Using Special Right Triangles Worksheet | PDF Printable Geometry ... - Free Printable

Explanation:
Let’s solve each section step by step.

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Section A: Fill in the exact trig values using the special right triangles.

We have two special right triangles:

1. 30°–60°–90° triangle: sides are in ratio
Short leg : Long leg : Hypotenuse = 1 : √3 : 2
From the diagram:
- Opposite 30° = 1 (short leg)
- Opposite 60° = √3 (long leg)
- Hypotenuse = 2

So:
- sin 30° = opposite/hypotenuse = 1/2
- cos 30° = adjacent/hypotenuse = √3/2
- tan 30° = opposite/adjacent = 1/√3 = √3/3

For 60°:
- sin 60° = √3/2
- cos 60° = 1/2
- tan 60° = √3/1 = √3

2. 45°–45°–90° triangle: sides are in ratio
Leg : Leg : Hypotenuse = 1 : 1 : √2
From diagram: legs = 1, hypotenuse = √2

So:
- sin 45° = 1/√2 = √2/2
- cos 45° = 1/√2 = √2/2
- tan 45° = 1/1 = 1

Now fill the table:

| | 0° | 30° | 45° | 60° | 90° |
|--------|----|---------|-----------|----------|-----|
| sin | 0 | 1/2 | √2/2 | √3/2 | 1 |
| cos | 1 | √3/2| √2/2 | 1/2 | 0 |
| tan | 0 | √3/3| 1 | √3 | — |

(Note: tan 90° is undefined — the black cell likely means “undefined” or “not applicable”.)

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Section B: Exact values

Use known values:

- sin 30° = 1/2
- cos 60° = 1/2
- tan 45° = 1
- cos 0° = 1
- sin 60° = √3/2
- cos 45° = √2/2
- tan 30° = √3/3
- tan 60° = √3
- sin 90° = 1

Now compute:

1) sin 30 + cos 60 = 1/2 + 1/2 = 1
2) tan 45 + cos 60 = 1 + 1/2 = 3/2
3) 4 sin 30 + cos 0 = 4·(1/2) + 1 = 2 + 1 = 3
4) 2 sin 60 = 2·(√3/2) = √3
5) 8 cos 45 = 8·(√2/2) = 4√2
6) (sin 45)² + 3 cos 60 = (√2/2)² + 3·(1/2) = (2/4) + 3/2 = 1/2 + 3/2 = 2
7) (3 tan 30)² + (tan 60)² = (3·√3/3)² + (√3)² = (√3)² + 3 = 3 + 3 = 6
8) 5 sin 60 − 3 cos 30 = 5·(√3/2) − 3·(√3/2) = (5√3 − 3√3)/2 = 2√3/2 = √3
9) 9 sin 30 − 3 sin 90 + 4(cos 45)²
= 9·(1/2) − 3·1 + 4·(√2/2)²
= 9/2 − 3 + 4·(2/4)
= 9/2 − 3 + 4·(1/2)
= 9/2 − 3 + 2
= 9/2 − 1 = 9/2 − 2/2 = 7/2

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Section C: Find missing angles using side ratios

Recall: In a right triangle,
- sin θ = opposite / hypotenuse
- cos θ = adjacent / hypotenuse
- tan θ = opposite / adjacent

All triangles are right triangles (right angle marked).

1) Triangle with hypotenuse = 6 cm, adjacent to angle a = 3 cm
So cos a = adjacent/hypotenuse = 3/6 = 1/2 → a = 60°
(From Section A: cos 60° = 1/2)

2) Triangle: hypotenuse = 10 cm, opposite to angle b = 5 cm
So sin b = 5/10 = 1/2 → b = 30°

3) Triangle: legs = 2 cm and ?; hypotenuse = 2√2 cm
Check: if legs are both 2 cm, then hypotenuse = √(2² + 2²) = √8 = 2√2 → yes, it's a 45°–45°–90° triangle.
Angle c is between leg 2 cm and hypotenuse, so adjacent = 2, hypotenuse = 2√2
cos c = 2 / (2√2) = 1/√2 = √2/2 → c = 45°

4) Triangle: vertical leg = 3√3 cm, horizontal leg = 3 cm
Angle d is at bottom left, between horizontal leg (adjacent = 3) and hypotenuse.
Opposite to d = 3√3, adjacent = 3
tan d = opposite/adjacent = (3√3)/3 = √3 → d = 60°
(Since tan 60° = √3)

5) Triangle: legs = 4√2 cm (vertical), ? (horizontal), hypotenuse = 8 cm
Check: (4√2)² + x² = 8² → 32 + x² = 64 → x² = 32 → x = √32 = 4√2
So it’s an isosceles right triangle → angles = 45°, 45°, 90°
Angle e is at bottom right, between horizontal leg and hypotenuse → adjacent = 4√2, hypotenuse = 8
cos e = 4√2 / 8 = √2/2 → e = 45°

6) Triangle: horizontal leg = 6√3 cm, hypotenuse = 12 cm
Angle f is at top right, between hypotenuse and horizontal leg? Wait — look: right angle is at left, so horizontal leg is adjacent to angle f, hypotenuse is 12, adjacent = 6√3
cos f = adjacent/hypotenuse = (6√3)/12 = √3/2 → f = 30°
(From Section A: cos 30° = √3/2)

Double-check all answers for consistency — they match standard special triangle values.

Final Answer:
Section A table filled as:
sin: 0, 1/2, √2/2, √3/2, 1
cos: 1, √3/2, √2/2, 1/2, 0
tan: 0, √3/3, 1, √3, undefined

Section B:
1) 1
2) 3/2
3) 3
4) √3
5) 4√2
6) 2
7) 6
8) √3
9) 7/2

Section C:
1) 60°
2) 30°
3) 45°
4) 60°
5) 45°
6) 30°