IB Mathematics AA Use of sine, cosine and tangent ratios Study Notes
IB Mathematics AA Use of sine, cosine and tangent ratios Study Notes
IB Mathematics AA Use of sine, cosine and tangent ratios Study Notes Offer a clear explanation of Use of sine, cosine and tangent ratios , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Use of sine, cosine and tangent ratios
Sine, Cosine, and Tangent Ratios
Sine, Cosine, and Tangent Ratios
In a right-angled triangle, the following trigonometric ratios are defined:
- Sine: \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)
- Cosine: \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- Tangent: \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
Where:
- Opposite is the side opposite the angle \( \theta \).
- Adjacent is the side next to the angle \( \theta \).
- Hypotenuse is the longest side (opposite the right angle).
These ratios help in calculating unknown sides or angles of right-angled triangles.
Example :
In a right-angled triangle, one angle is \( 30^\circ \) and the hypotenuse is 10 cm. Find the length of the opposite side.
▶️ Answer/Explanation
We use the sine ratio:
\( \sin 30^\circ = \frac{\text{opposite}}{10} \)
Since \( \sin 30^\circ = 0.5 \),
\( 0.5 = \frac{\text{opposite}}{10} \Rightarrow \text{opposite} = 10 \times 0.5 = 5 \text{ cm} \)
Example :
In a right-angled triangle, one angle is \( 60^\circ \) and the hypotenuse is 8 cm. Find the adjacent side.
▶️ Answer/Explanation
We use the cosine ratio:
\( \cos 60^\circ = \frac{\text{adjacent}}{8} \)
Since \( \cos 60^\circ = 0.5 \),
\( 0.5 = \frac{\text{adjacent}}{8} \Rightarrow \text{adjacent} = 8 \times 0.5 = 4 \text{ cm} \)
Example :
In a right-angled triangle, the opposite side is 4 cm and the adjacent side is 3 cm. Find the angle \( \theta \).
▶️ Answer/Explanation
We use the tangent ratio:
\( \tan \theta = \frac{4}{3} \)
So,
\( \theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ \)
The Sine Rule
The Sine Rule
The sine rule applies to any triangle (not just right-angled triangles) and relates the lengths of the sides to the sines of their opposite angles:
\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)
Where:
- a, b, c are the sides of the triangle
- A, B, C are the angles opposite these sides
The sine rule is used to:
- Find unknown sides when two angles and one side are known (AAS or ASA cases)
- Find unknown angles when two sides and a non-included angle are known (SSA case)
Example:
In triangle ABC:
- \( A = 40^\circ \)
- \( B = 70^\circ \)
- \( a = 7 \ \text{cm} \)
Find the length of side \( b \).
▶️ Answer/Explanation
Sine rule:
\( \frac{a}{\sin A} = \frac{b}{\sin B} \)
\( \frac{7}{\sin 40^\circ} = \frac{b}{\sin 70^\circ} \)
Calculate sines:
\( \frac{7}{0.6428} = \frac{b}{0.9397} \)
\( 10.89 = \frac{b}{0.9397} \Rightarrow b = 10.89 \times 0.9397 \approx 10.23 \ \text{cm} \)
The Cosine Rule
The Cosine Rule
The cosine rule applies to any triangle and relates the lengths of the sides of a triangle to the cosine of one of its angles:
\( c^2 = a^2 + b^2 – 2ab \cos C \)
Where:
- a, b, c are the sides opposite angles A, B, C respectively.
Other forms:
\( a^2 = b^2 + c^2 – 2bc \cos A \quad \text{and} \quad b^2 = a^2 + c^2 – 2ac \cos B \)
For finding angles:
\( \cos A = \frac{b^2 + c^2 – a^2}{2bc} \quad \cos B = \frac{a^2 + c^2 – b^2}{2ac} \quad \cos C = \frac{a^2 + b^2 – c^2}{2ab} \)
Example :
In triangle ABC:
- \( a = 8 \ \text{cm} \)
- \( b = 6 \ \text{cm} \)
- \( C = 60^\circ \)
Find side \( c \).
▶️ Answer/Explanation
Apply the cosine rule:
\( c^2 = a^2 + b^2 – 2ab \cos C \)
Substitute values:
\( c^2 = 8^2 + 6^2 – 2 \times 8 \times 6 \times \cos 60^\circ \)
Calculate:
\( c^2 = 64 + 36 – 96 \times 0.5 = 100 – 48 = 52 \) \( c = \sqrt{52} \approx 7.21 \ \text{cm} \)
Example :
In triangle ABC:
- \( a = 7 \ \text{cm} \)
- \( b = 5 \ \text{cm} \)
- \( c = 6 \ \text{cm} \)
Find angle \( A \).
▶️ Answer/Explanation
Apply the cosine rule for angle \( A \):
\( \cos A = \frac{b^2 + c^2 – a^2}{2bc} \)
Substitute values:
\( \cos A = \frac{5^2 + 6^2 – 7^2}{2 \times 5 \times 6} = \frac{25 + 36 – 49}{60} = \frac{12}{60} = 0.2 \)
Therefore:
\( A = \arccos 0.2 \approx 78.46^\circ \)
Area of a Triangle
Area of a Triangle
The area \( A \) of any triangle (not just right-angled) can be calculated using:
\( A = \frac{1}{2}ab \sin C \)
Where:
- a and b are two sides of the triangle
- C is the included angle between sides \(a\) and \(b\)
Example: Find the area of triangle ABC
- \( a = 10 \ \text{cm} \)
- \( b = 7 \ \text{cm} \)
- \( C = 45^\circ \)
▶️Answer/Explanation
Substitute into the formula:
\( A = \frac{1}{2} \times 10 \times 7 \times \sin 45^\circ \)
Since \(\sin 45^\circ = 0.7071\):
\( A = 35 \times 0.7071 = 24.75 \ \text{cm}^2 \)