Square Inscribed in a Circle Construction

Greetings math friends and welcome to MathSux! In this week’s post, we are going to take a step-by-step look at how a square inscribed in a circle construction works! Within this post we have a video, a GIF, and a step-by-step written explanation below, the choice of learning how to do this construction is up to you! Happy Calculating! 🙂

Square Inscribed in a Circle Construction

How to Construct a Square Inscribed in Circle:

Step 1: First, we are going to draw a circle using a compass (any size).

Step 2: Using a ruler, draw a diameter or straight line across the length of the circle, going through its midpoint.

Step 3: Next, open up the compass across the circle. Then take the point of the compass to one end of the diameter and swing the compass above the circle, creating an arc.

Step 4: Keeping that same length of the compass, go to the other side of the diameter and swing above the circle again making another arc until the two arcs intersect.

Step 5: Now, we are going to repeat steps 3 and 4, this time creating arcs below the circle using the same compass size.

Step 6: Connect the point of intersection above and below the circle using a ruler or straight edge. This creates a perpendicular bisector, cutting the diameter in half and forming a 90º angle.

Step 7: Lastly, we are going to use a ruler to connect each corner point to one another creating a square!

Constructions and Related Posts:

Looking to construct more than just a square inside a circle? Check out these related posts and step-by-step tutorials on geometry constructions below!

Construct an Equilateral Triangle

Perpendicular Line Segment through a Point

Angle Bisector

Construct a 45º angle

Altitudes of a Triangle (Acute, Obtuse, Right)

How to Construct a Parallel Line

Bisect a Line Segment

Best Geometry Tools!

Looking to get the best construction tools? Any compass and straight-edge will do the trick, but personally, I prefer to use my favorite mini math toolbox from Staedler. Stadler has a geometry math set that comes with a mini ruler, compass, protractor, and eraser in a nice travel-sized pack that is perfect for students on the go and for keeping everything organized….did I mention it’s only $7.99 on Amazon?! This is the same set I use for every construction video in this post. Check out the link below and let me know what you think!

Still got questions? No problem! Don’t hesitate to comment with any questions or concerns below. Also if you’re looking for a different type of construction you don’t yet see here, please let me know! Happy calculating! 🙂

Let’s be friends! Check us out on the following social media platforms for even more free MathSux practice questions and videos:

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Looking for something similar to square inscribed in a circle construction? Check out this post here on how to construct and equilateral triangle here! And if you’re looking for even more geometry constructions, check out the link here!

Simultaneous Equations: Algebra

Happy new year and welcome to Math Sux! In this post we are going to dive right into simultaneous equations and how to solve them three different ways! We will go over how to solve simultaneous equations using the (1) Substitution Method (2) Elimination Method and (3) Graphing Method. Each and every method leading us to the same exact answer! At the end of this post don’t forget to try the practice questions choosing the method that best works for you! Happy calculating! 🙂

What are Simultaneous Equations?

Simultaneous Equations are when two equations are graphed on a coordinate plane and they intersect at, at least one point.  The coordinate point of intersection for both equation is the answer we are trying to find when solving for simultaneous equations. There are three different methods for finding this answer:

We’re going to go over each method for solving simultaneous equations step by step with the example below:

Method #1: Substitution

The idea behind Substitution, is to solve for 1 variable first algebraically, and the plug this value back into the other equation solving for one variable.  Then solving for the remaining variable.  If this sounds confusing, don’t worry! We’re going to do this step by step:

Step 1: Let’s choose the first equation and move our terms around to solve for y.

Simultaneous Equations

Step 3: The equation is set up and ready to solve for x!

Simultaneous Equations

Step 4: All we need to do now, is plug x=3 into one of our original equations to solve for y.

Simultaneous Equations

Step 5: Now that we have solved for both x and y, we have officially found where these two simultaneous equations meet!

Simultaneous Equations

Method #2: Elimination

The main idea of Elimination is to add our two equations together to cancel out one of the variables, allowing us to solve for the remaining variable.  We do this by lining up both equations one on top of the other and adding them together.  If variables at first do not easily cancel out, we then multiply one of the equations by a number so it can. Check out how it’s done step by step below!

Step 1: First, let’s stack both equations one on top of the other to see if we can cancel anything out:

Simultaneous Equations

Step 2: Our goal is to get a 2 in front of y in the first equation, so we are going to multiply the entire first equation by 2.

Simultaneous Equations

Step 3: Now that we multiplied the entire first equation by 2, we can line up our two equations again, adding them together, this time canceling out the variable y to solve for x.

Simultaneous Equations

Step 4: Now, that we’ve found the value of variable x=3, we can plug this into one of our equations and solve for missing unknown variable y.

Simultaneous Equations

Step 5: Now that we have solved for both x and y, we have officially found where these two simultaneous equations meet!

This image has an empty alt attribute; its file name is Screen-Shot-2020-12-30-at-10.21.46-AM.png

Method #3: Graphing

The main idea of Graphing is to graph each a equation on a coordinate plane and then see at what point they intersect.  This is the best method to visualize and check our answer!

Step 1: Before we start graphing let’s convert each equation into y=mx+b (equation of a line) form.

Equation 1:

Equation 2:

Step 2: Now, let’s graph each line, y=3x-4 and y=-x+8, to see at what coordinate point they intersect.

Simultaneous Equations

Need to review how to draw an equation of a line? Check out this post here! Notice we got the same exact answer using all three methods (1) Substitution (2) Elimination and (3) Graphing.

Ready to try the practice problems on your own?! Check them out below!

Practice Questions:

Solve the following simultaneous equations for x and y.

Solutions:

  1. (1, 3)
  2. (4,5)
  3. (-1, -6)
  4. (3, -3)

Want more MathSux?  Don’t forget to check out our Youtube channel and more below! And if you have any questions, please don’t hesitate to comment below. Happy Calculating!

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If you are looking for a challenge, try solving three equations with three unknown variables, in this post here!

30 60 90 Triangle

Hi everyone and welcome to MathSux! In this post, we are going to break down 30 60 90 degree special right triangles. What is it? Where did it come from? What are the ratios of its side lengths and how do we use them? You will find all of the answers to these questions about this right angled triangle in this post. Also, don’t forget to check out the video below and practice questions at the end of this post to truly become a 30 60 90 special right triangle master! Happy calculating! 🙂

If you want to learn about the other special triangle, 45 45 90 triangle, check out this post here. And if you’re looking to make math suck just a little bit less? Subscribe to our Youtube channel for free math videos every week! 🙂

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30 60 90 Triangle Ratio:

30 60 90 triangle side lengths

Notice that the 30 60 90 triangle is made up of one right angle across from the hypotenuse (which is always going to be the longest side), a 60 degree angle with a longer leg on the opposite side, and a 30 degree angle measure across from the shorter leg.

What is a 30 60 90 Triangle and why is it “Special”?

The 30 60 90 triangle is a special right triangle because it forms an equilateral triangle when a mirror image of itself is drawn. This means that all sides are equal when a mirror image of the triangle is drawn which allows us to find the ratio between each of the sides of the triangle by using the Pythagorean Theorem. Check it out below!

30 60 90 Special Triangles

Now let’s draw a mirror image of our triangle to create an equilateral triangle.  Next, we can label the length of the new side opposite 30 degrees (the shorter leg) “a,” and add this new mirror image length with the original we had to get, a total of a+a=2a.

30 60 90 Special Triangles
30 60 90 Special Triangles

If we look at our original 30 60 90 triangle, we now have the following values for each side based on our equilateral triangle. Notice we still need to find the length value of the longer leg, opposite angle 60 degrees.

30 60 90 Special Triangles
Pythagorean Theorem Formula
Fill in values from our triangle longer leg
Distribute the exponent and start solving for our “?” by subtracting a^2 from both sides.
Now take the square root of both sides to solve for “?.”	We have found a solution! Our missing side length is equal to a√3.

To sum up, we have applied the Pythagorean Theorem formula, filled in values found in our triangle for each length, distributed the exponent and subtracted a2 from both sides, took the square root, and finally found our solution for our ratio for the longer leg opposite 60 degrees.

Now we can re-label our triangle, knowing the length of the hypotenuse in relation to the two legs. This creates a ratio that applies to all 30 60 90 triangles!

30 60 90 triangle side lengths

How do I use this ratio?

30 60 90 triangle side lengths

Knowing the above ratio, allows us to find any length of any and every 30 60 90 triangle, when given the value of one of its sides. Let’s tsee how this ratio works with some examples.

Example #1:

30 60 90 triangle side lengths

Step 1: First let’s look at our ratio and compare it to our given triangle.

30 60 90 triangle side lengths

Step 2: Notice we are given the value of a, which is equal to 4, knowing this we can now fill in each length of our triangle based on the ratio of a 30 60 90 triangle.

30 60 90 triangle side lengths
30 60 90 triangle side lengths

Now let’s look at another Example where we are given the length of the hypotenuse and need to find the values of the other two missing sides.

Example #2:

30 60 90 triangle side lengths

Step 1: First let’s look at our ratio and compare it to our given triangle.

30 60 90 triangle side lengths

Step 2: Notice we are given the value of the hypotenuse, 2a=20. Knowing this we can find the value of a by dividing 20 by 2 to get a=10. Once we have the value of a=10, we can easily find the length of the last, longer leg based on the 30 60 90 ratio:

30 60 90 triangle side lengths
30 60 90 triangle side lengths

Now for our last Example, we will see how to find the value of the shorter leg and hypotenuse, when we are given the side length of the longer leg across from 60º and need to find the other two missing sides.

Example #3:

30 60 90 triangle side lengths

Step 1: First let’s look at our ratio and compare it to our given triangle.

30 60 90 triangle side lengths

Step 2: In this case, we need to use little algebra to find the value of a, using the ratio for 30 60 90 triangles.

30 60 90 triangle side lengths
We set up this equation, knowing that 9 is across from 60 degrees, and that a√3=9. 	To solve for a, we divide square root both sides by √3. 	We get a solution a=9/√3! Yay! We cannot have a radical in the denominator, so we rationalize the denominator by multiplying the numerator and denominator by √3.

Now that we have one piece of the puzzle, the value of a, let’s fill in the value of the shorter leg of our triangle below:

Finally, let’s find the value of the length of the hypotenuse, which is equal to 2a in our ratio. Knowing the length of all sides, we can fill in the lengths of each side of our triangle for our solution.

Think you are ready to master these types of questions on your own? Try the practice problems below!

Practice Questions:

Find the value of the missing sides of each 30 60 90 degree triangle.

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating! 🙂

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Related Trigonometry Posts:

The Unit Circle

Basic Right Triangle Trigonometric Ratios (SOH CAH TOA)

4545 90 Special Triangles

Factoring Trigonometric Functions

Graphing Trigonometric Functions

Trig Identities

Transforming Trig Functions

Law of Cosines

Law of Sines

NYE Ball Fun Facts: Volume & Combinations

Greetings math friends, and Happy New Year! In today’s post we’re going do something a little different and take a look at the math behind the very famous and very shiny New Year’s Eve Ball that drops down every year at midnight.  We’ll break down the shape, the volume, and the number of those dazzling Waterford crystals (and no this post isn’t sponsored) and look at some NYE Ball Fun Facts.

NYE Ball Fun Facts

Shape: Geodesic Sphere

Yes, apparently the shape of the New Year’s ball is officially called a “Geodesic Sphere.”  It is 12 feet in diameter and weighs 11,875 pounds.

Volume (Estimate): 288π ft3

If we wanted to estimate the volume of the New Year’s Ball we would could use the formula for volume of a sphere:

NYE Ball Fun Facts

Number of Waterford Crystals: 2,688

Talk about the ultimate shiny bauble! The NYE ball lights up the night with all 2,688 crystals in the shape of different sized triangles, each with heights of 5.75 inches or 4.75 inches.

NYE Ball Fun Facts

Number of Lights: 48 light emitting diodes (LED’s)

On each triangle, there are 48 LEDs: 12 red, 12 blue, 12 green, and 12 white, for a total of 32,256 LEDs on the entire NYE ball itself.

NYE Ball Fun Facts

Permutations and Combinations:

Permutations: With this many lights and colors, there are over a billion potential permutations of colors on the entire NYE ball.

Combinations: Let’s break down one triangle with 48 LED lights each with 12 red, 12 blue, 12 green, and 12 white LEDs. How many possible combinations of lights are possible if we were to choose 7 blue, 5 red, 10 green, and 1 white turned on all at the exact same time?

We end up with the combination formula below:

NYE Ball Fun Facts

That means that there are 496,793,088 possible ways that 7 blue lights, 5 red lights, 10 green lights, and 1 white light can be lit up on a triangle that is part of the entire NYE ball!

Interested in more NYE fun facts?  Check out the sources of this article here.

NYE Fact Sheet from: timessquarenyc.org

NYE Ball picture: Timesquareball.net

If you like finding the volume of the NYE ball maybe, you’ll want to find the volume of the Hudson Yards Vessal in NYC here.  Happy calculating and Happy New Year from MathSux!

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Transforming Trig Functions: Amplitude, Frequency, Period, Phase Shifts

Hi everyone, and welcome to MathSux! In this post, we are going to break down transforming trig functions by identifying their amplitude, frequency, period, horizontal phase shift, and vertical phase shifts. Fear not! Because we will break down what each of these terms (amplitude, frequency, period, horizontal phase shift and vertical phase shifts) mean and how to find them when looking at a trigonometric function and then apply each of these changes step by step to our graph. In this particular post, we will be transforming and focusing on a cosine function, but keep in mind that the same rules apply for transforming sine functions as well (example shown below in practice).

And if you’re ready for more, check out the video and the practice problems below, happy calculating! 🙂

*For a review on how to derive the basic Trig functions (y=sinx, y=cosx, and y=tanx), click here.

What are the Different Parts of a Trig Function?

When transforming trig functions, there are several things to look out for, let’s take a look at what each part of a trig function represents below:

Trig functions

Amplitude: The distance (or absolute value) between the x axis and the highest point on the graph.

Frequency: This is the number of cycles that happen between 0 and 2π. (Α “cycle” in this case is the number of “s” cycles for the sine function).

Period: The x-value/length of one cycle. (Α “cycle” in this case is the number of “s” cycles for the sine function). This is found by looking at the graph and seeing where the first cycle ends, or, by using the formula:       

Horizontal Shift: When a trigonometric function is moved either left or right along the x axis.

Vertical Shift: When a trigonometric function is moved either up or down along the y axis.

Let’s try an Example, by graphing the following trig function step by step by identifying the amplitude, frequency, period, vertical shift, and horizontal phase shift.

Step 1: First let’s label and identify all the different parts of our trig function and what each part represents.

Trig functions

Step 2: Now let’s transform our trig graph one step at a time.  First, let’s start graphing y=cos(x) without any transformations, the basic graph.

Trig functions
Graph y=cos(x)

Step 3: Next, let’s add our amplitude of 2, otherwise known as the height, or distance to the x-axis.  To do this our highest and lowest points on the y-axis will now be moved to 2 and -2 respectively. 

Trig functions
Re-draw y=cos(x), with an amplitude of 2 to get y=2cos(x)

Step 4: Next, we can apply our horizontal shift to the left by (π/2) or 90º.  To do this, we need to look at where negative (π/2) is on our graph at (-π/2) and move our entire graph over to start at this new point, “shifting” over each coordinate point by (π/2) along the x axis.

Trig functions
Shift our graph y=2cos(x) over by 90º to the left to get y=2cos(x+(π/2))

Step 5: For our last transformation, we have a vertical phase shift up 1 unit.  All this means is that we are going to shift our entire graph up by 1 unit along the y axis.

Trig function graph
phase shift
vertical shift
Shift our entire graph y=2cos(x+(π/2)) up one unit along the y-axis to get y=2cos(x+(π/2))+1

Think you are ready to try graphing trig functions and identifying the amplitude, frequency, period, vertical phase shift, and horizontal phase shift? Check out the practice questions and answers below!

Practice Questions:

When you’re ready check out the function transformations solutions below:

Solutions:

Still, got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for an in-depth explanation. Happy calculating! 🙂

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Related Trigonometry Posts:

The Unit Circle

Basic Right Triangle Trigonometric Ratios (SOH CAH TOA)

4545 90 Special Triangles

30 60 90 Special Triangles

Graphing Trigonometric Functions

Trig Identities

Factoring Trigonometric Functions

Law of Cosines

Law of Sines

Graphing Trig Functions: Algebra 2/Trig.

Hi everyone and welcome to MathSux! This post is going to help you pass Algebra 2/Trig. In this post, we are going to apply our knowledge of the unit circle and trigonometry and apply it to graphing trig functions y=sin(x), y=cos(x), and y=tan(x). We have been using trig functions in basic trigonometry, but today, we will see what each function (sin, cos, and tan) look like graphed on a coordinate plane.

If you are already familiar with the basics of graphing trig functions and want to learn how to transform them, including knowing how to identify the amplitude, period frequency, vertical shift, and horizontal phase shift, then please check out this post here on transforming trigonometric functions.

If you have any questions, don’t hesitate to comment or check out the video below. Thanks for stopping by and happy calculating! 🙂

How do we get coordinate points for graphing Trigonometric Functions?

For deriving our trigonometric function graphs [y=sin(x), y=cos(x), and y=tan(x)] we are going to write out our handy dandy Unit Circle. By looking at our unit circle and remembering that coordinate points are in (cos(x), sin(x)) form and that tanx=(sin(x))/(cos(x)) we will be able to derive each and every trig graph!

*Note below is the unit circle we are going to reference to find each value, for an in-depth explanation of the unit circle, check out this link here.

How to Graph y=sin(x)?

While Graphing Sine, you may notice that the sine function curves creates what looks like an “S” shape. As they say, “S” is for Sine, this is the easiest way to remember what the sine function looks like! Check it out below.

Graphing Trig Functions

Now that we know what the sine function looks like once it’s graphed, let’s see why it looks that way in the first place by deriving each coordinate point with the help of the unit circle.

Step 1: We are going to derive each degree value for sin by looking at the unit circle. These will be our coordinates for graphing y=sin(x). *For a review on how to get these values, check out the link here explaining the unit circle.

Step 2: Now we need to convert all the x-values from degrees to radians.  Fear not because this can be done easily with a simple formula!

      To convert degrees à radians, just use the formula below:

Step 3: Now that we have our coordinate points and converted degrees to radians, we can draw out our function y=sin(x) on the coordinate plane! 

Graphing Trig Functions

Now we will follow the same process for graphing y=cos(x) and y=tan(x).

How to Graph y=cos(x)?

While Graphing Cosine, you may notice that the cosine function creates what looks like a “V” shape. I always like to think “V” is for victory, but of course, in this case, it is for cosine, that is the easiest way to remember what this function looks like! Check it out below.

Graphing Trig Functions

Now that we know what the cosine function looks like once graphed, let’s see why it looks this way in the first place by deriving each coordinate point with the help of the unit circle.

Step 1: We are going to derive each degree value for cos by looking at the unit circle. These will be our coordinates for graphing y=cos(x). *For a review on how to get these values, check out the link here explaining the unit circle.

Graphing Trig Functions

Step 2: Now we need to convert all the x-values from degrees to radians.

Graphing Trig Functions

Step 3: Now that we have our coordinate points and converted degrees to radians, we can draw out our function y=cos(x) on the coordinate plane! 

Graphing Trig Functions

What do you think would happen if we were to vertical shift our entire function up by one unit? Or move our entire function to the left by 90 degrees with a horizontal phase shift? Check out what happens here!

How to Graph y=tan(x)?

While Graphing Tangent, you may notice that the tangent function looks totally different than sine or cosine. Check it out below.

Step 1: We are going to derive each degree value for the tangent function by looking at the unit circle. In order to derive values for tan(x), we need to remember that tan(x)=sin(x)/cos(x). Once found, these will be our coordinates for graphing y=tan(x). *For a review on how to get these values, check out the link here explaining the unit circle.

Graphing Trig Functions

Step 2: Now we need to convert all the x-values from degrees to radians.

Graphing Trig Functions

Step 3: Now that we have our coordinate points and converted degrees to radians, we can draw out our function y=tan(x) on the coordinate plane! 

Graphing Trig Functions

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for an in-depth explanation. Happy calculating! 🙂

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Ready for the next lesson? Now that you know the basics of graphing the sine, cosine, and tangent functions, learn how to transform each trig function on a coordinate plane by identifying the amplitude, period frequency, vertical shift, and horizontal phase shift of each function, by checking out this post here on transforming trigonometric functions. When you’re ready for more, check out the related trig posts below!

Related Trigonometry Posts:

The Unit Circle

Basic Right Triangle Trigonometric Ratios (SOH CAH TOA)

4545 90 Special Triangles

30 60 90 Special Triangles

Law of Sines

Transforming Trig Functions

Factoring Trig Functions

Law of Cosines

Trig Identities

What is a Geometric Sequence?

Geometric Sequence Formula:

an=a1r(n-1)

a1 = First Term

r=Common Ratio (Number Multiplied/Divided by each successive term in sequence)

n= Term Number in Sequence

Hi everyone and welcome to Mathsux! In this post, we are going to answer the question, what is a geometric sequence (otherwise known as a geometric progression)? We will accomplish this by learning how to identify a geometric sequence, then we will break down the geometric sequence formula an=a1r(n-1), and solve two different types of examples. As always if you want more questions, check out the video below and the practice problems at the end of this post. Happy calculating! 🙂

What are Geometric Sequences?

Geometric sequences are a sequence of numbers that form a pattern when the same number is either multiplied or divided to each subsequent term. Take a look at the example of a geometric sequence below:

Example:

geometric progression

Notice we are multiplying 2 by each term in the sequence above. If the pattern were to continue, the next term of the sequence above would be 64. This is a geometric sequence!

In this geometric sequence, it is easy for us to see what the next term is, but what if we wanted to know the 15th term?  Instead of writing out and multiplying our terms 15 times, we can use a shortcut, and that’s where the Geometric Sequence formula comes in handy!

Geometric Sequence Formula:

Take a look at the geometric sequence formula below, where each piece of our formula is identified with a purpose.

an=a1r(n-1)

a1 = The first term is always going to be that initial term that starts our geometric sequence. In this case, our sequence is 4,8,16,32, …… so our first term is the number 4.

r= One key thing to notice about the formula below that is unique to geometric sequences is something called the Common Ratio. The common ratio is the number that is multiplied or divided to each consecutive term within the sequence.

n= Another interesting piece of our formula is the letter n, this always stands for the term number we are trying to find. A great way to remember this is by thinking of the term we are trying to find as the nth term, which is unknown.

geometric sequences

Now that we broke down our geometric sequence formula, let’s try to answer our original question below:

Example #1: Common ratio r>1

Step 1: First let’s identify the common ratio between each previous and subsequent term of the sequence. Notice each term in the sequence is multiplied by 2 (as we identified earlier in this post). Therefore, our common ratio for this sequence is 2.

geometric progression

Step 2: Next, let’s write the geometric sequence formula and identify each part of our formula (First Term=4, Term number=15, common ratio=2).

geometric sequences

Step 3: Now let’s fill in our formula and solve with the given values.

geometric sequences

Let’s look at another example where, the common ratio is a bit different, and instead of multiplying a number, this time we are going to be dividing the same number from each subsequent term, (this can also be thought of as multiplying by a common ratio that is a fraction):

Example #2: Common ratio 0<r<1

Step 1: First let’s identify the common ratio between each number in the sequence. Notice each term in the sequence is divided by 2 (or multiplied by 1/2 that way it is shown below).

geometric progression

Step 2: Next, let’s write the geometric sequence formula and identify each part of our formula (First Term=1000, Term number=10, common ratio=1/2).

geometric sequences

Step 3: Next let’s fill in our formula and solve with the given values.

Think you are ready to practice solving geometric sequences on your own? Try the following practice questions with solutions below:

Practice Questions:

  1. Find the 12th term given the following sequence: 1250, 625, 312.5, 156.25, 78.125, ….
  2. Find the 17th term given the following sequence: 3, 9, 27, 81, 243,…..
  3. Find the 10th term given the geometric sequence: 5000, 1250, 312.5, 78.125 …..
  4. Shirley has $100 that she deposits in the bank. She continues to deposit twice the amount of money every month. How much money will she deposit in the twelfth month at the end of the year?

Solutions:

Fun Fact!

Did you know that the geometric sequence formula can be considered an explicit formula? An explicit formula means that even though we do not know the other terms of a sequence, we can still find the unknown value of any term within the given sequence. For example, in the first example we did in this post (example #1), we wanted to find the value of the 15th term of the sequence. We were able to do this by using the explicit geometric sequence formula, and most importantly, we were able to do this without finding the first 14 previous terms one by one…life is so much easier when there is an explicit geometric sequence formula in your life!

Other examples of explicit formulas can be found within the arithmetic sequence formula and the harmonic series.

Related Posts:

Looking to learn more about sequences? You’ve come to the right place! Check out these sequence resources and posts below. Personally, I recommend looking at the finite geometric sequence or infinite geometric series posts next!

Arithmetic Sequence

Recursive Formula

Finite Arithmetic Series

Finite Geometric Series

Infinite Geometric Series

Golden Ratio in the Real World

Fibonacci Sequence

Still, got questions? No problem! Don’t hesitate to comment below or reach out via email. And if you would like to see more MathSux content, please help support us by following ad subscribing to one of our platforms. Thanks so much for stopping by and happy calculating!

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Rotations about a Point: Geometry

rotations about a point

Happy Wednesday math friends! In this post we’re going to dive into rotations about a point! In this post we will be rotating points, segments, and shapes, learn the difference between clockwise and counterclockwise rotations, derive rotation rules, and even use a protractor and ruler to find rotated points. The fun doesn’t end there though, check out the video and practice questions below for even more! And as always happy calculating! 🙂

What are Rotations?

Rotations are a type of transformation in geometry where we take a point, line, or shape and rotate it clockwise or counterclockwise, usually by 90º,180º, 270º, -90º, -180º, or -270º.

A positive degree rotation runs counter clockwise and a negative degree rotation runs clockwise.  Let’s take a look at the difference in rotation types below and notice the different directions each rotation goes:

rotations 90 degrees

How do we rotate a shape?

There are a couple of ways to do this take a look at our choices below:

  1. We can visualize the rotation or use tracing paper to map it out and rotate by hand.
  2. Use a protractor and measure out the needed rotation.
  3. Know the rotation rules mapped out below.  Yes, it’s memorizing but if you need more options check out numbers 1 and 2 above!

Rotation Rules:

rotations 90 degrees

Where did these rules come from?

To derive our rotation rules, we can take a look at our first example, when we rotated triangle ABC 90º counterclockwise about the origin. If we compare our coordinate point for triangle ABC before and after the rotation we can see a pattern, check it out below:

rotations

The rotation rules above only apply to those being rotated about the origin (the point (0,0)) on the coordinate plane.  But points, lines, and shapes can be rotates by any point (not just the origin)!  When that happens, we need to use our protractor and/or knowledge of rotations to help us find the answer. Let’s take a look at the Examples below:

Example #1:

rotations

Step 1: First, let’s look at our point of rotation, notice it is not the origin we rotating about but point k!  To understand where our triangle is in relation to point k, let’s draw an x and y axes starting at this point:

rotations

Step 2: Now let’s look at the coordinate point of our triangle, using our new axes that start at point k.

Step 2: Next, let’s take a look at our rule for rotating a coordinate -90º and apply it to our newly rotated triangles coordinates:

rotations

Step 3: Now let’s graph our newly found coordinate points for our new triangle .

rotations about a point

Step 4: Finally let’s connect all our new coordinates to form our solution:

rotations about a point

Another type of question with rotations, may not involve the coordinate plane at all! Let’s look at the next example:

Example #2:

rotations about a point

Step 1: First, let’s identify the point we are rotating (Point M) and the point we are rotating about (Point K).

rotations about a point

Step 2: Next we need to identify the direction of rotation.  Since we are rotating Point M 90º, we know we are going to be rotating this point to the left in the clockwise direction.

Step 3: Now we can draw a line from the point of origin, Point K, to Point M.

rotations about a point

Step 4: Now, using a protractor and ruler, measure out 90º, draw a line, and notice that point L lands on our 90º line. This is our solution! (Note: For help on how to use a protractor, check out the video above).

rotations about a point

Ready for more? Check out the practice questions below to master your rotation skills!

Practice Questions:

  1. Point B is rotated -90º about the origin. Which point represents newly rotated point B?    

2. Triangle ABC is rotated -270º about point M.  Show newly rotated triangle ABC as A prime B prime C prime.

3. Point G is rotated about point B by 180º. Which point represents newly rotated point B?

rotations about a point

4.  Segment AB is rotated 270º about point K.  Show newly rotated segment AB.

Solutions:

Still got questions on how to rotations about a point? No problem! Don’t hesitate to comment with any questions or check out the video above for even more examples. Happy calculating! 🙂

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Looking for more Transformations? Check out the related posts below!

Translations

Dilations

Reflections

The Unit Circle: Algebra 2/Trig.

Greetings math friends! In today’s post we’re going to go over some unit circle basics. We will find the value of trigonometric functions by using the unit circle and our knowledge of special triangles. For even more practice questions and detailed info., don’t forget to check out the video and examples at the end of this post. Keep learning and happy calculating! 🙂

What is the Unit Circle?

The Unit Circle is a circle where each point is 1 unit away from the origin (0,0).  We use it as a reference to help us find the value of trigonometric functions.

The Unit Circle

Notice the following things about the unit circle above:

  1. Degrees follow a counter-clockwise pattern from 0 to 360 degrees.
  2. Values of cosine are represented by x-coordinates.
  3. Values of sine are represented by y-coordinates.
  4. Using the unit circle we can find the degree and radian value of trigonometric functions (SOH CAH TOA). Check out the example below!

What’s the big deal with Quadrants?

Within a coordinate plane there are 4 quadrants numbered I, II, III, and IV used throughout all of mathematics. Within these quadrants there are different trigonometric functions that are positive to each unique quadrant.  This will be important when solving questions with reference angles later in this post. Check out which trig functions are positive in each quadrant below:

The Unit Circle
The Unit Circle

Now let’s look at some examples on how to find trigonometric functions using our circle!

The Unit Circle

Negative Degree Values:

The unit circle also allows us to find negative degree values which run clockwise, check it out below!

The Unit Circle

Knowing that negative degrees run clockwise, we can now find the value of trigonometric functions with negative degree values.

The Unit Circle

How to find trig ratios with 30º, 45º and 60º ?

Instead of memorizing much, much more of the unit circle, there’s a trick to memorizing two simple special triangles for answering these types of questions. The 45º 45º 90º  special triangle and the  30º 60º 90º special triangle. (Why does this work? These special triangles can also be derived and found on the unit circle).

special triangles

Using the above triangles and some basic trigonometry in conjugation with the unit circle, we can find so many more angles, take a look at the example below:

Since we need to find the value of tan(45º) , we will use the 45º, 45º, 90º  special triangle.

special triangles

For our last question, we are going to need to combine our knowledge of unit circles and special triangles:

-> In order to do this, we must first look at where our angle falls on the unit circle.  Notice that the angle 135º is encompassed by the pink lines and falls in quadrant 2.

The Unit Circle

-> Since our angle falls in the second quadrant where only the trig function sin is positive.  Since we are finding an angle with the function cosine, we know the solution will be negative.

-> Now we need to find something called a reference angle.  Which is what those θ, 180°-θ, θ-180°, 360°-θ and  symbols represent towards the center of the unit circle.  Using these symbols will help us find the value of cos(135º). 

Because the angle we are trying to find,135º , falls in the second quadrant, that means we are going to use the reference angle that falls in that quadrant 180º-θ theta, using the angle we are given as θ.

The Unit Circle

-> Now we can re-write and solve our trig equations using our newly found reference angle, 45º.

Now we are going to use our  45º 45º 90º special triangle and SOH CAH TOA to evaluate our trig function.  For a review on how to use SOH CAH TOA, check out this link here.

special triangles

When you’re ready, try the problems on your own below!

Practice Questions:

Solve the following trig functions using a unit circle and your knowledge of special triangles:

The Unit Circle

Solutions:

The Unit Circle

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above for even more examples. Happy calculating! 🙂

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Related Trigonometry Posts:

Law of Sines

Basic Right Triangle Trigonometric Ratios (SOH CAH TOA)

4545 90 Special Triangles

30 60 90 Special Triangles

Graphing Trig Functions

Transforming Trig Functions

Factoring Trig Functions

Law of Cosines

Trig Identities

Arithmetic Sequence Formula:

an=a1+(n-1)d

a1=First Term

n=Term Number in Sequence

d=Common Difference (Number Added/Subtracted to each Term in Sequence)

Hi everyone and welcome to Mathsux! In this post, we’re going to go over arithmetic sequences (otherwise known as arithmetic progression). We’ll identify what arithmetic sequences are, break down each part of the arithmetic sequence formula an=a1+(n-1)d, and solve two different types of examples. As always if you want more questions, check out the video below and the practice problems at the end of this post. Happy calculating! 🙂

What are Arithmetic Sequences?

Arithmetic sequences are a sequence of numbers that form a pattern when the same number is either added or subtracted to each successive term. Take a look at the example of an arithmetic sequence below:

arithmetic sequences

Notice the pattern? We are adding the number 2 to each term in the sequence above. If the pattern were to continue, the next term of the sequence above would be 10+2 which gives us 12. This is an arithmetic sequence!

In the above sequence, it’s easy for us to identify what the next term in the sequence would be, but what happens if we were asked to find the 123rd term of an arithmetic sequence?  That’s where the Arithmetic Sequence Formula would come in handy!

Arithmetic Sequence Formula:

Take a look at the arithmetic sequence formula below, where each piece of our formula is identified with a purpose.

an=a1+(n-1)d

a1= The first term is always going to be that initial term that starts our arithmetic sequence. In this case, our sequence is 4,6,8,10, …… so our first term is the number 4.

n= Another interesting piece of our formula is the letter n, this always stands for the term number we are trying to find. A great way to remember this is by thinking of the term we are trying to find as the nth term, which is unknown.

d = One key thing to notice about the formula below that is unique to arithmetic sequences is something called the Common Difference. The common difference is the number that is added or subtracted to each consecutive term within the sequence.

explicit formula

Now that we know the arithmetic sequence formula, let’s try to answer our original question below:

arithmetic progression

Step 1: First let’s identify the common difference between each previous and subsequent term of the sequence. Notice each term in the sequence is being added by 2 (like we identified earlier in this post). Therefore, our common difference for this sequence is 2.

constant difference

Step 2: Next, let’s write the arithmetic sequence formula and identify each part of our formula (First Term=4, Term number=123, common difference=2).

arithmetic sequence formula

Step 3: Fill in our formula and solve with the given values.

math tutors

Now let’s look at another example where we subtract the same number from each term in the sequence, making the common difference negative.

arithmetic progression

Step 1: First let’s identify the common difference between each previous term and each subsequent term of the sequence. Notice each term in the sequence is being subtracted by 3. Therefore, our common difference for this sequence is -3, negative, because we are subtracting.

common difference

Step 2: Next, let’s write the arithmetic sequence formula and identify each part of our formula (First Term=100, Term number=12, common difference=-3).

arithmetic sequences

Step 3: Finally, let’s fill in our formula and solve with the given values.

Think you are ready to practice solving arithmetic sequences on your own? Try the following practice questions with solutions below:

Practice Questions:

  1. Find the 123rd term given the following sequence: 8, 12, 16, 20, 24, ….
  2. Find the 117th term given the following sequence: 2, 2.5, 3, 3.5, …..
  3. Find the 52nd term given arithmetic sequence: 302, 300, 298, …..
  4. A software engineer charges $100 for the first hour of consulting and $50 for each additional hour.  How much would 500 hours of the consultation cost?

Solutions:

Still got questions? No problem! Don’t hesitate to comment with any questions or check out the video above. Happy calculating! 🙂

Fun Fact!

Did you know that the arithmetic sequence formula can be considered an explicit formula? An explicit formula means that even though we do not know the other terms of a sequence, we can still find the unknown value of any term within the given sequence. For example, in the first example we did in this post (example #1), we wanted to find the value of the 123rd term of the sequence. We were able to do this by using the explicit arithmetic sequence formula, and most importantly, we were able to do this without finding the first 122 previous terms one by one…life is so much easier when there is an explicit arithmetic sequence formula in your life!

Other examples of explicit formulas can be found within the geometric sequence formula and the harmonic series.

Related Posts:

Looking to learn more about sequences? You’ve come to the right place! Check out these sequence resources and posts below. Personally, I recommend looking at the geometric sequence or finite arithmetic series posts next!

Geometric Sequence

Recursive Formula

Finite Arithmetic Series

Finite Geometric Series

Infinite Geometric Series

Golden Ratio in the Real World

Fibonacci Sequence

Still, got questions? No problem! Don’t hesitate to comment below or reach out via email. And if you would like to see more MathSux content, please help support us by following ad subscribing to one of our platforms. Thanks so much for stopping by and happy calculating!

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