous Case (SSA): When given acute angle $A$ and adjacent side $b$, the critical altitude is $h = b \sin A$. Comparing opposite side $a$ to $h$:
    • $a < h$: Side $a$ is too short to reach the opposite line $\implies$ No triangle ($0$ solutions).
    • $a = h$: Side $a$ reaches the base perpendicularly $\implies$ $1$ Right triangle.
    • $h < a < b$: Side $a$ can swing to intersect the base at two distinct points $\implies$ $2$ Distinct triangles: $$\angle B_1 = \sin^{-1}\left(\frac{b \sin A}{a}\right) \quad (\text{acute}), \qquad \angle B_2 = 180^\circ - B_1 \quad (\text{obtuse})$$
    • $a \ge b$: Only the acute triangle is possible $\implies$ $1$ Triangle.
The Ambiguous Case (SSA): Two Triangles when h < a < b A C b h = b sin A B₁ (acute) a B₂ (obtuse) a
3.2

The Cosine Rule: Sides & Angles

  • Finding an Unknown Side (given SAS): $$\mathbf{c^2 = a^2 + b^2 - 2ab \cos C}$$ $$\mathbf{a^2 = b^2 + c^2 - 2bc \cos A}, \qquad \mathbf{b^2 = a^2 + c^2 - 2ac \cos B}$$ Notice: If $C = 90^\circ$, $\cos(90^\circ) = 0$, reducing directly to Pythagoras $c^2 = a^2 + b^2$!
  • Finding an Unknown Angle (given SSS): $$\mathbf{\cos C = \frac{a^2 + b^2 - c^2}{2ab}}$$
  • No Ambiguity in the Cosine Rule: Because the inverse cosine function $\cos^{-1}(x)$ has a unique output between $0^\circ$ and $180^\circ$:
    • If $\cos C > 0$, angle $C$ is strictly acute.
    • If $\cos C < 0$, angle $C$ is strictly obtuse.
    This makes the Cosine Rule the safest choice for determining whether a triangle contains an obtuse angle.
3.3

Area of a Triangle Using the Sine Formula

  • The Trigonometric Area Formula: The area of any triangle equals half the product of two side lengths and the sine of their included angle: $$\mathbf{\text{Area} = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B}$$
  • Derivation: In $\triangle ABC$, perpendicular altitude $h = b \sin A$. Substituting into standard area $\frac{1}{2} \cdot \text{base} \cdot h = \frac{1}{2} \cdot c \cdot (b \sin A) = \frac{1}{2}bc \sin A$.
  • Area of a Parallelogram: $$\text{Area} = ab \sin \theta$$ Since a parallelogram consists of two congruent triangles sharing angle $\theta$.
3.4

Real-World Navigation & Triangulation

  • Triangulation Strategy:
    1. Always draw a clear, large diagram including North arrows at all named stations.
    2. Use parallel line angle properties (alternate interior angles, co-interior angles summing to $180^\circ$) to determine interior triangle angles.
    3. Decide between Sine Rule and Cosine Rule based on given data (SAS or SSS $\implies$ Cosine Rule; AAS or ASA $\implies$ Sine Rule).

Worked Examples (Criterion A & C)

Example 1 Resolving the Ambiguous Case of the Sine Rule

Problem: In $\triangle ABC$, $\angle A = 32^\circ$, $b = 12\text{ cm}$, and $a = 8\text{ cm}$.
(a) Show that two possible triangles can be formed.
(b) Calculate both possible values for angle $B$ and side $c$.

Step 1: Check the altitude boundary $h$ $$h = b \sin A = 12 \sin(32^\circ) \approx 12 \times 0.52992 = 6.36\text{ cm}$$

Since $h < a < b$ ($6.36 < 8 < 12$), two distinct triangles exist.

Step 2: Find the two possible values of angle $B$ $$\frac{\sin B}{b} = \frac{\sin A}{a} \implies \sin B = \frac{12 \sin(32^\circ)}{8} \approx \frac{6.359}{8} \approx 0.79488$$

Acute solution $B_1$: $B_1 = \sin^{-1}(0.79488) \approx \mathbf{52.6^\circ}$.

Obtuse solution $B_2$: $B_2 = 180^\circ - 52.6^\circ = \mathbf{127.4^\circ}$.

Step 3: Calculate corresponding angle $C$ and side $c$ for both cases

Case 1 ($B_1 = 52.6^\circ$): $C_1 = 180^\circ - (32^\circ + 52.6^\circ) = 95.4^\circ$.

$$c_1 = \frac{a \sin C_1}{\sin A} = \frac{8 \sin(95.4^\circ)}{\sin(32^\circ)} \approx \frac{8(0.99556)}{0.52992} \approx \mathbf{15.0\text{ cm}}$$

Case 2 ($B_2 = 127.4^\circ$): $C_2 = 180^\circ - (32^\circ + 127.4^\circ) = 20.6^\circ$.

$$c_2 = \frac{a \sin C_2}{\sin A} = \frac{8 \sin(20.6^\circ)}{\sin(32^\circ)} \approx \frac{8(0.35184)}{0.52992} \approx \mathbf{5.31\text{ cm}}$$ Criterion A: Complete identification of dual solutions with rigorous boundary checking
Example 2 Marine Triangulation using the Cosine Rule

Problem: Port $A$ and Port $B$ are $45\text{ km}$ apart, with $B$ on a bearing of $120^\circ$ from $A$. A yacht departs Port $A$ on a bearing of $050^\circ$ and sails $30\text{ km}$ to position $P$.
(a) Find the interior angle $\angle PAB$.
(b) Calculate the distance from the yacht at $P$ to Port $B$.
(c) Find the area of the water triangular region $\triangle PAB$.

Step 1: Calculate interior angle $\angle PAB$ $$\angle PAB = 120^\circ - 050^\circ = \mathbf{70^\circ}$$
Step 2: Apply the Cosine Rule to find distance $PB$ $$PB^2 = AP^2 + AB^2 - 2(AP)(AB) \cos(\angle PAB)$$ $$PB^2 = 30^2 + 45^2 - 2(30)(45) \cos(70^\circ) = 900 + 2025 - 2700(0.34202)$$ $$PB^2 = 2925 - 923.45 = 2001.55 \implies PB = \sqrt{2001.55} \approx \mathbf{44.7\text{ km}}$$
Step 3: Calculate the area of $\triangle PAB$ $$\text{Area} = \frac{1}{2}(AP)(AB) \sin(70^\circ) = \frac{1}{2}(30)(45) \sin(70^\circ) = 675 \times 0.93969 \approx \mathbf{634\text{ km}^2}$$ Criterion C & D: Authentic maritime bearing diagramming and multi-step trigonometric resolution

MYP Extended Practice Exercises: Sine & Cosine Rules

Tier 1 (Standard Non-Right Triangles), Tier 2 (Ambiguous Cases & Bearings), and Tier 3 (Extended Proofs & Geometric Optimization).

Question 1 Tier 1: Standard 3 Marks

In $\triangle ABC$, $a = 15\text{ cm}$, $\angle A = 48^\circ$, and $\angle B = 65^\circ$. Find the length of side $b$ to 3 significant figures.

View Full Method Markscheme
[M1] Apply Sine Rule: $\frac{b}{\sin B} = \frac{a}{\sin A} \implies \frac{b}{\sin(65^\circ)} = \frac{15}{\sin(48^\circ)}$.
[M1] Isolate $b$: $b = \frac{15 \sin(65^\circ)}{\sin(48^\circ)}$.
[A1] Calculate: $b = \frac{15(0.90631)}{0.74314} \approx \mathbf{18.3\text{ cm}}$.
Question 2 Tier 1: Standard 3 Marks

In $\triangle ABC$, $a = 7\text{ cm}$, $b = 9\text{ cm}$, and included angle $\angle C = 54^\circ$. Find the length of side $c$.

View Full Method Markscheme
[M1] Apply Cosine Rule: $c^2 = a^2 + b^2 - 2ab \cos C = 7^2 + 9^2 - 2(7)(9) \cos(54^\circ)$.
[M1] Evaluate: $c^2 = 49 + 81 - 126(0.58779) = 130 - 74.06 = 55.94$.
[A1] Take square root: $c = \sqrt{55.94} \approx \mathbf{7.48\text{ cm}}$.
Question 3 Tier 1: Standard 3 Marks

Find the area of a triangular garden with sides $14\text{ m}$ and $18\text{ m}$ enclosing an angle of $42^\circ$.

View Full Method Markscheme
[M1] Apply area formula: $\text{Area} = \frac{1}{2}ab \sin C = \frac{1}{2}(14)(18) \sin(42^\circ)$.
[M1] Multiply: $126 \sin(42^\circ) \approx 126 \times 0.66913$.
[A1] Calculate: $\text{Area} \approx \mathbf{84.3\text{ m}^2}$.
Question 4 Tier 2: Intermediate 4 Marks

A triangle has side lengths $a = 6\text{ cm}$, $b = 8\text{ cm}$, and $c = 12\text{ cm}$. Calculate the size of the largest interior angle to 1 decimal place.

View Full Method Markscheme
[M1] The largest angle lies opposite the longest side ($c = 12\text{ cm}$), which is angle $C$.
[M1] Apply angle form of Cosine Rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{6^2 + 8^2 - 12^2}{2(6)(8)}$.
[M1] Evaluate: $\cos C = \frac{36 + 64 - 144}{96} = \frac{-44}{96} \approx -0.45833$.
[A1] Negative cosine indicates an obtuse angle: $C = \cos^{-1}(-0.45833) \approx \mathbf{117.3^\circ}$.
Question 5 Tier 2: Intermediate 4 Marks

Determine how many distinct triangles can be formed with $\angle A = 40^\circ$, side $b = 10\text{ cm}$, and side $a = 5\text{ cm}$. Justify your answer mathematically.

View Full Method Markscheme
[M1] Calculate critical altitude $h$: $h = b \sin A = 10 \sin(40^\circ)$.
[M1] Evaluate: $h \approx 10 \times 0.64279 = 6.43\text{ cm}$.
[R1] Compare side $a$ to altitude: $a = 5\text{ cm}$, so $a < h$ ($5 < 6.43$).
[A1] Side $a$ is too short to reach the opposite side. Therefore, $\mathbf{0\text{ triangles (no triangle)}}$ can be formed.
Question 6 Tier 2: Intermediate 4 Marks

The area of a triangle with sides $8\text{ cm}$ and $11\text{ cm}$ is $35\text{ cm}^2$. Find the two possible measures for the included angle $\theta$.

View Full Method Markscheme
[M1] Use area formula: $35 = \frac{1}{2}(8)(11) \sin \theta = 44 \sin \theta$.
[M1] Solve for $\sin \theta$: $\sin \theta = \frac{35}{44} \approx 0.79545$.
[A1] Acute solution: $\theta_1 = \sin^{-1}(0.79545) \approx \mathbf{52.7^\circ}$.
[A1] Supplementary obtuse solution: $\theta_2 = 180^\circ - 52.7^\circ = \mathbf{127.3^\circ}$.
Question 7 Tier 2: Intermediate 4 Marks

Two hiker rescue stations $A$ and $B$ are $16\text{ km}$ apart, with $B$ due East of $A$. A smoke flare is spotted at position $S$ on a bearing of $035^\circ$ from $A$ and $310^\circ$ from $B$. Find the distance from station $A$ to the flare.

View Full Method Markscheme
[M1] Bearing from $A$ to East line ($B$) is $090^\circ$. So $\angle SAB = 90^\circ - 35^\circ = 55^\circ$.
[M1] At $B$, West line towards $A$ is $270^\circ$. Bearing to $S$ is $310^\circ$, so $\angle SBA = 310^\circ - 270^\circ = 40^\circ$.
[M1] Angle at flare $S$: $\angle ASB = 180^\circ - (55^\circ + 40^\circ) = 85^\circ$.
[A1] Apply Sine Rule: $\frac{AS}{\sin(40^\circ)} = \frac{16}{\sin(85^\circ)} \implies AS = \frac{16 \sin(40^\circ)}{\sin(85^\circ)} \approx \frac{16(0.64279)}{0.99619} \approx \mathbf{10.3\text{ km}}$.
Question 8 Tier 3: Extended 5 Marks

In a quadrilateral $ABCD$, $AB = 6\text{ cm}$, $BC = 8\text{ cm}$, $CD = 7\text{ cm}$, $DA = 9\text{ cm}$, and $\angle ABC = 90^\circ$.
(a) Find the length of diagonal $AC$.
(b) Calculate angle $\angle ADC$ to 1 decimal place.
(c) Find the total area of the quadrilateral.

View Full Method Markscheme
[M1] Part (a): Right triangle $\triangle ABC$: $AC = \sqrt{6^2 + 8^2} = \sqrt{100} = \mathbf{10\text{ cm}}$.
[M1] Part (b): In $\triangle ADC$, sides are $AD = 9, CD = 7, AC = 10$. Apply Cosine Rule: $$\cos(\angle ADC) = \frac{9^2 + 7^2 - 10^2}{2(9)(7)} = \frac{81 + 49 - 100}{126} = \frac{30}{126} \approx 0.2381$$
[A1] $\angle ADC = \cos^{-1}(0.2381) \approx \mathbf{76.2^\circ}$.
[M1] Part (c): Area $(\triangle ABC) = \frac{1}{2}(6)(8) = 24\text{ cm}^2$. Area $(\triangle ADC) = \frac{1}{2}(9)(7) \sin(76.2^\circ) \approx 31.5(0.9711) \approx 30.6\text{ cm}^2$.
[A1] Total area: $24 + 30.6 = \mathbf{54.6\text{ cm}^2}$.
Question 9 Tier 3: Extended 5 Marks

Prove that for any triangle with sides $a, b, c$ and semi-perimeter $s = \frac{a+b+c}{2}$: $$1 + \cos C = \frac{2s(s - c)}{ab}$$

View Full Method Markscheme
[M1] Substitute the Cosine Rule $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ into the left side: $$1 + \cos C = 1 + \frac{a^2 + b^2 - c^2}{2ab} = \frac{2ab + a^2 + b^2 - c^2}{2ab}$$
[M1] Group the perfect square trinomial in the numerator: $$a^2 + 2ab + b^2 - c^2 = (a + b)^2 - c^2$$
[M1] Apply difference of squares: $$(a + b)^2 - c^2 = [(a + b) - c][(a + b) + c]$$
[M1] Recall $a + b + c = 2s$, so $(a + b - c) = (a + b + c) - 2c = 2s - 2c = 2(s - c)$.
[A1] Numerator becomes: $[2(s - c)][2s] = 4s(s - c)$. Thus: $$\frac{4s(s - c)}{2ab} = \mathbf{\frac{2s(s - c)}{ab}} = \text{RHS} \quad \blacksquare$$
Question 10 Tier 3: Extended 5 Marks

Two radio towers $A$ and $B$ are separated by a river. An engineer at point $C$ on the same side as $A$ measures distances $CA = 250\text{ m}$ and $CB = 420\text{ m}$, and angle $\angle ACB = 64^\circ$.
(a) Find the direct distance between the two towers $AB$.
(b) Find the shortest distance from tower $B$ to the straight survey line $CA$ extended.

View Full Method Markscheme
[M1] Part (a): Apply Cosine Rule: $$AB^2 = 250^2 + 420^2 - 2(250)(420) \cos(64^\circ) = 62500 + 176400 - 210000(0.43837)$$
[M1] $AB^2 = 238900 - 92058 = 146842 \implies AB = \sqrt{146842} \approx \mathbf{383\text{ metres}}$ (3 s.f.).
[M1] Part (b): The shortest distance from $B$ to line $CA$ is the perpendicular altitude $h$ dropped from $B$ onto $CA$.
[M1] In right triangle containing angle $\angle ACB$: $h = CB \sin(\angle ACB) = 420 \sin(64^\circ)$.
[A1] Calculate: $h \approx 420 \times 0.89879 \approx \mathbf{377\text{ metres}}$ (3 s.f.).

Non-Right Trigonometry Formulas

  • The Sine Rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
  • The Cosine Rule (Side): $c^2 = a^2 + b^2 - 2ab \cos C$
  • The Cosine Rule (Angle): $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}$
  • Trigonometric Area: $\text{Area} = \dfrac{1}{2}ab \sin C$
  • Ambiguous Case Criterion: $h = b \sin A$. If $h < a < b$, two triangles exist.

Top Common MYP Pitfalls

1. Overlooking the Ambiguous Case in SSA

Assuming $\sin^{-1}$ always gives the only angle. In SSA with $a < b$, you must check whether $180^\circ - \theta$ yields a valid second triangle!

2. Order of Operations in Cosine Rule

Evaluating $a^2 + b^2 - 2ab \cos C$ as $(a^2 + b^2 - 2ab) \times \cos C$. Multiplication $2ab \cos C$ MUST be computed before subtraction!

3. Non-Included Angle in Area Formula

Using $\frac{1}{2}ab \sin A$. The angle MUST be the included angle between the two given sides (between $a$ and $b$ is angle $C$)!

4. Negative Cosine Confusion

Panicking when $\cos C$ is negative in the Cosine Rule. A negative cosine simply means angle $C$ is obtuse ($> 90^\circ$).

Interactive Mini Test: Sine & Cosine Rules

Answer the questions to see your score and feedback.

Question 1 of 5 Rule Selection Strategy

Which rule should be chosen first to find side $c$ given $a = 8$, $b = 11$, and included angle $C = 62^\circ$?

Question 2 of 5 Triangle Area Calculation

What is the exact area of a triangle with sides $6\text{ cm}$ and $10\text{ cm}$ and included angle $30^\circ$?

Question 3 of 5 Ambiguous Case Condition

In $\triangle ABC$, $\angle A = 30^\circ$ and $b = 10$. For which value of side $a$ do exactly TWO distinct triangles exist?

Question 4 of 5 Obtuse Angle Recognition

If applying the Cosine Rule yields $\cos C = -0.5$, what is the measure of angle $C$?

Question 5 of 5 Parallelogram Area

A parallelogram has adjacent sides of $8\text{ cm}$ and $12\text{ cm}$ and an interior angle of $45^\circ$. What is its area?