IB MYP 4-5 Maths- Sine rule and cosine rule - Study Notes - New Syllabus
IB MYP 4-5 Maths- Sine rule and cosine rule – Study Notes
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- Sine rule and cosine rule
IB MYP 4-5 Maths- Sine rule and cosine rule – Study Notes – All topics
Sine rule and cosine rule
Sine Rule
The Sine Rule relates the sides and angles of any triangle (not necessarily right-angled):
\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)
where:
- \( a, b, c \) are the sides of the triangle
- \( A, B, C \) are the angles opposite those sides
When to use the Sine Rule?
- When you know:
- AAS (two angles and one side), or
- ASA (two angles and included side).
Example :
In triangle ABC, \( A = 40^\circ, B = 75^\circ, a = 8 \text{ cm} \). Find side \( b \).
▶️ Answer/Explanation
Step 1: Find \( C \): \( C = 180^\circ – (40^\circ + 75^\circ) = 65^\circ \)
Step 2: Apply Sine Rule: \( \frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow \frac{8}{\sin 40^\circ} = \frac{b}{\sin 75^\circ} \)
\(\sin 40^\circ \approx 0.6428, \sin 75^\circ \approx 0.9659\).
\( \frac{8}{0.6428} = \frac{b}{0.9659} \Rightarrow b = \frac{8 \times 0.9659}{0.6428} \approx 12.02 \text{ cm} \)
Example :
In triangle ABC, \( a = 12 \text{ cm}, b = 18 \text{ cm}, A = 50^\circ \). Find angle B.
▶️ Answer/Explanation
Apply Sine Rule: \( \frac{\sin A}{a} = \frac{\sin B}{b} \Rightarrow \frac{\sin 50^\circ}{12} = \frac{\sin B}{18} \)
\(\sin 50^\circ \approx 0.7660\).
\( \frac{0.7660}{12} = \frac{\sin B}{18} \Rightarrow \sin B = \frac{18 \times 0.7660}{12} = 1.149 \)
\( \sin B \) cannot be greater than 1, so this configuration is impossible (no such triangle exists).
Observation: Always check feasibility of triangle when solving!
Cosine Rule
The Cosine Rule is used in triangles (not necessarily right-angled) when we know:
- SAS (two sides and included angle) → to find the third side, or
- SSS (all three sides) → to find an angle.
Cosine Rule Formulas:
- For a side: \( a^2 = b^2 + c^2 – 2bc\cos A \) Similarly: \( b^2 = a^2 + c^2 – 2ac\cos B,\; c^2 = a^2 + b^2 – 2ab\cos C \)
- For an angle: \( \cos A = \frac{b^2 + c^2 – a^2}{2bc} \) Similarly: \( \cos B = \frac{a^2 + c^2 – b^2}{2ac},\; \cos C = \frac{a^2 + b^2 – c^2}{2ab} \)
When to use Cosine Rule?
- When triangle is SAS or SSS (but not AAS/ASA → use Sine Rule instead).
Example :
In triangle ABC, \( A = 60^\circ, b = 8 \text{ cm}, c = 10 \text{ cm} \). Find side \( a \).
▶️ Answer/Explanation
Apply Cosine Rule: \( a^2 = b^2 + c^2 – 2bc\cos A \)
\( a^2 = 8^2 + 10^2 – 2(8)(10)\cos 60^\circ \)
\( \cos 60^\circ = 0.5 \).
\( a^2 = 64 + 100 – 160(0.5) = 164 – 80 = 84 \)
\( a = \sqrt{84} \approx 9.17 \text{ cm} \)
Example :
In triangle ABC, \( a = 7 \text{ cm}, b = 9 \text{ cm}, c = 10 \text{ cm} \). Find angle A.
▶️ Answer/Explanation
Apply Cosine Rule for angle: \( \cos A = \frac{b^2 + c^2 – a^2}{2bc} \)
\( \cos A = \frac{9^2 + 10^2 – 7^2}{2(9)(10)} = \frac{81 + 100 – 49}{180} = \frac{132}{180} \approx 0.7333 \)
\( A = \cos^{-1}(0.7333) \approx 43.1^\circ \)
Applications
Area of a Triangle using Sine Rule
The area of a triangle can be calculated using the formula:
\( \text{Area} = \frac{1}{2}ab\sin C \)
Where:
- \( a \) and \( b \) are two sides of the triangle
- \( C \) is the included angle between sides \( a \) and \( b \)
Other forms:
\( \text{Area} = \frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B \)
This is possible because any two sides and the included angle can be used.
Example :
Find the area of a triangle with sides \( a = 8 \text{ cm}, b = 10 \text{ cm} \), and included angle \( C = 60^\circ \).
▶️ Answer/Explanation
\( \text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(8)(10)\sin 60^\circ \)
\( \sin 60^\circ = 0.866 \)
\( \text{Area} = 40 \times 0.866 = 34.64 \text{ cm}^2 \)
Example :
A triangular plot of land has two sides measuring 50 m and 70 m, and the angle between them is 120°. Find its area.
▶️ Answer/Explanation
\( \text{Area} = \frac{1}{2}(50)(70)\sin 120^\circ \)
\( \sin 120^\circ = \sin 60^\circ = 0.866 \)
\( \text{Area} = 1750 \times 0.866 = 1515.5 \text{ m}^2 \)
Ambiguous Case in Sine Rule (SSA)
The ambiguous case occurs when applying the Sine Rule to find angles in a triangle when two sides and a non-included angle are known (SSA condition).
“Ambiguous”?
- The sine function is positive in both the first and second quadrants (i.e., for angles between 0° and 180°).
- This means that if \( \sin \theta = k \), then: \( \theta = \arcsin(k) \quad \text{or} \quad \theta’ = 180^\circ – \arcsin(k) \)
- This can lead to two possible triangles → two solutions (or sometimes one or none).
Possible Outcomes:
- No Solution: If the given side is too short to form a triangle.
- One Solution: When the given angle is 90° or the opposite side is longer than the other given side.
- Two Solutions: When both an acute and obtuse angle satisfy the sine condition.
Steps:
- Apply the Sine Rule: \( \frac{\sin A}{a} = \frac{\sin B}{b} \)
- Find the possible value of unknown angle using \( \arcsin \)
- Check if the second possible angle (180° − found angle) is valid by ensuring the sum of angles ≤ 180°.
Example :
In triangle ABC, \( A = 40^\circ \), \( a = 8 \text{ cm} \), and \( b = 10 \text{ cm} \). Find angle B.
▶️ Answer/Explanation
Apply Sine Rule:
\( \frac{\sin B}{b} = \frac{\sin A}{a} \Rightarrow \frac{\sin B}{10} = \frac{\sin 40^\circ}{8} \)
\( \sin B = 10 \times \frac{\sin 40^\circ}{8} = 1.25 \times 0.6428 = 0.8035 \)
\( B = \arcsin(0.8035) \approx 53^\circ \)
Second possible angle: \( B’ = 180^\circ – 53^\circ = 127^\circ \)
Check:
- If \( B = 53^\circ \), then \( C = 180 – 40 – 53 = 87^\circ \)
- If \( B = 127^\circ \), then \( C = 180 – 40 – 127 = 13^\circ \)
Two solutions: \( B = 53^\circ \) or \( B = 127^\circ \).
Example :
In triangle XYZ, \( X = 50^\circ \), \( x = 5 \text{ cm} \), and \( y = 15 \text{ cm} \). Find angle Y.
▶️ Answer/Explanation
\( \frac{\sin Y}{15} = \frac{\sin 50^\circ}{5} \Rightarrow \sin Y = 15 \times \frac{\sin 50^\circ}{5} = 3 \times 0.766 = 2.298 \)
This is impossible since \( \sin Y \leq 1 \). Therefore, No solution exists.