📐 Level 1 - Topic 6: Special Triangles (30°-60°-90° & 45°-45°-90°) 🌟

1) Special Right Triangles - Why "Special"? ✨

In geometry and trigonometry, certain right triangles appear frequently and have very predictable properties. These are known as special right triangles. The two most important types are the 30°-60°-90° triangle and the 45°-45°-90° triangle.

Special Right Triangles are right triangles with specific angle measures that lead to simple and predictable ratios of their side lengths and trigonometric ratios. The two main types are:

30°-60°-90° Triangle: Angles are \( 30^\circ \), \( 60^\circ \), and \( 90^\circ \).
45°-45°-90° Triangle: Angles are \( 45^\circ \), \( 45^\circ \), and \( 90^\circ \) (also known as an isosceles right triangle).

Why are they "Special"?

Exact Ratios: The ratios of their side lengths are simple and involve square roots of small whole numbers (like \( \sqrt{2} \) and \( \sqrt{3} \)).
Exact Trigonometric Values: We can find the exact values (without needing calculators) for trigonometric functions of \( 30^\circ \), \( 45^\circ \), and \( 60^\circ \). These angles are very common in problems.
Geometric Significance: They arise naturally in geometric constructions and have symmetrical properties. For example, a 45-45-90 triangle is half of a square, and a 30-60-90 triangle is half of an equilateral triangle.

2) The 30°-60°-90° Triangle - Side Ratios and Trigonometric Values 📐

Let's first explore the 30°-60°-90° triangle. The key property of these triangles is the ratio of their side lengths.

Side Length Ratios in a 30°-60°-90° Triangle:

In a 30°-60°-90° triangle, the side lengths are in the ratio \( 1 : \sqrt{3} : 2 \), where:

The side opposite the \( 30^\circ \) angle (shortest leg) has length \( 1x \) (we can say \( x \)).
The side opposite the \( 60^\circ \) angle (longer leg) has length \( \sqrt{3}x \).
The hypotenuse (opposite the \( 90^\circ \) angle) has length \( 2x \).

You can visualize this. If the shortest leg is 1 unit, then the longer leg is \( \sqrt{3} \) units, and the hypotenuse is 2 units. This ratio holds for *any* 30-60-90 triangle, regardless of its size. [*(Consider adding a diagram here showing a 30-60-90 triangle with sides labeled as x, sqrt(3)x, 2x)*]

Trigonometric Ratios for \( 30^\circ \) and \( 60^\circ \):

Using these side ratios and the definitions of sine, cosine, and tangent:

For \( 30^\circ \) angle:
- Opposite side = \( x \), Adjacent side = \( \sqrt{3}x \), Hypotenuse = \( 2x \)
- \( \sin(30^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{x}{2x} = \frac{1}{2} \)
- \( \cos(30^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{3}x}{2x} = \frac{\sqrt{3}}{2} \)
- \( \tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{\sqrt{3}x} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \) (rationalized)
For \( 60^\circ \) angle:
- Opposite side = \( \sqrt{3}x \), Adjacent side = \( x \), Hypotenuse = \( 2x \)
- \( \sin(60^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\sqrt{3}x}{2x} = \frac{\sqrt{3}}{2} \)
- \( \cos(60^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{x}{2x} = \frac{1}{2} \)
- \( \tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sqrt{3}x}{x} = \sqrt{3} \)

3) The 45°-45°-90° Triangle - Side Ratios and Trigonometric Values 📐

Now let's consider the other special right triangle: the 45°-45°-90° triangle (isosceles right triangle).

Side Length Ratios in a 45°-45°-90° Triangle:

In a 45°-45°-90° triangle, the side lengths are in the ratio \( 1 : 1 : \sqrt{2} \), where:

The two legs (opposite the \( 45^\circ \) angles) have length \( 1x \) (we can say \( x \)).
The hypotenuse (opposite the \( 90^\circ \) angle) has length \( \sqrt{2}x \).

If each leg is 1 unit long, then the hypotenuse is \( \sqrt{2} \) units. Again, this ratio holds for all 45-45-90 triangles. [*(Consider adding a diagram here showing a 45-45-90 triangle with sides labeled as x, x, sqrt(2)x)*]

Trigonometric Ratios for \( 45^\circ \):

Using these side ratios and the definitions of sine, cosine, and tangent (note that for \( 45^\circ \), the opposite and adjacent sides are equal):

For \( 45^\circ \) angle:
- Opposite side = \( x \), Adjacent side = \( x \), Hypotenuse = \( \sqrt{2}x \)
- \( \sin(45^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{x}{\sqrt{2}x} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) (rationalized)
- \( \cos(45^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{x}{\sqrt{2}x} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \) (rationalized)
- \( \tan(45^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{x} = 1 \)

4) Summary of Trigonometric Ratios for Special Angles 🌟

Let's summarize the exact trigonometric ratios for the special angles \( 30^\circ \), \( 45^\circ \), and \( 60^\circ \):

Angle \( \theta \)	\( \sin(\theta) \)	\( \cos(\theta) \)	\( \tan(\theta) \)
\( 30^\circ \)	\( \frac{1}{2} \)	\( \frac{\sqrt{3}}{2} \)	\( \frac{\sqrt{3}}{3} \)
\( 45^\circ \)	\( \frac{\sqrt{2}}{2} \)	\( \frac{\sqrt{2}}{2} \)	\( 1 \)
\( 60^\circ \)	\( \frac{\sqrt{3}}{2} \)	\( \frac{1}{2} \)	\( \sqrt{3} \)

Memorization Tip: Notice the pattern in sine values for 30°, 45°, 60°: \( \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2} \). The numbers under the square root increase (1, 2, 3). Cosine values are in reverse order of sine values for these angles. Tangent can be derived from sine/cosine or remembered directly. *Practice recalling these values frequently!*

5) Using Special Triangles - Examples 🎯

Example 1: Finding a Side in a 30-60-90 Triangle

In a 30°-60°-90° triangle, the hypotenuse is 12 cm. Find the lengths of the other two sides.

Solution:

In a 30-60-90 triangle, side ratios are \( x : \sqrt{3}x : 2x \). Hypotenuse corresponds to \( 2x \).
Given hypotenuse = 12 cm, so \( 2x = 12 \Rightarrow x = 6 \) cm.
Side opposite \( 30^\circ \) angle (shorter leg) = \( x = 6 \) cm.
Side opposite \( 60^\circ \) angle (longer leg) = \( \sqrt{3}x = 6\sqrt{3} \) cm.

Example 2: Finding a Side in a 45-45-90 Triangle

In a 45°-45°-90° triangle, one leg is 7 inches long. Find the lengths of the other leg and the hypotenuse.

Solution:

In a 45-45-90 triangle, side ratios are \( x : x : \sqrt{2}x \). Legs are equal in length, and correspond to \( x \).
Given leg = 7 inches, so \( x = 7 \) inches.
Other leg (also opposite \( 45^\circ \)) = \( x = 7 \) inches.
Hypotenuse = \( \sqrt{2}x = 7\sqrt{2} \) inches.

Example 3: Using Trigonometric Ratio and Special Angle

A right triangle has an angle of \( 60^\circ \) and the adjacent side to this angle is 5 meters. Find the length of the hypotenuse.

Solution:

We are given adjacent side and angle \( 60^\circ \), we need to find hypotenuse. Use cosine ratio.
\( \cos(60^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \)
We know \( \cos(60^\circ) = \frac{1}{2} \) and Adjacent = 5 m.
So, \( \frac{1}{2} = \frac{5}{\text{Hypotenuse}} \). Solving for Hypotenuse, we get Hypotenuse = \( 5 \times 2 = 10 \) meters.

6) Practice Questions 🎯

6.1 Fundamental – Special Triangle Ratios

1. What are the angles in a 30°-60°-90° triangle? What are the ratios of its side lengths?

2. What are the angles in a 45°-45°-90° triangle? What are the ratios of its side lengths?

3. What are the exact values of \( \sin(30^\circ), \cos(30^\circ), \tan(30^\circ) \)?

4. What are the exact values of \( \sin(45^\circ), \cos(45^\circ), \tan(45^\circ) \)?

5. What are the exact values of \( \sin(60^\circ), \cos(60^\circ), \tan(60^\circ) \)?

6. In a 30°-60°-90° triangle, if the shorter leg is 8 cm, find the lengths of the longer leg and the hypotenuse.

7. In a 45°-45°-90° triangle, if the hypotenuse is \( 5\sqrt{2} \) inches, find the lengths of the legs.

8. Use a trigonometric ratio and the special angle value to find the hypotenuse of a right triangle if one angle is \( 30^\circ \) and the side opposite to it is 4 cm.

9. A ramp is designed to have a \( 45^\circ \) angle of elevation. If the horizontal distance covered by the ramp is 10 feet, what is the vertical height of the ramp? (Use trigonometric ratio and special angle).

10. True or False: In a 30°-60°-90° triangle, the hypotenuse is always exactly twice the length of the longer leg. Explain.

6.2 Challenging – Applications & Extensions 💪🚀

1. Derive the side length ratios of a 30°-60°-90° triangle by starting with an equilateral triangle and bisecting one of its angles.

2. Derive the side length ratios of a 45°-45°-90° triangle by starting with a square and drawing a diagonal.

3. A rhombus has angles of \( 60^\circ \) and \( 120^\circ \). The shorter diagonal is 10 cm. Find the length of the sides and the longer diagonal of the rhombus. (Hint: Rhombus diagonals bisect angles and are perpendicular).

4. An isosceles triangle has a vertex angle of \( 120^\circ \) and the equal sides are of length 8 cm. Find the length of the base of the triangle. (Hint: Split the isosceles triangle into two right triangles).

5. (Conceptual) Imagine you have to explain special right triangles to someone who has never seen them before. How would you visually demonstrate or explain why the side ratios are always constant for 30-60-90 and 45-45-90 triangles, regardless of their size?

7) Summary 🎉

Special Right Triangles: 30°-60°-90° and 45°-45°-90° triangles. They have predictable side ratios and trigonometric values.
30°-60°-90° Triangle Ratios: \( 1 : \sqrt{3} : 2 \) (opposite 30° : opposite 60° : hypotenuse).
45°-45°-90° Triangle Ratios: \( 1 : 1 : \sqrt{2} \) (legs : hypotenuse).
Memorize Values: Know exact \( \sin, \cos, \tan \) for \( 30^\circ, 45^\circ, 60^\circ \). Refer to the summary table.
Applications: Special triangles simplify problems involving these angles. Useful in geometry, trigonometry, and applications.

Fantastic! You've now mastered special right triangles and their trigonometric ratios! These triangles are powerful tools that will appear again and again in trigonometry and related fields. Practice working with these ratios and visualizing these triangles – you'll find they make many calculations much easier! 🌟📐🚀