9.2 Advanced Trigonometry
Beyond Right-Angled Triangles
The Sine Rule and Cosine Rule work for any triangle, not just right-angled ones. They're essential for solving triangles where SOH CAH TOA won't work.
The Sine Rule and Cosine Rule work for any triangle, not just right-angled ones. They're essential for solving triangles where SOH CAH TOA won't work.
⚡ The Two Rules:
Sine Rule
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
Use when you have: 2 angles + 1 side, or 2 sides + 1 non-included angle
Cosine Rule
$$a^2 = b^2 + c^2 - 2bc \cos A$$
Use when you have: 3 sides, or 2 sides + included angle
Which Rule to Use?
Sine Rule: When you have angle-side-angle patterns
Cosine Rule: When you have side-side-side or side-angle-side patterns
Sine Rule: When you have angle-side-angle patterns
Cosine Rule: When you have side-side-side or side-angle-side patterns
Example 1: Sine Rule (Finding a Side)
A triangle has angle A = 40°, angle B = 60°, and side a = 8 cm. Find side b.
Formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
Solution:
$\frac{8}{\sin 40°} = \frac{b}{\sin 60°}$
$\frac{8}{0.6428} = \frac{b}{0.8660}$
$b = \frac{8 \times 0.8660}{0.6428} = 10.78$ cm
Answer: Side b = 10.78 cm
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
Solution:
$\frac{8}{\sin 40°} = \frac{b}{\sin 60°}$
$\frac{8}{0.6428} = \frac{b}{0.8660}$
$b = \frac{8 \times 0.8660}{0.6428} = 10.78$ cm
Answer: Side b = 10.78 cm
Example 2: Sine Rule (Finding an Angle)
A triangle has sides a = 7 cm, b = 9 cm, and angle A = 35°. Find angle B.
Formula (rearranged):
$$\sin B = \frac{b \sin A}{a}$$
Solution:
$\sin B = \frac{9 \times \sin 35°}{7}$
$\sin B = \frac{9 \times 0.5736}{7} = 0.7373$
$B = \sin^{-1}(0.7373) = 47.5°$
Answer: Angle B = 47.5°
Formula (rearranged):
$$\sin B = \frac{b \sin A}{a}$$
Solution:
$\sin B = \frac{9 \times \sin 35°}{7}$
$\sin B = \frac{9 \times 0.5736}{7} = 0.7373$
$B = \sin^{-1}(0.7373) = 47.5°$
Answer: Angle B = 47.5°
Example 3: Cosine Rule (Finding a Side)
A triangle has sides b = 5 cm, c = 7 cm, and included angle A = 50°. Find side a.
Formula:
$$a^2 = b^2 + c^2 - 2bc \cos A$$
Solution:
$a^2 = 5^2 + 7^2 - 2(5)(7) \cos 50°$
$a^2 = 25 + 49 - 70 \times 0.6428$
$a^2 = 74 - 45 = 29$
$a = \sqrt{29} = 5.39$ cm
Answer: Side a = 5.39 cm
$$a^2 = b^2 + c^2 - 2bc \cos A$$
Solution:
$a^2 = 5^2 + 7^2 - 2(5)(7) \cos 50°$
$a^2 = 25 + 49 - 70 \times 0.6428$
$a^2 = 74 - 45 = 29$
$a = \sqrt{29} = 5.39$ cm
Answer: Side a = 5.39 cm
Example 4: Cosine Rule (Finding an Angle)
A triangle has all three sides: a = 8 cm, b = 6 cm, c = 10 cm. Find angle A.
Formula (rearranged):
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
Solution:
$\cos A = \frac{6^2 + 10^2 - 8^2}{2(6)(10)}$
$\cos A = \frac{36 + 100 - 64}{120} = \frac{72}{120} = 0.6$
$A = \cos^{-1}(0.6) = 53.1°$
Answer: Angle A = 53.1°
Formula (rearranged):
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
Solution:
$\cos A = \frac{6^2 + 10^2 - 8^2}{2(6)(10)}$
$\cos A = \frac{36 + 100 - 64}{120} = \frac{72}{120} = 0.6$
$A = \cos^{-1}(0.6) = 53.1°$
Answer: Angle A = 53.1°
💡 Sine & Cosine Rule Tips:
• Sine Rule: Links opposite sides and angles (a with A, b with B)
• Cosine Rule: Like Pythagoras with an extra term for non-right triangles
• Finding angles: Always use sin⁻¹ or cos⁻¹ at the end
• Cosine Rule: Like Pythagoras with an extra term for non-right triangles
• Finding angles: Always use sin⁻¹ or cos⁻¹ at the end
🎯 Sine and Cosine Practice
Area Formula
When you don't have the perpendicular height, you can use the ½absinC formula to find the area of any triangle.
When you don't have the perpendicular height, you can use the ½absinC formula to find the area of any triangle.
⚡ Area Formulas:
Basic Formula
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
When you have base and perpendicular height
½absinC Formula
$$\text{Area} = \frac{1}{2}ab \sin C$$
When you have two sides and the included angle
Heron's Formula
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$
where $s = \frac{a+b+c}{2}$
where $s = \frac{a+b+c}{2}$
When you have all three sides
Example 1: Using ½absinC
A triangle has sides a = 7 cm and b = 9 cm with an included angle C = 45°. Find the area.
Formula:
$$\text{Area} = \frac{1}{2}ab \sin C$$
Solution:
$\text{Area} = \frac{1}{2} \times 7 \times 9 \times \sin 45°$
$\text{Area} = \frac{1}{2} \times 7 \times 9 \times 0.7071$
$\text{Area} = 31.5 \times 0.7071 = 22.27$ cm²
Answer: Area = 22.27 cm²
$$\text{Area} = \frac{1}{2}ab \sin C$$
Solution:
$\text{Area} = \frac{1}{2} \times 7 \times 9 \times \sin 45°$
$\text{Area} = \frac{1}{2} \times 7 \times 9 \times 0.7071$
$\text{Area} = 31.5 \times 0.7071 = 22.27$ cm²
Answer: Area = 22.27 cm²
Example 2: Using Heron's Formula
A triangle has sides 5 cm, 6 cm, and 7 cm. Find its area.
Step 1: Calculate semi-perimeter (s)
$s = \frac{a + b + c}{2} = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9$ cm
Step 2: Apply Heron's formula
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{9(9-5)(9-6)(9-7)}$
$\text{Area} = \sqrt{9 \times 4 \times 3 \times 2}$
$\text{Area} = \sqrt{216} = 14.7$ cm²
Answer: Area = 14.7 cm²
Step 1: Calculate semi-perimeter (s)
$s = \frac{a + b + c}{2} = \frac{5 + 6 + 7}{2} = \frac{18}{2} = 9$ cm
Step 2: Apply Heron's formula
$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$
$\text{Area} = \sqrt{9(9-5)(9-6)(9-7)}$
$\text{Area} = \sqrt{9 \times 4 \times 3 \times 2}$
$\text{Area} = \sqrt{216} = 14.7$ cm²
Answer: Area = 14.7 cm²
Example 3: Problem Solving
A triangular garden has sides of 12 m and 15 m with an angle of 60° between them. What is the area of the garden?
Using ½absinC:
$\text{Area} = \frac{1}{2} \times 12 \times 15 \times \sin 60°$
$\text{Area} = \frac{1}{2} \times 12 \times 15 \times 0.8660$
$\text{Area} = 90 \times 0.8660 = 77.94$ m²
Answer: The garden has an area of 77.94 m²
Using ½absinC:
$\text{Area} = \frac{1}{2} \times 12 \times 15 \times \sin 60°$
$\text{Area} = \frac{1}{2} \times 12 \times 15 \times 0.8660$
$\text{Area} = 90 \times 0.8660 = 77.94$ m²
Answer: The garden has an area of 77.94 m²
🎯 Area of Triangle Practice
cm²
cm²
Working in Three Dimensions
We can extend Pythagoras and trigonometry to solve problems in 3D space, such as finding diagonals in cuboids or angles in pyramids.
We can extend Pythagoras and trigonometry to solve problems in 3D space, such as finding diagonals in cuboids or angles in pyramids.
⚡ 3D Pythagoras:
3D Pythagoras
$$d^2 = l^2 + w^2 + h^2$$
Space diagonal of a cuboid
Key Strategy for 3D Problems:
1. Identify a right-angled triangle in the 3D shape
2. Use 2D Pythagoras or trigonometry on that triangle
3. Often need to solve in multiple steps
1. Identify a right-angled triangle in the 3D shape
2. Use 2D Pythagoras or trigonometry on that triangle
3. Often need to solve in multiple steps
Example 1: Cuboid Space Diagonal
A box has dimensions 4 cm × 3 cm × 2 cm. Find the length of the space diagonal (corner to opposite corner).
Formula:
$$d^2 = l^2 + w^2 + h^2$$
Solution:
$d^2 = 4^2 + 3^2 + 2^2$
$d^2 = 16 + 9 + 4$
$d^2 = 29$
$d = \sqrt{29} = 5.39$ cm
Answer: The space diagonal is 5.39 cm
$$d^2 = l^2 + w^2 + h^2$$
Solution:
$d^2 = 4^2 + 3^2 + 2^2$
$d^2 = 16 + 9 + 4$
$d^2 = 29$
$d = \sqrt{29} = 5.39$ cm
Answer: The space diagonal is 5.39 cm
Example 2: Pyramid Angle
A square-based pyramid has a base of 6 cm × 6 cm and vertical height 4 cm. Find the angle between a slant edge and the base.
Step 1: Find half the base diagonal
Base diagonal $= 6\sqrt{2} = 8.49$ cm
Half diagonal $= 4.24$ cm
Step 2: Use trigonometry in the vertical triangle
$\tan(\theta) = \frac{\text{height}}{\text{half diagonal}} = \frac{4}{4.24} = 0.943$
$\theta = \tan^{-1}(0.943) = 43.3°$
Answer: The angle is 43.3°
Step 1: Find half the base diagonal
Base diagonal $= 6\sqrt{2} = 8.49$ cm
Half diagonal $= 4.24$ cm
Step 2: Use trigonometry in the vertical triangle
$\tan(\theta) = \frac{\text{height}}{\text{half diagonal}} = \frac{4}{4.24} = 0.943$
$\theta = \tan^{-1}(0.943) = 43.3°$
Answer: The angle is 43.3°
Example 3: Flagpole Problem
A 10 m tall flagpole stands vertically. A wire is attached from the top of the pole to a point on the ground 8 m away from the base. What angle does the wire make with the ground?
Solution:
This forms a right-angled triangle:
• Opposite (height) = 10 m
• Adjacent (ground distance) = 8 m
$\tan(\theta) = \frac{10}{8} = 1.25$
$\theta = \tan^{-1}(1.25) = 51.3°$
Answer: The wire makes an angle of 51.3° with the ground
Solution:
This forms a right-angled triangle:
• Opposite (height) = 10 m
• Adjacent (ground distance) = 8 m
$\tan(\theta) = \frac{10}{8} = 1.25$
$\theta = \tan^{-1}(1.25) = 51.3°$
Answer: The wire makes an angle of 51.3° with the ground
💡 3D Problem Tips:
• Draw it out: Sketch the 3D shape and identify 2D triangles
• Two-step process: Often need to find one length before finding another
• Right angles are key: Look for vertical lines and base edges
• Use 3D Pythagoras: For space diagonals in cuboids
• Break it down: Complex 3D problems = multiple 2D triangles
• Two-step process: Often need to find one length before finding another
• Right angles are key: Look for vertical lines and base edges
• Use 3D Pythagoras: For space diagonals in cuboids
• Break it down: Complex 3D problems = multiple 2D triangles
🎯 Some Problems
Real Life Uses:
• Architecture: Calculating roof angles and support structures
• Engineering: Designing bridges and frameworks
• Navigation: GPS and 3D positioning
• Computer Graphics: 3D modeling and game development
• Construction: Building scaffolding and cranes
• Aviation: Flight paths and angles of ascent/descent
• Architecture: Calculating roof angles and support structures
• Engineering: Designing bridges and frameworks
• Navigation: GPS and 3D positioning
• Computer Graphics: 3D modeling and game development
• Construction: Building scaffolding and cranes
• Aviation: Flight paths and angles of ascent/descent