---
excerpt: "The unit circle is a useful tool for understanding and calculating the values of the trigonometric functions for any angle. It is also a key concept in understanding more advanced mathematical topics such as complex numbers and calculus. In this lesson, wwe will explore the properties of the unit circle and provide some worked examples of its use cases."
keywords: "unit circle, unit circle chart, quadrants, Unit Circle Radians, Unit Circle in radians, blank unit circle, unit circle quiz, unit circle table"
---

<DynamicUnitCircle/>
## The Unit Circle

The unit circle is a circle of radius 1 unit and center at the origin (0,0) of a coordinate plane. Its x-intercepts are (1, 0) and (-1, 0), and the y-intercepts are (0, 1) and (0, -1).
To understand the properties of a unit circle, it is helpful to consider a line drawn from the origin to an arbitrary point (x,y) in the circle. This line is referred to as the terminal side, while the angle from the x-axis towards the Terminal side in the counterclockwise direction is known as the Central Angle as shown in the diagram.

![A unit circle with terminal side and central angle labelled::300px](/lessons/unit-circle/unit_circle.png)

The coordinates of the arbitrary point are given by the trigonometric ratios sine and cosine. The x-coordinate is equal to the cosine of the central angle, and the y-coordinate is the sine of the angle. These ratios are defined in terms of the lengths of the sides of a right triangle, just like the Pythagorean theorem.

<Snippet ribbon="Example">
<Question tag="Example 1. "> The Sine and Cosine value of $45\degree$</Question>
<Answer tag="Answer:">
Consider a 45-degree angle with its terminal side on the unit circle. If we draw a right triangle with this angle, the hypotenuse will have length 1 (since it is a radius of the unit circle), and the other two sides will have identical lengths (since the triangle is a 45-45-90 triangle).

![A unit circle with 45 degrees central angle::300px](/lessons/unit-circle/unit_circle_45.png)

The Pythagorean theorem tells us that the sum of the squares of the other two sides is equal to the square of the hypotenuse, so by applying the Pythagorean theorem we get:

$$
x^2 + y^2 = 1^2
$$

Since, x = y, we can replace y with x or vice versa:

$$
2x^2 = 1
$$

Solving for x we get:

$$
x =  \frac{1}{\sqrt{2}}
$$

hence

$$
y = \frac{1}{\sqrt{2}}
$$

Therefore, the sine and cosine of a 45-degree angle are both equal to $\frac{1}{\sqrt{2}}$.

</Answer>
</Snippet>

## Unit Circle in Radians

A radian is the size of the central angle in a unit circle that makes an arc length of 1. An angle of pi radians in a unit circle makes an arc length of pi, which is half the circumference of the unit circle.
![Unit Circle Radians of angle 1 rad and pi rad::250px](/lessons/unit-circle/unit_circle_rad.png)
Radians are a more natural unit of measurement for angles when working with trigonometric ratios and circular functions because they take the relationship between the angle and the arc length of a circle into account.

Below is a unit circle table that indicates standard angles (in radians) and the corresponding sin, cos and tan ratios.
[image of unit circle table]
![A unit circle table with ratios of standard angles::250px](/lessons/unit-circle/unit_circle_table.png)

## The Four Quadrants

The unit circle is studied by dividing it into four quadrants along the x- and y-axes as it gives more insight into the behavior of trigonometric ratios. These quadrants are labeled I, II, III, and IV, starting in the upper right and moving counterclockwise.

![The four quadrants in a unit circle::300px](/lessons/unit-circle/quadrants.png)

The trigonometric ratios of angles in a particular quadrant share common characteristics. For example, the sine value of any central angle lying in the 2nd quadrant would be positive, whereas the cosine values in the 2nd quadrant would always be negative. You can derive such characteristics for the other quadrants if you recall that the sine value of the angle is the y-coordinate, and the cosine value is the x-coordinate of a point on the unit circle.
![Signs in quadrants::250px](/lessons/unit-circle/signs_in_quadrants.png)

Even sine and cosine values of angles between different quadrants share characteristics if they have the same reference angle. A reference angle is the smallest angle the terminal side makes with the x-axis.

We can generalize this for an angle with a reference angle $\theta$ as follows:

<center>
  <FancyTable1>
  | Angle      | Quadrant | sine |  cosine    |
  | :---        |    :----:   |    :----:   |          ---: |
  | $\theta$      | I       | $\sin\theta$       | $\cos\theta$   |
  | $\pi - \theta$   | II       | $\sin\theta$       | $-\cos\theta$   |
  | $\pi + \theta$   | III        | $-\sin\theta$       | $-\cos\theta$      |
  | $2\pi -\theta$   | IV        | $-\sin\theta$       | $\cos\theta$      |
  | $2\pi+\theta$   | I        | $\sin\theta$       | $\cos\theta$      |
  </FancyTable1>
</center>

<Snippet ribbon="Example">
<Question tag="Example 2."> Obtain $\sin240\degree$ $\left(\frac{4\pi}{3} \text{rad}\right)$ using the value of  $\sin60\degree$ $\left(\frac{\pi}{3} \text{rad}\right)$</Question>
<Answer tag="Answer:">
The angles $60\degree$ and $240\degree$ lie on separate quadrants yet they share the same reference angle of 60$\degree$. So let's apply the relationships mentioned in the table earlier.
![unit circle chart::300px](/lessons/unit-circle/unit_circle_ref_angles.png)
$$
\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt3}{2}
$$

$$
\sin\left(\frac{4\pi}{3}\right) =  \sin\left(\pi + \frac{\pi}{3} \right)
$$

$$
\text{Since we know that:}
$$

$$
\sin\left(\pi + \theta\right) =  -\sin\left(\theta \right)
$$

$$
\text{Hence},
$$

$$
\sin\left(\pi + \frac{\pi}{3} \right) = -\sin\left(\frac{\pi}{3}\right)
$$

$$
\text{Therefore, } \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt3}{2}
$$

</Answer>
</Snippet>

## Applications of the unit circle

One of the primary uses of the unit circle is to evaluate and graph circular functions, such as sines and cosines. These functions are periodic, meaning that they repeat over a certain interval. The interval for sines and cosines is 360 degrees or 2π radians. The values of these functions can be determined using the coordinates of points on the unit circle. The interactive unit circle show right on top of this page demonstrates how the properties of a unit circle extend to periodic motions such as those of planets, pendulums, etc.

<Snippet ribbon="Question">
<Question tag="Unit Circle Quiz."> Complete the blank unit circle chart, using your knowledge of quadrants and the values filled on the chart.
![A blank unit circle chart::400px](/lessons/unit-circle/blank_unit_circle.png)</Question>
<Answer tag="Solution:">
Here's how you'd work out the missing ratios for the angle $\frac{5\pi}{4}$ rad.

<strong>Step 1.</strong> Identify the reference angle

As $\frac{5\pi}{4}$ lies on the 3rd quadrant its reference angle is determined by subtracting it by $\pi$.

$$
\frac{5\pi}{4} - \pi = \frac{5\pi}{4} -\frac{4\pi}{4}
$$

$$
\frac{5\pi}{4} - \pi = \frac{\pi}{4}
$$

$$
\text{Hence the reference angle is } \frac{\pi}{4} \text{ rad}


$$

<strong>Step 2.</strong> Infer the desired coordinates from the reference angle.

$$
\text{The coordinates of } \frac{\pi}{4} \text{ : } \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)
$$

The magnitude of the x and y coordinates of $ \frac{5\pi}{4} $ would be identical but the signs would be negative as it lies in the 3rd quadrant.

$$
\text{Therefore},
$$

$$
\text{The coordinates of } \frac{5\pi}{4} \text{ : } \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)
$$

You can use a similar approach to figure out the rest of the missing coordinates and obtain the complete unit circle chart.
![A complete unit circle chart::400px](/lessons/unit-circle/unit_circle_chart_complete.png)

</Answer>
</Snippet>


The unit circle

---
excerpt: "Triangles come up in a lot of real-life applications such as GPS, surveying, 3d modeling, game graphics, and more. There are infinitely many triangle formations, yet we find many patterns and proportionalities among their properties. This article explains the various theorems and formulas used to extract properties of triangles supported by visualizations and examples."
---

## The Triangle

The triangle is one of the simplest polygons there is. It is a polygon enclosed by three straight segments. A triangle has three vertices, three sides, and three angles. The length of sides and the size of angles of a triangle are related. We can use these relationships to determine the lengths of sides, angles, area, and more. 

![An obtuse angle triangle, a right triangle, and an acute angle triangle::300px](/lessons/triangle/triangles.png)

## The Right Triangle

The right triangle is the most frequently studied type of triangle. It is at the core of trigonometric ratios and the Pythagorean theorem. Some of the common properties of right triangles are:

<NumberedList>
  <li>
    <div>
      One angle is $90\degree$
    </div>
  </li>
  <li>
    <div>
      The sum of the two remaining angles is $90\degree$
    </div>
  </li>
  <li>
    <div>
      The longest side is known as the hypotenuse
    </div>
  </li>
  <li>
    <div>
      The right angle is always on the opposite of the hypotenuse
    </div>
  </li>
</NumberedList>

![Two right triangles with hypotenuse labelled::350px](/lessons/triangle/right_triangles.png)

## The area of a Right Triangle

The formula for the area of a right triangle is 
$$
\frac{1}{2} \times \text{base} \times \text{height}
$$ 
It is actually derived from the area of a rectangle. Let’s use an example to demonstrate the relationship between the area of a rectangle and a triangle.

<Snippet ribbon="Example">
<Question tag="Example 1. "> How to find the area of a right triangle of base 6 cm and height 4 cm</Question>
<Answer tag="Answer:">
If we enclose the right triangle in a 4x6 rectangle we get two identical right triangles, each of base 6 cm and height 4 cm.

![A 4x6 rectangle divided into right triangles across its diagonal::250px](/lessons/triangle/area_proof_1.png)

Since the area of a rectangle is the product of its width and length, we get:
$$
\text{Area of the Rectangle} = 6\text{cm} \times 4\text{cm}
$$
$$
\text{Area of the Rectangle} = 24 \text{ cm}^2
$$
The two triangles that make up the entire rectangle should have an area of 24 $\text{cm}^2$. Hence, the area of each right triangle would be half of 24 $\text{cm}^2$.
$$
\text{Area of one right triangle} = \frac{1}{2}\times 24
$$
$$
\text{Area of one right triangle} = 12 cm^2
$$

Hence, the area of a right triangle with base 6cm and height 4cm is 12 $\text {cm}^2$

</Answer>
</Snippet>

The area of an arbitrary right triangle will always be half the area of the rectangle that encloses it.

## Area of a Non Right Triangle

Determining the area of non right triangles has significant use cases in construction, architectural design, and even estimating areas of non-regular polygons. The relationship between the area of the rectangle and triangle illustrated in the previous example holds for non-right triangles too. 

<strong>Proof:</strong>
Shown below is a non right triangle

![A non right triangle ABC::250px](/lessons/triangle/non_right_triangle.png)
Let’s enclose the triangle ABC inside a rectangle AEFC.
You may notice that the triangle BCF(shaded in red) and BCD are congruent, while triangle ABE (shaded in green) and triangle ABD are congruent. 
![Non right triangle enclosed in rectangle::250px](/lessons/triangle/non_right_area_proof.png)
From visual inspection, we can conclude that the area of ABC is equal to the sum of the areas of triangle ABE and BCF. Hence, triangle ABC occupies an area that is half of the entire rectangle AEFC. 
$$
\text{Area of } AEFC = AC\times AE
$$
$$
\text{Area of } AEFC = AC\times h
$$
Therefore, 
$$
\text{Area of  triangle } ABC = \frac{1}{2}\times AC \times h
$$
Did you notice that this is similar to the formula of the area of a right triangle?
The formula for the area of a right triangle is a specific case of the area of a non right triangle. 

Therefore the general formula fit for both right and non right triangles can be given as:  
$$
\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{perpendicular height}
$$

<Snippet ribbon="Question">
<Question tag="Question 1."> Find the area of the following triangle ![An obtuse angle triangle::200px](/lessons/triangle/question_1.png)</Question>
<Answer tag="Solution:">
<strong>Step 1.</strong> Identify the base and perpendicular height.

$$
\text{Length of base (AB) = } 7 \text{ cm}
$$
$$
\text{Perpendicular height} = 5 \text{ cm}
$$

<strong>Step 2.</strong> Apply the formula for the area of a non right triangle

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{perpendicular height}
$$
$$
\text{Area} = \frac{1}{2} \times 7 \times 5
$$
$$
\text{Area} = 17.5 \text{ cm}^2
$$

</Answer>
</Snippet>

When the perpendicular height is unknown, the alternate formula $\text{Area} = \frac{1}{2} \times a \times b \times \sin(C) $ is beneficial.

This formula is also known as the Side-Angle-Side area formula and is useful because it applies to both right and non right triangles. 

Note that when C = $90\degree$, the formula resolves to $\frac{1}{2}ab$, the formula for the area of a right triangle discussed earlier.


Let’s put the universal formula into action using an example. 

<Snippet ribbon="Example">
<Question tag="Example 2. "> Find the area of a non right triangle when the height isn’t given![A acute angle triangle::250px](/lessons/triangle/example_2.png)</Question>
<Answer tag="Answer:">
When applying the SAS area formula, identifying the parameters is essential. 
<NumberedList>
<li>
<strong>a</strong> denotes the side opposite to the angle A
</li>
<li>
<strong>b</strong> denotes the side opposite angle B
</li>
<li>
<strong>c</strong> denotes the side opposite angle C
</li>
</NumberedList>

According to the definitions
$
\text{a} = 10.3 \text{ cm },\text{ b} = 8.4\text{ cm }, \text{c} = 7.2 \text{ cm ,}
\text{ C} = 44\degree
$
$$
\text{Area of triangle } ABC = \frac{1}{2} \times a \times b \times \sin(C)
$$

$$
\text{Area of triangle } ABC = \frac{1}{2} \times 10.3 \times 8.4 \times \sin(44\degree)
$$
$$
\text{Area of triangle } ABC = 29.9 \text{ cm}^2
$$


</Answer>
</Snippet>

## Pythagoras theorem

Also known as the right angle formula, the Pythagoras theorem describes the mathematical relationship between the lengths of the three sides of a right triangle. 

The famous Greek philosopher Pythagoras discovered this mathematical relationship, and the following shows the geometric proof he used to derive the theorem.

![Geometric proof of the pythagoras theorem::250px](/lessons/triangle/pythagorean_theorem.png)

The above diagram shows that the sum of the areas of the two squares of lengths AB and AC is equal to the sum of the area of the square of length equivalent to the hypotenuse of the right triangle. 

For a triangle of sides a, b, and c, where c is the hypotenuse, the Pythagorean theorem formula can be given as:
$$
a^2 + b^2 = c^2
$$
<Snippet ribbon="Example">
<Question tag="Example 3. "> A cable car moves from point H to I. During the journey, it descends a height of 75 m and covers a horizontal distance of 100 m. Find the length of the cable.![A cable car descending from a hill::250px](/lessons/triangle/example_3.png)</Question>
<Answer tag="Answer:">
Triangle IKH is a right triangle as line HK is vertical, and IK is horizontal. 

To find the length of HI we can apply the Pythagorean theorem.

$$
a^2 + b^2 = c^2
$$
$$
75^2 + 100^2 = c^2
$$
$$
5625 + 10000 = c^2
$$
$$
c^2 = 15625
$$
$$
c = \sqrt{15625}
$$
$$
c = 125
$$
Hence the length of the cable must be 125 meters.


</Answer>
</Snippet>

## Triangle Properties Quiz

<Snippet ribbon="Question">
<Question tag="Triangle Properties Quiz."> Determine the area of the triangle given below
![An obtuse angle triangle with certain side lengths labelled](/lessons/triangle/quiz_image.png)</Question>
<Answer tag="Solution:">
Here's how you'd work out the area of non right triangle ABC.
The length of base AB is given as 7cm. The perpendicular height from AB to the vertex C is not given. Hence it needs to be calculated.


<strong>Step 1.</strong> Use the Pythagorean theorem to determine h

![An obtuse angle triangle with perpendicular height marked](/lessons/triangle/quiz_step_1.png)

BCD is right triangle, and the sides BC and BD are:
$$
BC = 5 \text{ cm}
$$
$$
BD = 10\text{ cm} - 7 \text{ cm} = 3 \text{ cm}
$$
We can apply the Pythagorean theorem to the right triangle BCD.
$$
h^2 + BD^2 = BC^2
$$
$$
h^2 = BC^2 - BD^2
$$
$$
h^2 = 5^2 - 3^2
$$
$$
h^2 = 25 - 9
$$
$$
h^2 = 16
$$
$$
h = 4 \text{ cm}
$$
<strong>Step 2.</strong> Workout the area using the non right triangle area formula.

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{perpendicular height}
$$
$$
\text{Area} = \frac{1}{2} \times AB \times h
$$
$$
\text{Area} = \frac{1}{2} \times 7 \text{ cm} \times 4 \text{ cm}
$$
$$
\text{Area} = 14 \text{ cm}^2
$$
</Answer>
</Snippet>


Lessons

The unit circle

The triangle