Plane Geometry and Trigonometry

Angles, triangles, circles, and basic trigonometry for ACT

Plane Geometry and Trigonometry

Angles

Types:

  • Acute: <90°< 90°
  • Right: =90°= 90°
  • Obtuse: >90°> 90° and <180°< 180°
  • Straight: =180°= 180°

Complementary angles: Add to 90°90° Supplementary angles: Add to 180°180°

Triangles

Angle Sum

All triangles: angles sum to 180°180°

Pythagorean Theorem

For right triangles: a2+b2=c2a^2 + b^2 = c^2 (where cc is the hypotenuse)

Special Right Triangles

45-45-90: sides: x,x,x2\text{sides: } x, x, x\sqrt{2}

30-60-90: sides: x,x3,2x\text{sides: } x, x\sqrt{3}, 2x

Area

A=12bhA = \frac{1}{2}bh

Circles

Circumference: C=2πr=πdC = 2\pi r = \pi d

Area: A=πr2A = \pi r^2

Arc length: Arc=θ360°×2πr\text{Arc} = \frac{\theta}{360°} \times 2\pi r

Trigonometry

SOH-CAH-TOA

sinθ=oppositehypotenuse\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}

cosθ=adjacenthypotenuse\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

tanθ=oppositeadjacent\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

Common Values

| Angle | sin | cos | tan | |-------|-----|-----|-----| | 30°30° | 12\frac{1}{2} | 32\frac{\sqrt{3}}{2} | 13\frac{1}{\sqrt{3}} | | 45°45° | 22\frac{\sqrt{2}}{2} | 22\frac{\sqrt{2}}{2} | 11 | | 60°60° | 32\frac{\sqrt{3}}{2} | 12\frac{1}{2} | 3\sqrt{3} |

Law of Sines (for any triangle)

asinA=bsinB=csinC\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

ACT Tips

  • Memorize special right triangles! They appear often
  • Draw diagrams if not provided
  • Use calculator for trig values (set to degrees!)
  • Common Pythagorean triples: 3-4-5, 5-12-13, 8-15-17

📚 Practice Problems

1Problem 1easy

Question:

In a right triangle, one leg is 3 and the other leg is 4. What is the hypotenuse?

💡 Show Solution

Solution:

Use Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2 32+42=c23^2 + 4^2 = c^2 9+16=c29 + 16 = c^2 25=c225 = c^2 c=5c = 5

Answer: 55

ACT Tip: This is the 3-4-5 Pythagorean triple - memorize it!

2Problem 2medium

Question:

A circle has a diameter of 10 cm. What is its area?

💡 Show Solution

Solution:

Given diameter d=10d = 10 cm, so radius r=5r = 5 cm

Area formula: A=πr2A = \pi r^2 A=π(5)2A = \pi (5)^2 A=25π cm2A = 25\pi \text{ cm}^2

Answer: 25π25\pi cm² (or approximately 78.54 cm²)

ACT Tip: Leave answer in terms of π\pi unless told to approximate!

3Problem 3hard

Question:

In a right triangle, if the angle is 30°30° and the hypotenuse is 12, what is the length of the side opposite the 30°30° angle?

💡 Show Solution

Solution:

This is a 30-60-90 triangle with ratio x:x3:2xx : x\sqrt{3} : 2x

The hypotenuse is 2x=122x = 12 x=6x = 6

The side opposite 30°30° is the shortest side = xx

Answer: 66

Alternative using trig: sin30°=opposite12\sin 30° = \frac{\text{opposite}}{12} 12=opposite12\frac{1}{2} = \frac{\text{opposite}}{12} opposite=6\text{opposite} = 6

ACT Tip: Recognize special triangles to solve faster!