IB MYP 4-5 Maths- Trigonometric ratios in right angled triangles- Study Notes - New Syllabus
IB MYP 4-5 Maths- Trigonometric ratios in right angled triangles – Study Notes
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- Trigonometric ratios in right angled triangles
IB MYP 4-5 Maths- Trigonometric ratios in right angled triangles – Study Notes – All topics
Trigonometric Ratios in Right-Angled Triangles
Trigonometric Ratios in Right-Angled Triangles
Trigonometric ratios relate the angles of a right-angled triangle to the ratios of its sides. These ratios are used to calculate unknown sides or angles in a right-angled triangle.
Sides of the Triangle:
- Hypotenuse: The longest side, opposite the right angle.
- Opposite side: The side opposite the angle under consideration.
- Adjacent side: The side next to the angle under consideration (but not the hypotenuse).
Sine and Cosine on the Unit Circle
The unit circle is a circle with a radius of 1 unit centered at the origin (0, 0) in the coordinate plane. Any point \( P(x, y) \) on the unit circle satisfies:
\( x^2 + y^2 = 1 \)
For an angle \( \theta \) measured from the positive x-axis:
- \( x = \cos \theta \) (horizontal coordinate)
- \( y = \sin \theta \) (vertical coordinate)
This means cosine represents the x-coordinate, and sine represents the y-coordinate of a point on the unit circle.
Key Features:
- Radius = 1 unit.
- Angle \( \theta \) is measured counterclockwise from the positive x-axis.
- Coordinates of any point on the unit circle: \( (\cos \theta, \sin \theta) \).
Trigonometric Ratios:
- Sine: \( \sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}} \)
- Cosine: \( \cos \theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}} \)
- Tangent: \( \tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}} \)
Mnemonic: SOH-CAH-TOA
Sine = Opposite / Hypotenuse
Cosine = Adjacent / Hypotenuse
Tangent = Opposite / Adjacent
Special Angle Trigonometric Values:
| Angle (°) | \( \sin \theta \) | \( \cos \theta \) | \( \tan \theta \) |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | \( \dfrac{1}{2} \) | \( \dfrac{\sqrt{3}}{2} \) | \( \dfrac{1}{\sqrt{3}} \) |
| 45° | \( \dfrac{1}{\sqrt{2}} \) | \( \dfrac{1}{\sqrt{2}} \) | 1 |
| 60° | \( \dfrac{\sqrt{3}}{2} \) | \( \dfrac{1}{2} \) | \( \sqrt{3} \) |
| 90° | 1 | 0 | Undefined |
Steps to Solve Trigonometric Problems:
- Identify the angle given or to be found.
- Label the sides: Opposite, Adjacent, Hypotenuse.
- Choose the correct ratio (SOH, CAH, or TOA) based on the sides you know and the side you need to find.
- Write the equation and solve for the unknown.
- If finding an angle, use the inverse function: \( \sin^{-1}, \cos^{-1}, \tan^{-1} \).
Example :
A right-angled triangle has an angle of 30° and the adjacent side is 8 cm. Find the hypotenuse.
▶️ Answer/Explanation
Given: Adjacent = 8 cm, angle = 30°, find hypotenuse.
Use \( \cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} \):
\( \cos 30° = \dfrac{8}{h} \Rightarrow 0.866 = \dfrac{8}{h} \Rightarrow h = \dfrac{8}{0.866} \approx \boxed{9.24 \text{ cm}} \)
Example :
A right triangle has an angle of 45° and the hypotenuse is 10 cm. Find the opposite side.
▶️ Answer/Explanation
Given: Hypotenuse = 10 cm, angle = 45°.
Use \( \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} \):
\( \sin 45° = \dfrac{\text{opposite}}{10} \Rightarrow 0.707 = \dfrac{\text{opposite}}{10} \Rightarrow \text{opposite} = 10 \times 0.707 = \boxed{7.07 \text{ cm}} \)
Example :
A right triangle has opposite = 5 cm and adjacent = 12 cm. Find the angle θ.
▶️ Answer/Explanation
Use \( \tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12} \).
\( \tan \theta = 0.4167 \Rightarrow \theta = \tan^{-1}(0.4167) \approx \boxed{22.6°} \).
Example :
A ladder 6 m long leans against a wall making an angle of 60° with the ground. How high does it reach on the wall?
▶️ Answer/Explanation
Opposite side = height, hypotenuse = 6 m, angle = 60°.
Use \( \sin 60° = \dfrac{\text{height}}{6} \Rightarrow 0.866 = \dfrac{\text{height}}{6} \Rightarrow \text{height} = 6 \times 0.866 = \boxed{5.20 \text{ m}} \).
Angle of Elevation and Angle of Depression
Angle of Elevation and Angle of Depression
These terms are used when measuring angles of view from a horizontal line:
1. Angle of Elevation:
- The angle formed between the horizontal line of sight and the line of sight looking up at an object.
- Example: Looking up at the top of a tree from the ground.
2. Angle of Depression:
- The angle formed between the horizontal line of sight and the line of sight looking down at an object.
- Example: Looking down at a car from the top of a building.
Key Points:
- Angles of elevation and depression are always measured from the horizontal.
- Horizontal lines are assumed to be parallel, so the angle of elevation from one point equals the angle of depression from the other point.
- Problems involving these angles usually form right-angled triangles.
Trigonometric Ratios Used:
- \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)
- \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
Example :
A person is standing 30 m away from a building. The angle of elevation to the top of the building is 35°. Find the height of the building (to 1 decimal place).
▶️ Answer/Explanation
Triangle formed: Opposite = height of building, Adjacent = 30 m, Angle = 35°.
Use: \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
\( \tan 35^\circ = \frac{h}{30} \Rightarrow h = 30 \times \tan 35^\circ \)
\( h = 30 \times 0.700 = 21.0 \, \text{m (approx)} \)
Final Answer: \( 21.0 \, \text{m} \)
Example :
From the top of a lighthouse 40 m high, the angle of depression to a boat in the sea is 25°. Find the distance of the boat from the base of the lighthouse (to the nearest metre).
▶️ Answer/Explanation
Triangle formed: Opposite = 40 m, Adjacent = distance (d), Angle = 25°.
Use: \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
\( \tan 25^\circ = \frac{40}{d} \Rightarrow d = \frac{40}{\tan 25^\circ} \)
\( d = \frac{40}{0.466} \approx 85.8 \, \text{m} \)
Final Answer: \( 86 \, \text{m (approx)} \)