---
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>


The triangle

---
excerpt: "The cone is a fundamental shape with a range of applications in various fields, from construction to aerodynamics. There are various types of cones and different measures used to describe a cone. Whether you're a student, engineer, or just interested in the geometry of the natural world, this article will help you learn the skills of working out the properties of right circular cones and other variants."
description: "Learn the basics of the cone in geometry with Tutordb's comprehensive lesson. Understand its properties, formulas, and applications with easy-to-follow explanations, examples, and exercises. Start mastering geometry today!"
keywords: ["Cone", "Cone Geometry", "Cone Properties", "Cone Formulas"]
---

The cone is a primitive three-dimensional shape that appears in nature and has a wide range of applications in construction, aerodynamics, and various industrial fields. Understanding the properties of cones is essential to professionals and enthusiasts alike, as cones have relationships with other 3D objects such as pyramids, cylinders, and spheres. In this article, we will explore cones and their variants and learn how to determine the properties of cones, including surface area and volume. Through step-by-step examples and clear explanations, you will gain a thorough understanding of this fundamental shape and its practical uses.

![Examples of cones in real world::200px](/lessons/cone/image_of_cones_and_real_world_cone_objects.png)

## The Right Circular Cone

The cone can be described as a 3D shape that tapers smoothly from a flat circular base to an apex. The principle parameters that are used to describe a cone are the radius of the base, the perpendicular height from the base to the apex, and slant length (the distance from a point on the edge of the base to the apex).

The right circular cone - If the axis perpendicular to the base of the cone through the center directly passes through the apex, which can also be stated as the line connecting the apex and center of the base being perpendicular to the surface of the base, if this is true then the cone is considered as a right circular cone.
![The right circular cone::300px](/lessons/cone/right_circular_cone.png)

## Other Variations of the conventional cone
<NumberedList>

  <li>
    <div>
      A truncated cone can be described as a 3D object that has a flat base that is tapered smoothly to another flat face. It’s easy to imagine this as slicing a cone with a plane parallel to its base across a given height and taking away the smaller cone formed above the slice.
![The truncated cone with labels of surfaces and dimensions::200px](/lessons/cone/3d_truncated_cone_with_surfaced_labeled.png)
    </div>
  </li>
  <li>
    <div>
     The oblique cone - On contrary to the right circular cone, if the line connecting the apex and center of the base is at an angle other than 90 degrees, then it is an oblique cone.
![The oblique cone::200px](/lessons/cone/oblique_cone.png)
    </div>
  </li>
  <li>
    <div>
      The elliptical cone is another variant of the cone, where the distinction is in the shape of the base. Rather than the circular base of the conventional cone, the elliptical cone, as its name suggests, has an ellipse as its base.
    </div>
  </li>
</NumberedList>

## The area of a cone

A cone’s surface is made up of two major surfaces, a flat circular surface at the base, and a curved surface. Finding the area of the flat surface can be done easily by applying the formula for the area of a circle, but the curved surface is more challenging.
![A right circular cone 3D depiction with surfaces colored::400px](/lessons/cone/cone_area_3D.png)
Since it is difficult to work with curved surfaces it’s quite common to cut and flatten out the curved surface when figuring out the area. The diagram below shows both surfaces of a right circular cone laid out on a 2D plane.
![Surfaces of a right circular cone 2D and 3D depiction::400px](/lessons/cone/cone_area_surfaces.png)
You may notice that when the curved surface is flattened out (shown by the blue shape) it is a sector of a circle of radius equal to the slant length of the cone, and the perimeter is equivalent to the circumference of the base ($2\pi\text{r}$).
Let's walkthrough how to calculate the total surface area of a cone step by step using an example.

<Snippet ribbon="Example">
<Question tag="Example 1. "> Determine the surface area of a right circular cone in terms of $\pi$ ![An Example for calculating surface area of a cone::200px](/lessons/cone/cone_with_radius_and_slant_height_labeled.png)</Question>
<Answer tag="Answer:">
<strong>Step 1:</strong> Calculate the area of the base.

The base is a circle of radius 5cm, hence:

$$
\text{Area} = \pi\times\text{5}^2
$$

$$
\text{Area} = 25\pi\text{ cm}^2
$$

<strong>Step 2:</strong> Calculate the area of the curved surface.

As shown previous the 2D net of athe curved surface is a sector of radius l and perimeter $2\pi\text{r}$.
First we need to build an equation that relates the measures r and l to the area of the sector. We can use ratios for this.

$$
\frac{\text{Area of Sector}}{\text{Area of circle of radius l}} = \frac{\text{Circumference of Sector}}{\text{Circumference of whole circle}}
$$

$$
\frac{\text{A}}{\pi\text{l}^2} = \frac{2\pi\text{r}}{2\pi\text{l}}
$$

Simplifying this we get:

$$
\text{A} = \pi\text{r}\text{l}
$$

Substitute r & l in the equation to obtain the curved surface area

$$
\text{A}_\text{curved} = \pi\times\text{5}\times\text{13}
$$

$$
\text{A}_\text{curved} = 65\pi \text{ cm}^2
$$

<strong>Step 3:</strong> Calculate the total surface area.

$$
\text{Surface Area} = \text{Curved surface area } + \text{Base area}
$$

Substitute results from Step 1 and 2.

$$
\text{Surface Area} = 65\pi + 25\pi
$$

$$
\text{Surface Area} = 90\pi \text{cm}^2
$$

</Answer>
</Snippet>

## How to find the volume of a cone

The formula for the volume of a cone is given by $\frac{1}{3}\pi\text{r}^2\text{h}$, where <strong>r</strong> is the radius of the base and <strong>h</strong> is the perpendicular height from the base to the apex.
As with any other formula, memorizing this may be challenging but it may help to think of the volume of the cone as a ratio to the volume of a cylinder. Since the cylinder is a 3D object with a uniform circular cross-section across its height, its volume is given by the $\text{Area of cross section}\times\text{height}$, i.e $\pi\text{r}^2\text{h}$. A cone with an identical base and altitude would have a volume that is $\frac{1}{3}$ of that of the cylinder.

<Snippet ribbon="Example">
<Question tag="Example 2. "> Determine the volume of a right circular cone ![An Example for applying volume of a cone formula::200px](/lessons/cone/cone_with_height_and_radius_labeled.png)</Question>
<Answer tag="Answer:">
<strong>Step 1:</strong> Identify the measures of the cone
The two measures that can be extracted from the diagram are the radius of the base(r) and the perpendicular height (h).
$$
\text{h} = 9\text{cm}
$$
$$
\text{r} = 5\text{cm}
$$

<strong>Step 2:</strong> Apply the formula
Since we have all the parameters required we can dirctly plug it in the formula for formula for volume of a right circular cone.

$$
\text{V} = \frac{1}{3}\pi\text{r}^2\text{h}
$$

$$
\text{V} = \frac{1}{3}\pi\times 5^2\times 9
$$

Simplifying this we get:

$$
V = 75\pi \text{ cm}^3
$$

$$
V = 235.62\text{ cm}^3 (\text{nearest }2 \text{ d.p})
$$

</Answer>
</Snippet>

## The volume of a truncated cone

![A labeled truncated cone::200px](/lessons/cone/truncated_cone.png)
Apart from the usual measures of a cone, the truncated cone has an extra measure which is the radius of its top surface which is essential to determine the volume of the truncated cone.
The formula that can be used is:

$$
\text{Volume} = \frac{1}{3}\pi\times\left(\text{r}^2 + \text{r}\cdot \text{R} + \text{R}²\right)\cdot\text{h}
$$

Note that the height measured here is the perpendicular distance from the bottom surface to the top surface.

<Snippet ribbon="Question">
<Question tag="Question 1."> A metal manufacturing company is designing a cone shaped machine part. The cone is made up of two layers of metal, metal A and metal B.Metal A will fill up a third of the height and the rest by Metal B. What is the volume of each metal required to manufacture a single machine part? ![Diagram of machine part with dimensions marked::200px](/lessons/cone/image_of_cone_with_two_layers.png)</Question>
<Answer tag="Answer:">
First let's focus on finding the volume of metal A required. The shape formed by metal A is a truncated cone.
 
<strong>Step 1:</strong> Determine the measures of the truncated cone.

The diagram below illustrates the truncated cone fromed by metal A.
![2D section of cone-shaped toy::200px](/lessons/cone/2d_cross_section_of_truncated_cone_formed_from_a_cone.png)
Upon inspection we can determine the following measures:

$$
\text{R} = 6\text{ cm}
$$

The height (h) of the truncated cone is $\frac{1}{3}$ of the height of the toy, hence:

$$
\text{h} = \frac{1}{3}\times 12\text{ cm}
$$

$$
\text{h} = 4 \text{ cm}
$$

We can use proportions of lengths in a shape to work out the radius r.Just like the heights, the ratio between R and r would be 3:1, therefore:

$$
r = \frac{1}{3}\times\text{R}
$$

$$
r = \frac{1}{3}\times\text{6}
$$

$$
r = 2 \text{ cm}
$$

Now we know all the measures required to calculate the volume of the truncated cone.

<strong>Step 2:</strong> Calculate volume of metal A

$$
\text{Volume of Truncated Cone} = \frac{1}{3}\pi\times\left(\text{r}^2 + \text{r}\cdot \text{R} + \text{R}²\right)\cdot\text{h}
$$

$$
\text{V}_A = \frac{1}{3}\pi\times\left(\text{2}^2 + \text{2}\cdot \text{6} + \text{6}²\right)\cdot\text{4}
$$

$$
\text{V}_A = 217.82\text{cm}^3
$$

<strong>Step 3: Calculate the volume of Metal B</strong>

Here we can either directly apply the formula for volume of a right circular cone to the small cone formed by metal B, or deterine the volume of the entire toy and subtract it by the result obtained in Step 2. let's try the former.

The measures of the cone formed by Metal B are:

$$
\text{r} = 2\text{ cm}
$$

$$
\text{h} = 12 - 4 = 8\text{ cm}
$$

Applying the formula we get:

$$
\text{V}_B = \frac{1}{3}\pi\text{r}^2\cdot\text{h}
$$

$$
\text{V}_B = \frac{1}{3}\pi\text{2}^2\cdot\text{8}
$$

$$
\text{V}_B = 33.51\text{ cm}^3
$$

So, the company will need 217.82$\text{cm}^3$ of Metal A and 33.51$\text{cm}^3$ of Metal B to make each machine part.
</Answer>
</Snippet>

## Alternative formulas for the volume of a right circular cone

You may not necessarily always have access to the desired measures of a cone. For example, if you are given the slant height instead of the height, we can’t directly apply the formula of the volume of a cone discussed previously.
![Cone with labeled Radius, Height, Angle and Slant Height::200px](/lessons/cone/cone_labeled.png)
Here is a list of some of the commonly used alternatives:

<center>
  <FancyTable1>
  | Given Parameters      |Alternate Formulae for Volume of a right circular cone |  
  | :---        |       ---: |
  | Slant Height(c) and Radius(r)     |<center> $\frac{1}{3}\pi\text{r}^2\cdot\sqrt{\text{c}^2 - \text{r}^2}$ </center>     |
  | Slant Height(c) and Height(h)   | <center> $\frac{1}{3}\pi\left(\text{c}^2 - \text{h}^2\right)\cdot\text{h}$  </center>|
  | Slant Height(c) and Semi Vertical Angle $\theta$  | <center>$\frac{1}{3}\pi\text{c}^3\cdot\cos{\theta}\cdot\sin^2({\theta})$</center> |
  | Semi Vertical Angle($\theta$) and Radius(r)   | <center>$\frac{1}{3\tan{\theta}}\cdot\pi\text{r}^3$ </center>  |
  | Semi Vertical Angle($\theta$) and Height(h)   | <center>$\frac{1}{3}\cdot\pi\text{h}^3\cdot\tan^2({\theta})$ </center>  |
  </FancyTable1>
</center>
Although these formulas seem to be quite different from one another, they can be derived through the original formula by applying basic trigonometric ratios.

<Snippet ribbon="Question">
<Question tag="Quiz on the cone"> 
A pencil manufacturer has worked out that for every 10 minutes of writing, their pencil loses about 0.04mm^3 of graphite. Assuming the tip of the pencil is cone-shaped with an cone angle of $15\degree$, work out the decrease in height of the tip after writing for 1 hour in mm.
![Pencil with cone-shaped tip zoomed in::200px](/lessons/cone/cone_shape_in_pencil_tip_with_height_marked.png)
</Question>
<Answer tag="Solution:">
Since in this question we are trying to determine the height of the cone, let's make h the subject of the formula for the right circular cone.

<strong>Step 1.</strong> Rearrange the formula of the volume of a cone to make h(height) the subject

$$
\text{V} = \frac{1}{3}\pi\text{r}^2\cdot\text{h}
$$

Since the radius (r) is not available, we have to substitute r in terms of h and the semi vertical angle of the cone.

By applying the tan ratio to the right angle triangle:

$$
\text{r} = \text{h}\tan{\theta}
$$

Replacing r in the formula for V:

$$
\text{V} = \frac{1}{3}\pi\text{h}^3\tan^2({\theta})
$$

Now let's make h the subject:

$$
\text{h} = \sqrt[3]{\frac{3V}{\pi\tan^2({\theta})}}
$$

<strong>Step 2.</strong> Determine the volume of graphite lost altogether.

The rate of graphite loss is :

$$
\text{Rate of graphite loss} = \frac{0.04}{10} \text{ mm}^3\text{/min}
$$

$$
\text{Rate of graphite loss} = 0.004 \text{ mm}^3\text{/min}
$$

Hence, if the pencil is used for 1 hour (60 minutes):

$$
\text{Volume of grpahite lost} = 0.004*60
$$

$$
\text{Volume of graphite lost} = 0.24 \text{ mm}^3
$$

<strong>Step 3.</strong> Substitute the measures of the cone into the formula obtained in Step 1

$$
\text{h} = \sqrt[3]{\frac{3V}{\pi\tan^2({\theta})}}
$$

$$
\text{h} = \sqrt[3]{\frac{3\times 0.24}{\pi\tan^2({15\degree})}}
$$

$$
\text{h} \approx 1.47\text{ mm}
$$

</Answer>
</Snippet>


Learn the basics of the cone in geometry with Tutordb's comprehensive lesson. Understand its properties, formulas, and applications with easy-to-follow explanations, examples, and exercises. Start mastering geometry today!

The Cone

---
excerpt: "Vertical angles are a fundamental concept in geometry that play an important role in many applications. Understanding the properties of vertical angles can also help in solving various problems in geometry, such as proving the congruency of triangles and finding missing angles in a diagram. Therefore, learning about vertical angles is essential for anyone studying geometry or pursuing a career in a related field."
---

## Angles formed by intersecting lines

One important concept in geometry is the angles formed by intersecting lines. Some various characteristics and relationships can be identified between the angles formed by intersecting lines. This article will explore the specific types of angles and their properties. Through clear examples and step-by-step explanations, you will learn how to identify and analyze these angles and apply this knowledge to solve challenging problems.

## The Linear Pair

When two lines intersect, the angles that are adjacent to each other and share a common vertex are known as linear pairs.
![Linear pair of angles in intersecting lines::250px](/lessons/intersecting_lines/linear_pairs.png)
When there are several intersecting lines, we can identify several pairs of angles that are adjacent to each other, but not all of these pairs can be identified as linear pairs. Linear pairs are adjacent angles that form a straight line, i.e they share one common line and one other uncommon line.
![Linear pairs labelled in 3 intersecting lines::250px](/lessons/intersecting_lines/linear_pairs_2.png)
This is described mathematically using the linear pair theorem. The linear pair theorem says that the measures of the two angles that make up a linear pair sum up to 180 degrees.

<Snippet ribbon="Example">
<Question tag="Example 1."> Identify the measures of all the angles formed by the intersecting lines![Two intersecting lines with one labelled angle::220px](/lessons/intersecting_lines/example_1.png)</Question>
<Answer tag="Answer:">
<strong>Step 1:</strong> Identify Linear pairs

Since $\angle$BOD is known, it’s best to try and identify the linear pairs that consist of $\angle$BOD.
We an see that $\angle$BOC is adjacent to $\angle$BOD, hence it forms a linear pair.
Similarly $\angle$AOD is adjacent to $\angle$BOD, and forms a straight line, hence $\angle$AOD and $\angle$BOD form a linear pair as well.

<strong>Step 2:</strong> Use the linear pair theorem to build relationships between angles

The linear pair theorem states that the sum of measures of angles of a linear pair is 180 degrees. We can apply this theorem to the linear pairs identified in step 1.

$$
\angle BOC + \angle BOD = 180\degree
$$

$$
\angle AOD + \angle BOD = 180\degree
$$

<strong>Step 3:</strong> Solve the equations

By substituting for the measure of BOD (43 $\degree$) we can calculate both $\angle$BOC and $\angle$AOD as follows.

$$
\angle BOC = 180\degree - \angle BOD
$$

$$
\angle BOC = 137 \degree
$$

$$
\angle AOD = 180\degree - \angle BOD
$$

$$
\angle AOD = 137 \degree
$$

We’ve figured out 2 of the three unknown angles. Finally we can use the fact that the angles around a point is 360 to determine the size of AOC.

$$
\angle AOC + \angle BOC + \angle AOD + \angle BOD = 360\degree
$$

$$
\angle AOC = 360\degree - \angle BOC - \angle AOD - \angle BOD
$$

$$
\angle AOC = 360\degree - 137\degree - 137\degree - 43\degree
$$

$$
\angle AOC = 43\degree
$$

</Answer>
</Snippet>

## The Linear Pair Perpendicular Theorem

The linear pair perpendicular theorem is an extension of the linear pair theorem, and it states that if the angles of a linear pair are of equal size, then they are both right angles. The converse of the theorem would show that the angles formed at the intersection of two perpendicular lines are linear pairs.

![Two perpendicular lines intersecting::200px](/lessons/intersecting_lines/perpendicular_lines.png)

## Vertical Angles

Vertical angles are a pair of non-adjacent angles formed when two lines intersect. They are opposite to each other, and their measures are always equal. Vertical angles can be seen in various shapes, polygons, geometrical objects, etc. and it is used to deduce other theorems such as the triangle sum theorem and exterior angle theorem.
![Two intersecting lines with vertical angles marked::200px](/lessons/intersecting_lines/vertical_angles.png)

## Proof of the vertical angle theorem

To prove the vertical angle theorem, i.e the measure of vertical angles are equal, we can consider a pair of intersecting lines as shown below.

![Two intersecting lines::200px](/lessons/intersecting_lines/proof_1.png)

Using the definition of vertical angles we can first identify a pair of vertical angles.
Next, let's figure out how to answer the question “Are vertical angles congruent?”.

First consider angles $\angle$BOC and $\angle$BOD, which form a linear pair. Since they are a linear pair, the following statement is true.

$$
\angle BOC + \angle BOD = 180\degree
$$

Similarly for angles BOD and AOD we can build a relationship:

$$
\angle AOD + \angle BOD = 180\degree
$$

By combining statement 1 and statement 2, we can show that:

$$
\angle BOC - \angle AOD = 0\degree
$$

$$
\angle BOC = \angle AOD
$$

$\angle$BOC and $\angle$AOD are vertical angles, and the above proof shows they are congruent.

Similarly, by selecting a different linear pair you can also prove the vertical angle theorem for the vertical angles $\angle AOC$ and $\angle BOD$ as well.

<Snippet ribbon="Example">
<Question tag="Example 2."> Determine the measure of $\alpha$![Two intersecting lines with one labelled angle::200px](/lessons/intersecting_lines/example_2.png)</Question>
<Answer tag="Solution:">
<strong>Step 1.</strong> Identify the vertical angles

There are several vertical angle pairs formed by these intersecting lines. But to solve for alpha, we have to choose a pair that includes alpha and some known angles. According to that the vertical angles $\angle$AOE and $\angle$BOF are suitable.

<strong>Step 2.</strong> Apply the vertical angle theorem

From the vertical angle theorem we know that vertical angles are congruent. Hence, the relationship between the vertical angles chosen in step 1 can be given as:

$$
\angle AOE = \angle BOF
$$

<strong>Step 3:</strong> Substitute and solve for $\alpha$

$$
\angle AOE = \angle BOF
$$

$$
90\degree + \alpha = 119\degree
$$

$$
\alpha =  119\degree - 90\degree
$$

$$
\alpha = 29 \degree
$$

</Answer>
</Snippet>

<Snippet ribbon="Question">
<Question tag="Quiz on angles formed by intersecting lines"> 
Find the missing angles
![Three intersecting lines with two labelled angle::200px](/lessons/intersecting_lines/quiz.png)
</Question>
<Answer tag="Solution:">

<strong>Step 1.</strong> Identify vertical angle pairs

$\angle$EOB and $\angle$AOF are vertical angles, and so are $\angle$DOF and $\angle$COE. Therefore using the vertical angle theorem we can state that:

$$
\angle EOB = \angle AOF
$$

$$
\angle DOF = \angle COE,
$$

Hence $\angle$ COE and $\angle$ AOF are of size 50$\degree$ and 68$\degree$ respectively.

<strong>Step 2.</strong> Isolate the variable

$\angle$BOC and $\angle$BOD form a linear pair, and so does $\angle$COF and $\angle$DOF. The linear pair theorem can be applied to these angles.

$$
\angle BOC + \angle BOD = 180\degree -->(1)
$$

Since $\angle$ BOC = $\angle$ BOE + $\angle$ COE, we can rewrite the equation as follows:

$$
\angle BOE + \angle COE + \angle BOD = 180\degree -->(1')
$$

<strong>Step 3.</strong> Substitute the angles found in Step 1 in the equations derived in Step 2

For equation 1:

$$
68\degree + 50\degree + \angle BOD = 180\degree
$$

$$
118\degree + \angle BOD = 180\degree
$$

$$
\angle BOD = 62 \degree
$$

As BOD and AOC are vertical angles, they are congruent. Hence, AOC is 62 degrees.

<center>
  <FancyTable1>
  | Angle      |Measure|  
  | :---        | ---: |
  | $\angle$BOD |62$\degree$ |
  | $\angle$AOC |62$\degree$ |
  | $\angle$ AOF|68$\degree$|
  | $\angle$ COE|50$\degree$ |
  </FancyTable1>
</center>
</Answer>
</Snippet>


Lessons

The unit circle

The triangle

The Cone

What are intersecting lines?