Trigonometry Basics
by 박구글1 mo ago
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trigonometry
sin
cos
tan
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1. What Is Trigonometry?

Trigonometry is the study of relationships between the angles and side lengths of triangles. The word comes from Greek roots meaning “triangle measurement.”
Trigonometry is used in many real situations:
  • Measuring heights of buildings or trees
  • Navigation (ships, airplanes, GPS)
  • Engineering and construction
  • Computer graphics and game design
  • Physics (waves, motion, forces)
In Grade 10, trigonometry usually begins with right triangles (triangles with one 90 ∘ 90 ∘ 90^\circ angle).

1.1. Basic triangle vocabulary

A triangle has:
  • Sides (three line segments)
  • Angles (three interior angles)
A right triangle has:
  • One right angle: 90 ∘ 90 ∘ 90^\circ
  • Two acute angles (less than 90 ∘ 90 ∘ 90^\circ), and they always add up to 90 ∘ 90 ∘ 90^\circ.

1.2. The hypotenuse

In a right triangle, the side opposite the right angle is the hypotenuse.
  • The hypotenuse is always the longest side.

1.3 Opposite and adjacent (relative to an angle)

When working with trigonometry, we usually focus on one acute angle, often labeled θ θ \theta.
Relative to θ θ \theta:
  • The opposite side is the side directly across from angle θ θ \theta.
  • The adjacent side is the side next to angle θ θ \theta that is not the hypotenuse.
It is important to remember:
  • “Opposite” and “adjacent” depend on which angle you choose.
  • The hypotenuse is always the hypotenuse.

2. The Three Main Trigonometric Ratios

For right triangles, trigonometry begins with three ratios:
  • Sine
  • Cosine
  • Tangent
These are defined using the side lengths of a right triangle.
Let θ θ \theta be an acute angle in a right triangle.

2.1 Sine

sin ⁡ ( θ ) = opposite hypotenuse sin ⁡ ( θ ) = opposite hypotenuse \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}

2.2 Cosine

cos ⁡ ( θ ) = adjacent hypotenuse cos ⁡ ( θ ) = adjacent hypotenuse \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}

2.3 Tangent

tan ⁡ ( θ ) = opposite adjacent tan ⁡ ( θ ) = opposite adjacent \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}

2.4 A memory aid: SOH-CAH-TOA

A common way to remember the ratios is:
  • SOH: Sine = Opposite / Hypotenuse
  • CAH: Cosine = Adjacent / Hypotenuse
  • TOA: Tangent = Opposite / Adjacent

3. Using Trig Ratios in Right Triangles

3.1 Step-by-step problem method

When solving right-triangle trig problems:
  1. Draw a diagram (if one is not given).
  2. Label the angle θ θ \theta.
  3. Identify opposite, adjacent, and hypotenuse relative to θ θ \theta.
  4. Choose the correct trig ratio (sin, cos, or tan).
  5. Substitute values and solve.

3.2 Example 1: Find a missing side using sine

A right triangle has hypotenuse 10 cm 10 cm 10\,\text{cm} and an angle θ = 30 ∘ θ = 30 ∘ \theta = 30^\circ. Find the side opposite θ θ \theta.
Use sine:
sin ⁡ ( 30 ∘ ) = opposite 10 sin ⁡ ( 30 ∘ ) = opposite 10 \sin(30^\circ) = \frac{\text{opposite}}{10}
Since sin ⁡ ( 30 ∘ ) = 1 2 sin ⁡ ( 30 ∘ ) = 1 2 \sin(30^\circ)=\frac{1}{2}:
1 2 = opposite 10 1 2 = opposite 10 \frac{1}{2} = \frac{\text{opposite}}{10}
Multiply both sides by 10 10 10:
opposite = 5 cm opposite = 5 cm \text{opposite} = 5\,\text{cm}

3.3 Example 2: Find a missing side using cosine

A right triangle has adjacent side 12 m 12 m 12\,\text{m} (next to θ θ \theta) and angle θ = 60 ∘ θ = 60 ∘ \theta = 60^\circ. Find the hypotenuse.
Use cosine:
cos ⁡ ( 60 ∘ ) = 12 hypotenuse cos ⁡ ( 60 ∘ ) = 12 hypotenuse \cos(60^\circ) = \frac{12}{\text{hypotenuse}}
Since cos ⁡ ( 60 ∘ ) = 1 2 cos ⁡ ( 60 ∘ ) = 1 2 \cos(60^\circ)=\frac{1}{2}:
1 2 = 12 hypotenuse 1 2 = 12 hypotenuse \frac{1}{2} = \frac{12}{\text{hypotenuse}}
Swap and solve:
hypotenuse = 24 m hypotenuse = 24 m \text{hypotenuse} = 24\,\text{m}

3.4 Example 3: Find an angle using tangent

A right triangle has opposite side 8 cm 8 cm 8\,\text{cm} and adjacent side 6 cm 6 cm 6\,\text{cm}. Find θ θ \theta.
Use tangent:
tan ⁡ ( θ ) = 8 6 = 4 3 tan ⁡ ( θ ) = 8 6 = 4 3 \tan(\theta) = \frac{8}{6} = \frac{4}{3}
To find θ θ \theta, use an inverse tangent (calculator):
θ = tan − 1 ⁡ ( 4 3 ) θ = tan − 1 ⁡ 4 3 \theta = \tan^{-1}\left(\frac{4}{3}\right)
Approximate:
θ ≈ 53.1 ∘ θ ≈ 53.1 ∘ \theta \approx 53.1^\circ

4. Inverse Trigonometric Functions

When you know a trig ratio value and need the angle, you use inverse trig functions:
  • sin − 1 ⁡ ( x ) sin − 1 ⁡ ( x ) \sin^{-1}(x) or arcsin ⁡ ( x ) arcsin ⁡ ( x ) \arcsin(x)
  • cos − 1 ⁡ ( x ) cos − 1 ⁡ ( x ) \cos^{-1}(x) or arccos ⁡ ( x ) arccos ⁡ ( x ) \arccos(x)
  • tan − 1 ⁡ ( x ) tan − 1 ⁡ ( x ) \tan^{-1}(x) or arctan ⁡ ( x ) arctan ⁡ ( x ) \arctan(x)
Example:
If sin ⁡ ( θ ) = 0.6 sin ⁡ ( θ ) = 0.6 \sin(\theta)=0.6, then
θ = sin − 1 ⁡ ( 0.6 ) ≈ 36.9 ∘ θ = sin − 1 ⁡ ( 0.6 ) ≈ 36.9 ∘ \theta = \sin^{-1}(0.6) \approx 36.9^\circ

4.1 Calculator notes

  • Make sure your calculator is in degree mode when working with degrees.
  • Write answers to a reasonable precision (often nearest tenth of a degree or nearest hundredth for lengths, depending on the question).

5. Special Right Triangles

Some triangles have angles and side ratios that are especially important.

5.1 The 45 ∘ 45 ∘ 45^\circ- 45 ∘ 45 ∘ 45^\circ- 90 ∘ 90 ∘ 90^\circ triangle

This is an isosceles right triangle.
If each leg is length 1 1 1, then hypotenuse is:
1 2 + 1 2 = 2 1 2 + 1 2 = 2 \sqrt{1^2+1^2}=\sqrt{2}
So the side ratio is:
1 : 1 : 2 1 : 1 : 2 1 : 1 : \sqrt{2}
Trig values:
sin ⁡ ( 45 ∘ ) = 1 2 = 2 2 , cos ⁡ ( 45 ∘ ) = 2 2 , tan ⁡ ( 45 ∘ ) = 1 sin ⁡ ( 45 ∘ ) = 1 2 = 2 2 , cos ⁡ ( 45 ∘ ) = 2 2 , tan ⁡ ( 45 ∘ ) = 1 \sin(45^\circ)=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2},\quad \cos(45^\circ)=\frac{\sqrt{2}}{2},\quad \tan(45^\circ)=1

5.2 The 30 ∘ 30 ∘ 30^\circ- 60 ∘ 60 ∘ 60^\circ- 90 ∘ 90 ∘ 90^\circ triangle

Side ratios:
1 : 3 : 2 1 : 3 : 2 1 : \sqrt{3} : 2
(Where 2 2 2 is the hypotenuse, 1 1 1 is opposite 30 ∘ 30 ∘ 30^\circ, and 3 3 \sqrt{3} is opposite 60 ∘ 60 ∘ 60^\circ.)
Trig values:
sin ⁡ ( 30 ∘ ) = 1 2 , cos ⁡ ( 30 ∘ ) = 3 2 , tan ⁡ ( 30 ∘ ) = 1 3 = 3 3 sin ⁡ ( 30 ∘ ) = 1 2 , cos ⁡ ( 30 ∘ ) = 3 2 , tan ⁡ ( 30 ∘ ) = 1 3 = 3 3 \sin(30^\circ)=\frac{1}{2},\quad \cos(30^\circ)=\frac{\sqrt{3}}{2},\quad \tan(30^\circ)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}
sin ⁡ ( 60 ∘ ) = 3 2 , cos ⁡ ( 60 ∘ ) = 1 2 , tan ⁡ ( 60 ∘ ) = 3 sin ⁡ ( 60 ∘ ) = 3 2 , cos ⁡ ( 60 ∘ ) = 1 2 , tan ⁡ ( 60 ∘ ) = 3 \sin(60^\circ)=\frac{\sqrt{3}}{2},\quad \cos(60^\circ)=\frac{1}{2},\quad \tan(60^\circ)=\sqrt{3}

6. The Pythagorean Theorem and Trigonometry

Before trigonometry, you may have learned the Pythagorean Theorem:
a 2 + b 2 = c 2 a 2 + b 2 = c 2 a^2+b^2=c^2
In a right triangle:
  • c c c is the hypotenuse
  • a a a and b b b are the legs

6.1 How they connect

  • The Pythagorean theorem helps you find a missing side when you know two sides.
  • Trigonometry helps you find a missing side when you know one side and one angle.
Example:
If a right triangle has legs 9 9 9 and 12 12 12, then
9 2 + 12 2 = c 2 9 2 + 12 2 = c 2 9^2+12^2=c^2
81 + 144 = 225 = c 2 81 + 144 = 225 = c 2 81+144=225=c^2
c = 15 c = 15 c=15

7. Angles of Elevation and Depression (Real-World Trig)

Trigonometry is often used to find heights and distances.

7.1 Angle of elevation

An angle of elevation is the angle you look upward from a horizontal line.
Example idea: Standing on the ground, looking up at the top of a building.

7.2 Angle of depression

An angle of depression is the angle you look downward from a horizontal line.
Example idea: Looking down from a lighthouse to a boat.

7.3 Example: Measuring a building height

You stand 40 m 40 m 40\,\text{m} from a building. The angle of elevation to the top is 35 ∘ 35 ∘ 35^\circ. Approximate the building’s height (ignore your eye height).
Here, the height is the opposite side, and 40 40 40 m is the adjacent side.
Use tangent:
tan ⁡ ( 35 ∘ ) = height 40 tan ⁡ ( 35 ∘ ) = height 40 \tan(35^\circ) = \frac{\text{height}}{40}
height = 40 tan ⁡ ( 35 ∘ ) height = 40 tan ⁡ ( 35 ∘ ) \text{height} = 40\tan(35^\circ)
Approximate:
height ≈ 40 ( 0.700 ) ≈ 28.0 m height ≈ 40 ( 0.700 ) ≈ 28.0 m \text{height} \approx 40(0.700) \approx 28.0\,\text{m}

8. Solving Right Triangles

To solve a right triangle means to find all unknown sides and angles.
Since one angle is 90 ∘ 90 ∘ 90^\circ, you usually need:
  • One other angle
  • And at least one side length
Then you can find the remaining sides and angles.

8.1 Example: Solve a triangle

A right triangle has hypotenuse 20 20 20 and an angle θ = 25 ∘ θ = 25 ∘ \theta=25^\circ.
Find:
  • opposite side
  • adjacent side
  • the other acute angle
Other acute angle:
90 ∘ − 25 ∘ = 65 ∘ 90 ∘ − 25 ∘ = 65 ∘ 90^\circ-25^\circ = 65^\circ
Opposite side:
sin ⁡ ( 25 ∘ ) = opp 20 ⇒ opp = 20 sin ⁡ ( 25 ∘ ) sin ⁡ ( 25 ∘ ) = opp 20 ⇒ opp = 20 sin ⁡ ( 25 ∘ ) \sin(25^\circ)=\frac{\text{opp}}{20} \Rightarrow \text{opp} = 20\sin(25^\circ)
Adjacent side:
cos ⁡ ( 25 ∘ ) = adj 20 ⇒ adj = 20 cos ⁡ ( 25 ∘ ) cos ⁡ ( 25 ∘ ) = adj 20 ⇒ adj = 20 cos ⁡ ( 25 ∘ ) \cos(25^\circ)=\frac{\text{adj}}{20} \Rightarrow \text{adj} = 20\cos(25^\circ)
Approximations:
opp ≈ 20 ( 0.423 ) = 8.46 opp ≈ 20 ( 0.423 ) = 8.46 \text{opp} \approx 20(0.423)=8.46
adj ≈ 20 ( 0.906 ) = 18.12 adj ≈ 20 ( 0.906 ) = 18.12 \text{adj} \approx 20(0.906)=18.12
So the triangle is approximately:
  • opposite ≈ 8.46 ≈ 8.46 \approx 8.46
  • adjacent ≈ 18.12 ≈ 18.12 \approx 18.12
  • hypotenuse = 20 = 20 =20
  • angles: 25 ∘ , 65 ∘ , 90 ∘ 25 ∘ , 65 ∘ , 90 ∘ 25^\circ,\ 65^\circ,\ 90^\circ

9. Common Mistakes and How to Avoid Them

  1. Mixing up opposite and adjacent
    • Always label them relative to the chosen angle.
  2. Using the wrong trig ratio
    • Ask: which sides do I have, and which side do I need?
  3. Calculator in the wrong mode
    • Degree mode vs radian mode. Grade 10 problems are usually in degrees.
  4. Rounding too early
    • Keep extra digits during calculations; round at the end.
  5. Forgetting units
    • Length units: meters, centimeters, etc.
    • Angle units: degrees ( ∘ ∘ ^\circ)

10. Practice Questions

A. Identify sides

  1. In a right triangle, one acute angle is θ θ \theta. Write which side is always the hypotenuse.
  2. Explain how to decide which side is “opposite” θ θ \theta.

B. Use trig ratios (find a side)

  1. A right triangle has hypotenuse 15 15 15 and angle θ = 40 ∘ θ = 40 ∘ \theta=40^\circ. Find the opposite side.
opp = 15 sin ⁡ ( 40 ∘ ) opp = 15 sin ⁡ ( 40 ∘ ) \text{opp} = 15\sin(40^\circ)
  1. A right triangle has adjacent side 9 9 9 and angle θ = 28 ∘ θ = 28 ∘ \theta=28^\circ. Find the hypotenuse.
hyp = 9 cos ⁡ ( 28 ∘ ) hyp = 9 cos ⁡ ( 28 ∘ ) \text{hyp} = \frac{9}{\cos(28^\circ)}
  1. A right triangle has opposite side 7 7 7 and adjacent side 11 11 11. Find the angle.
θ = tan − 1 ⁡ ( 7 11 ) θ = tan − 1 ⁡ 7 11 \theta = \tan^{-1}\left(\frac{7}{11}\right)

C. Word problems

  1. You are 50 m 50 m 50\,\text{m} from a flagpole. The angle of elevation to the top is 22 ∘ 22 ∘ 22^\circ. Find the height.
height = 50 tan ⁡ ( 22 ∘ ) height = 50 tan ⁡ ( 22 ∘ ) \text{height} = 50\tan(22^\circ)
  1. From the top of a cliff, you look down at a boat. The angle of depression is 12 ∘ 12 ∘ 12^\circ. The cliff is 80 m 80 m 80\,\text{m} high. How far is the boat from the base of the cliff (horizontal distance)?
tan ⁡ ( 12 ∘ ) = 80 d ⇒ d = 80 tan ⁡ ( 12 ∘ ) tan ⁡ ( 12 ∘ ) = 80 d ⇒ d = 80 tan ⁡ ( 12 ∘ ) \tan(12^\circ)=\frac{80}{d}\Rightarrow d=\frac{80}{\tan(12^\circ)}

11. Summary

  • Trigonometry connects angles and side lengths in right triangles.
  • The three main ratios are:
sin ⁡ ( θ ) = opp hyp , cos ⁡ ( θ ) = adj hyp , tan ⁡ ( θ ) = opp adj sin ⁡ ( θ ) = opp hyp , cos ⁡ ( θ ) = adj hyp , tan ⁡ ( θ ) = opp adj \sin(\theta)=\frac{\text{opp}}{\text{hyp}},\quad \cos(\theta)=\frac{\text{adj}}{\text{hyp}},\quad \tan(\theta)=\frac{\text{opp}}{\text{adj}}
  • Use inverse trig functions to find angles:
θ = sin − 1 ⁡ ( x ) , θ = cos − 1 ⁡ ( x ) , θ = tan − 1 ⁡ ( x ) θ = sin − 1 ⁡ ( x ) , θ = cos − 1 ⁡ ( x ) , θ = tan − 1 ⁡ ( x ) \theta=\sin^{-1}(x),\quad \theta=\cos^{-1}(x),\quad \theta=\tan^{-1}(x)
  • Always draw and label the triangle carefully, and check calculator settings.