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CIE IGCSE Maths Study Notes

6.5.1 Sine and Cosine Rules

Understanding the sine and cosine rules is essential for solving problems involving non-right-angled triangles. These rules allow us to find unknown sides and angles in any triangle, expanding our problem-solving toolkit beyond the Pythagorean theorem, which applies only to right-angled triangles.

Introduction to the Sine Rule

The sine rule is a powerful tool for dealing with triangles where we don't have a right angle. It's especially useful when we have either two angles and one side (AAS or ASA) or two sides and a non-enclosed angle (SSA) information. The rule states:

asinA=bsinB=csinC\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

where aa, bb, and cc are the lengths of the sides of the triangle, and AA, BB, and CC are the opposite angles.

Sine Rule

Image courtesy of Cue Math

Example Using the Sine Rule

Given: a=7a = 7 cm, A=35°A = 35°, and B=60°B = 60°, find bb.

Solution:

1. Apply the sine rule:

asinA=bsinB\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

2. Substitute the known values:

7sin35°=bsin60°\dfrac{7}{\sin 35°} = \dfrac{b}{\sin 60°}

3. Solve the equation:

b=7sin 35°×sin 60°b = \dfrac{7}{\text{sin 35°}} \times {\text{sin 60°}}b=7 sin 60°sin 35°b = \dfrac{7 \text{ sin 60°}}{\text {sin 35°}}b10.57b \approx 10.57

Therefore, the value of b is approximately 10.5710.57 cm.

Introduction to the Cosine Rule

The cosine rule helps us find a side of a triangle when we know the other two sides and the included angle, or an angle when we know all three sides. It's particularly useful for triangles that don't fit the criteria for using the sine rule. The cosine rule is expressed as:

a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc \cos A

This can also be rearranged to solve for an angle AA:

A=cos1(b2+c2a22bc)A = \cos^{-1} \left( \dfrac{b^2 + c^2 - a^2}{2bc} \right)Cosine Rule

Example Using the Cosine Rule

Given: b=8b = 8 cm, c=6c = 6 cm, and A=50°A = 50°, find aa.

Solution:

1. Insert the known values into the cosine rule formula:

a2=82+622×8×6×cos50°a^2 = 8^2 + 6^2 - 2 \times 8 \times 6 \times \cos 50°

2. Solve for a2a^2:

a2=64+3696cos50°a^2 = 64 + 36 - 96 \cos 50°a2=10096cos50°a^2 = 100 - 96 \cos 50°

3. Calculate aa:

a2=10096cos50°\sqrt{a^2} = \sqrt{100 - 96 \cos 50°}a6.19a \approx 6.19

This process determines the length of side aa given the lengths of sides bb and cc, and the angle AA between them.

Practice Problems

Problem 1: Using the Sine Rule to Find an Angle

Given a triangle with sides a=9a = 9 cm, b=7b = 7 cm, and c=5c = 5 cm, and you know that angle A=70°A = 70°, find angle BB.

Solution:

1. Use the sine rule:

asinA=bsinB\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

2. Substitute known values:

9sin70°=7sinB\dfrac{9}{\sin 70°} = \dfrac{7}{\sin B}

3. Solve for sinB\sin B and then find BB.

9 sin B=7 sin 70°9 \text{ sin } B = 7 \text{ sin 70°}sin B=7 sin 70°9\text {sin } B = \dfrac {7 \text{ sin 70°}}{9}sin B0.7309\text {sin } B \approx 0.7309B=sin1(0.7309)B = \text{sin}^{-1} (0.7309)B46.96°B \approx 46.96°

Problem 2: Using the Cosine Rule to Find a Side

You have a triangle where b=10b = 10 cm, c=7c = 7 cm, and the angle between them A=65°A = 65°. Find the length of side aa.

Solution:

1. Apply the cosine rule:

a2=102+722×10×7×cos65°a^2 = 10^2 + 7^2 - 2 \times 10 \times 7 \times \cos 65°

2. Calculate a2a^2 and then aa.

a2=100+49140×cos65°a^2 = 100 + 49 - 140 \times \cos 65°a2=149140 cos 65°a^2 = 149 - 140 \text { cos 65°}a2=149140 cos 65°\sqrt {a^2} = \sqrt {149 - 140 \text{ cos 65°}}a9.48a \approx 9.48

Tips for Using the Sine and Cosine Rules

  • Identify whether to use the sine or cosine rule based on the given information.
  • Accuracy is key when using your calculator, ensure it's set to the correct mode (degrees or radians) based on the angle measurements you're working with.
  • Practice with different triangle configurations to become more comfortable with these rules.

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