Special Right Triangles and Special Angles in Trigonometry

Why “special” right triangles matter

Some angles occur so often that it is worth memorizing their trigonometric values exactly (not as decimals). The key idea is that in certain right triangles, the side lengths have simple ratios. Because sine, cosine, and tangent are ratios of sides, those trig values become exact numbers involving radicals.

The 45-45-90 triangle (isosceles right triangle)

Geometric construction (simple steps)

You can build a 45-45-90 triangle from a square.

• Draw a square with side length 1.
• Draw a diagonal from one corner to the opposite corner. The diagonal splits the square into two congruent right triangles.
• Each triangle has angles 45°, 45°, and 90° (because the square’s corners are 90° and the diagonal bisects the corner angles).

Find the side ratios

In one of the triangles, the legs are both 1. The hypotenuse is the diagonal of the square. Use the Pythagorean Theorem:

hypotenuse^2 = 1^2 + 1^2 = 2  =>  hypotenuse = √2

So the side ratio is:

• legs : hypotenuse = 1 : √2

Exact trig values from the ratios

Pick one acute angle (45°). Relative to that angle, the opposite leg = 1, adjacent leg = 1, hypotenuse = √2.

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• sin 45° = opposite/hypotenuse = 1/√2 = √2/2

• cos 45° = adjacent/hypotenuse = 1/√2 = √2/2

• tan 45° = opposite/adjacent = 1/1 = 1

Notice the symmetry: sine and cosine match because the legs match.

The 30-60-90 triangle

Geometric construction (simple steps)

You can build a 30-60-90 triangle by splitting an equilateral triangle.

• Draw an equilateral triangle with side length 2 (choosing 2 makes the arithmetic clean).
• From one vertex, draw a line straight down to the opposite side so it hits the midpoint. This creates two congruent right triangles.
• Each right triangle has angles 30°, 60°, and 90°.

Find the side ratios

In the equilateral triangle, all sides are 2. The altitude splits the base into two segments of length 1 and 1. In one right triangle:

• hypotenuse = 2 (one side of the equilateral triangle)
• short leg (adjacent to 60°, opposite 30°) = 1 (half the base)

Use the Pythagorean Theorem to find the long leg (the altitude):

long_leg^2 = 2^2 - 1^2 = 4 - 1 = 3  =>  long_leg = √3

So the side ratio is:

• short : long : hypotenuse = 1 : √3 : 2

Exact trig values from the ratios

For 30°: opposite = 1, adjacent = √3, hypotenuse = 2.

• sin 30° = 1/2

• cos 30° = √3/2

• tan 30° = 1/√3 = √3/3

For 60°: opposite = √3, adjacent = 1, hypotenuse = 2.

• sin 60° = √3/2

• cos 60° = 1/2

• tan 60° = √3/1 = √3

Reference table of exact trig values (special angles)

Use this as a no-calculator lookup for common angles.

Angle   sin        cos        tan  (exact)  (exact)  (exact) 30°     1/2        √3/2      √3/3 45°     √2/2      √2/2      1 60°     √3/2      1/2        √3

Radical simplification you will use constantly

Rationalizing a denominator

When a trig ratio gives something like 1/√2 or 1/√3, rewrite it with no radical in the denominator.

• 1/√2 = (1/√2)(√2/√2) = √2/2

• 1/√3 = (1/√3)(√3/√3) = √3/3

Common radical products

• √a · √b = √(ab)

• (√a)/(√b) = √(a/b) (then rationalize if needed)

• √12 = √(4·3) = 2√3

• √8 = √(4·2) = 2√2

Practice Set A: Find exact trig values (no calculator)

Use the special triangles and simplify fully.

• 1) sin 45°

• 2) cos 60°

• 3) tan 30°

• 4) sin 60°

• 5) cos 30°

• 6) tan 45°

Practice Set B: Match angles to ratios

Match each ratio to the correct angle (30°, 45°, or 60°). Each angle may be used more than once.

• 1) √3/2

• 2) 1/2

• 3) √2/2

• 4) √3

• 5) √3/3

• 6) 1

For each ratio, also state whether it could be a sine, cosine, or tangent value (some ratios fit more than one function depending on the angle).

Practice Set C: Simplify expressions with radicals using exact trig values

Rewrite each expression exactly and simplify.

• 1) 2 sin 30° + cos 60°

• 2) (sin 60°)(cos 30°)

• 3) tan 60° − tan 30°

• 4) (cos 45°)/(sin 45°)

• 5) (1)/(cos 30°) (simplify and rationalize if needed)

• 6) (√12)(sin 45°)

Diagram-based quiz: fill missing sides, then compute trig ratios

For each diagram, first fill in the missing side lengths using the special-triangle ratios, then compute the requested trig ratios exactly (simplify radicals).

Quiz 1: 45-45-90 triangle

A right triangle has a right angle at C. The legs are CA and CB. You are told CA = 7 and the triangle is 45-45-90.

• a) Find CB.

• b) Find hypotenuse AB.

• c) Using angle A (one of the 45° angles), compute sin A, cos A, tan A.

Quiz 2: 30-60-90 triangle (given hypotenuse)

A right triangle has angles 30°, 60°, 90°. The hypotenuse is 10.

• a) Find the short leg (opposite 30°).

• b) Find the long leg (opposite 60°).

• c) Compute sin 30°, cos 30°, tan 60° using your side lengths (then simplify to exact values).

Quiz 3: 30-60-90 triangle (given long leg)

A right triangle has angles 30°, 60°, 90°. The long leg (opposite 60°) is √75.

• a) Simplify √75.

• b) Use the 1 : √3 : 2 ratio to find the short leg and the hypotenuse.

• c) Compute sin 60°, cos 60°, tan 30° exactly.

Quiz 4: Mixed identification from a sketch description

You see a right triangle with one acute angle labeled 60°. The side adjacent to 60° is 5 (this is the short leg in a 30-60-90 triangle).

• a) Identify the triangle type and label which side is short, long, hypotenuse.

• b) Find the long leg and the hypotenuse.

• c) Compute sin 60° and cos 60° using your side lengths.

Answer-check format (use this to self-verify)

When you finish any problem in this chapter, check these two things:

• Did you use the correct special-triangle ratio? (45-45-90: 1,1,√2; 30-60-90: 1,√3,2)

• Did you simplify radicals and rationalize denominators where needed?