Trig - Finding unknown angles

So previously we have looked at using trigonometry to find unknown sides, but now we are going to be using trigonometry to find unknown angles.

Before we begin, we need to familiarise ourselves with a new type of trig ratio. We are already familiar with sin, cos, and tan. These are great for finding an unknown side length, but when the unknown is an angle, we need to use something different. For example, if we are asked to find Ө when:

sin Ө = 0.7432

To solve this we want Ө on one side of the equals sign and everything else on the other side. We have an issue, because we have sin with our Ө on one side of the equals sign. To move sin to the other side of the the equals sign, we can shift it over after we turn it into what we call "inverse sin" (which basically means "the opposite of sin"). This looks like sin^-1

So when we rearrange the equation above we get: 

Ө = sin^-10.7432

Which we can put in our calculators and solve. To put sin^-1 into your calculator, you usually just have to press the shift button on your calculator and then press the sin button. Some calculators may not have a shift button. These usually have a button that says "2nd-Func". Press this and then press the sin button. If your calculator has neither of these, your best bet would be to check an instruction manual or do a Google search by simply typing in your calculator brand and model number then "inverse sin". 

Now in this explanation I have only talked about sin, however the same applies for cos and tan. They both have inverses which are cos^-1 and tan^-1

Now we will look at some applications for these. We will often be faced with problems like the following: 

Find the unknown angle in the following triangle

 Now we deal in these in the same way we looked at when we were finding the unknown sides, by following the 5 steps.
So step 1 was to label the sides:

Step 2 was to find which sides we will be using. We are given values for our adjacent and opposite sides, so we will be using these. 

Step 3 was to work out which trig ratio we will be using. For this problem we will be using tan as that is the only one that uses the opposite and adjacent sides. 

Step 4 was to write and equation. This is where it gets slightly different to when we found the side lengths. 

We are trying to find the angle Ө, so to start with we have:

tan Ө = "something" 

To work out what this "something" is, we need to use those formula triangles we used when we were finding our side lengths. For this problem we will use this one: 

Now, we are trying to find tan Ө, we can cover up the T. By doing this we can find our "something". Covering it up we get: 

This means our something is "O over A" or basically opposite divided by adjacent. So we get:

tan Ө = O/A

Now we don't actually want tan Ө, all we want is Ө. So using the skills from the top of this post, we can rearrange our formula so it just has Ө on one side of the equals sign, and everything else on the other side. We get:

                                                                      Ө = tan^-1 O/A

Now step 5 is just substituting in values for O and A and solving for Ө. So if we look back up at the orginal question and substitute these values in we get: 

                                                                 Ө = tan^-1 38.15/23.45

If we put this in our calculators we should get and answer of 58.42..... degrees. Now if your calculator doesn't automatically put brackets in after you press the sin, cos or tan button, you will have to do this manually, so you do need brackets around the 38.15/23.45 

These skills should allow you to finish off the worksheet for trigonometry in the worksheets tab. As always, if you have any questions don't hesitate to see me or to comment on this post. 

Trigonometry - Finding Unknown Sides

Continuing with trigonometry, we looked at finding unknown side lengths. To start with, we need to familiarise ourselves with the following acronym:

SOH CAH TOA

This acronym stands for Sin, Opposite, Hypotenuse, Cos, Adjacent, Hypotenuse, Tan, Opposite, Adjacent. We use this to help us to remember what each of the trig ratios is (i.e. Sin is Opposite over Hypotenuse, which we can get from the order of the acronym). 

When we are finding side lengths, we want to rearrange this acronym into sets, with a set for each trig ratio. The best bet is to arrange them into triangles like so:

We can see that if we move left to right through these triangles we still have our acronym in its correct order. Now you are probably all wondering "why do we need to do this?". All will be revealed soon. 

Now when we are faced with a problem like so:

           Find the unknown value in the following problem

To solve problems like this one, we need to use 5 key steps. They are:

1. Label the sides (i.e. label the hypotenuse, the adjacent and the opposite sides)
2. Find which sides we are using for this problem (i.e. which sides are we given some information about)
3. Find which ratio requires us to use these sides (i.e. if we are using the opposite and the hypotenuse, then we will have to use the Sin ratio)
4. Write out the equation required to solve the problem
5. Substitute in our values and solve our equation. 

So using these steps, lets solve the problem above. 

Firstly we have to label our sides. If you are unfamiliar with how to do this, look at the posts below. When we do this for our example above we get: 

Now step two is to find which sides we are actually using for this problem. If we look at the triangle, we are given the length of the hypotenuse. We are also given an unknown value for the opposite side. We are not given any information regarding the adjacent side, so we won't be using this side. For this triangle the two sides we are using are the opposite side and the hypotenuse. 

Step 3 is to find which ratio uses these two particular sides. In this triangle we know it will be the Sin ratio, because the two sides we are using are the opposite side and the hypotenuse. 

Step 4 is to write out the equation. To do this we need to use the triangles we looked at above. Since we know we are using the Sin ratio (from step 3), we only need to look at the first of the 3 little triangles: 

We can use this triangle to help us write the equation. Since we are trying to find the value of the opposite side, we know we want the equation to be O = "something"

To work our what this "something" is, we use the triangle. Since we are finding the value for O (the opposite side), we cover this up in our triangle. We are now left with:

The "S" stands for "sin of an angle" and the "H" stands for "the length of the hypotenuse". Since they are next to one another, we need to multiply them together. This is the "something" in our equation. So finally, for this step, our equation is:

O = sin Ө x H

Step 5 is to substitute our values into this equation and solve it. For this triangle, our value for Ө is 40 (the size of the angle in degrees) and our value for H is 64cm (the length of the hypotenuse). So to solve this equation for O, we substitute these values into our equation and then use a calculator to solve it. 

When we substitute our values in we get:

O = sin (40) x 64

Putting this into our calculator we get a value of 41.1cm  to 1 decimal place for our opposite side. 

Looking at one more example, we may have: 

             Find the unknown for the following triangle:

Again, following the same 5 steps, we can easily solve this one. 

Step 1: Label the sides

Step 2: We will be using the opposite and the adjacent sides

Step 3: This means that we will be using a Tan ratio. 

Step 4: We will be using the following triangle:

We don't know our adjacent side, so we want the equation to be A = "something". To work out this "something" we cover up the part we want to find out, so in this case the A. 

We are left with O over T, or to be more specific "the length of our opposite side" over "tan of our angle". Now because they aren't next to one another like the previous example, we do not multiply. This time, they are over one another, so we are dividing. This gives us a final equation of:

A = O/tan Ө 

Step 5 gives us:

A = 38.2/tan (27)

If we put this into our calculator we get a final answer of 75.0cm to 1 decimal place for our adjacent side

This should give you enough information to work through the 4th set of problems on the worksheet (which can be found in the worksheets tab). If you have any questions please feel free to comment on here or send me an email. 

Trigonometry Lesson 2

Today we continued with Trigonometry. There were a lot of people away today on University/College open days, so we had a pretty easy day. That being said, a lot of people should be looking on here to catch up!

We mainly focused on how we use calculators today. To do this trigonometry we need to be using scientific calculators. These are basically any calculator that has a "sin" "cos" and "tan" button. While a lot of phones do now have these on their calculators, it is a lot easier and a lot safer to use a proper calculator.

When we first turn our calculator on we have to make sure it is in degrees mode. Since each calculator is slightly different, I suggest you look for instructions for your specific calculator as to how to do this. If you are having troubles, leave a comment or come and see me and I will try and work it out with you.

Once our calculator is in degrees mode, we can then start working through problems. For example, we may be asked to find sin 43^o 
To do this, we have to grab our calculator and press the "sin" button. We then enter the number of degrees. For this example we will enter "43". Some calculators will try and put this in brackets for you, {i.e. your calculator will read out sin(43 }, which is fine. You just have to make sure that you close these brackets by pressing the " ) " button. Once we have entered this into our calculator, hit the equals button. If you have done this correctly, you should get 0.6819983601

You may also be asked to find something like 6.7 cos 31<sup>o
This is basically asking us to find "6.7 lots of cos 31 degrees". To do this, we grab our calculators and put in "6.7" then hit the "multiply" button. We then press the "cos" button and then type in "31". If you need to, make sure you close the brackets. Then hit equals. You should get 5.743020915

You may also be asked to find 43/</sup>tan 21^o

To do this we want to grab the calculator again and put in "43" then press the "divide" button then hit "tan" then "21", again remembering to close your brackets if your calculator requires. Press equals and you should get 112.0188298




We also looked at finding unknowns in different trig problems. For example we may be able to find x in sin 37^o = x/9
To do this we want to rearrange the equation to have x by itself on one side of the equals sign, then everything else on the other side. To do this we want to multiply each side by 9 (i.e. the number on the bottom of the fraction). This will effectively cancel the 9 from the right hand side, and we will be left with:

9 sin 37^o = x

This leaves us with x by itself on one side of the equals sign and everything else on the other side so we can now just use our calculator to solve it. We can then easily solve for x by sticking the part on the left hand side into our calculator in the same way we did in the examples above. We will then get x = 5.416335208

Similar to this, we may also be asked to find cos 82^o = 14/x

Again, same as before we want to rearrange this so the x in by itself on one side of the equals sign and everything else is on the other side. This question is slightly different to the one above. If we look at the fraction on the right hand side, we see the the number is on the top and the letter is on the bottom. This is the opposite to the previous example above, where the letter was on top and the number was on the bottom. This means when we rearrange the equation, we will pretty much do the opposite of what we did above. In the example above, we moved the 9 to the left hand side and multiplied it. For this example we will move the 14 to the left had side as well, except because we are doing the opposite we will divide it this time. So, we will be left with:

14/cos 82^o = x

This leaves us with x by itself on one side of the equals sign and everything else on the other side so we can now just use our calculator to solve it. Again, we just put the part on the left hand side of the brackets into our calculator using the same technique as we did above. If we do this we should get     x = 100.5941515

This should give you the skills to work through the next two problems on the worksheet. This worksheet can be found under the "Worksheets" tab at the top of the page. As always, feel free to comment on this post if you have any questions or concerns. 

Also, if you do not have a scientific calculator as I mentioned above I would highly suggest you get one. You should be able to find one in Kmart or Officeworks or other places. I did find this one on eBay for a great price if you are happy to buy one online: 

http://www.ebay.com.au/itm/Scientific-Statistical-Calculator-Battery-Operated-240-Function-10-2-digits-New-/251909345045?pt=LH_DefaultDomain_15&hash=item3aa6f78f15

Test back + Intro to Trigonometry

Today we handed the tests back and had a look through those. If you would like to know your specific mark or have any questions regarding the test feel free to email me or comment on this post. I will be uploading a copy of my answers for the test if people are interested, however we are still waiting on students to complete tests, so once they have finished I will put the answers up.

We also look at a new topic today. We are now starting a unit on Trigonometry which is all to do with studying triangles. This unit was covered last year in your year 9 course, however we will look further into this as we progress through the unit. To start with today we just looked at the basic Trig (trig is short for trigonometric) ratios. These are our sine ratio, our cosine ratio, and our tangent ratios (often referred to as sin, cos, and tan ratios). These are as follows:

Remember that the funny little symbol θ simply means "angle", so tan θ means "the tan ratio of the angle is equal to". 

Now we can see that the ratios are all to do with different sides of the triangle. Each side has a special name. The most important side is the hypotenuse. This is the longest side of the triangle. The opposite side is the side opposite our angle θ. Finally, our adjacent side is the shorter side that touches, or is next to, our angle θ. 

So from this we can find our different trig ratios for different angles. For example, we may be asked to find different ratios for triangles like this: 

Now we may be asked to find sin α 

The first step we must take is to work our which side is which. I find the easiest side to find is always the hypotenuse. For this triangle, the hypotenuse is the 5m long side. The next easiest side to find is the opposite side. In this example we want to find the side opposite to the angle α. For this example, our opposite side is the 4m long side, leaving the adjacent to be the 3m long side. 

We then want to substitute our values in for the ratio required. So for this example we are using the sin ratio. The sin ratio is opposite/hypotenuse, so we can substitute our values in for that. 

So finally, we get sin α = 4/5, since our opposite side length is 4 and our hypotenuse side length is 5. 

Using the same triangle, we can find a variety of different trig ratios. Another example may be to find tan β. 

Again, our first step is to work out which side is which. The hypotenuse is always the longest side, so as with the previous example it will be the 5 m side. The next two sides do get a little tricky though. Since we have now changed the angle of reference (from α to β) the opposite and adjacent sides do change as well. Since we are now using β, we must look for a side opposite to this angle. The opposite side is now the 3 m side. This means that the adjacent side is the 4 m side. 

We then want to substitute our values in for the ratio required. So for this example we are using the tan ratio. The tan ratio is opposite/adjacent, so we can substitute our values in for that. 

So finally, we get tan β = 3/4, since our opposite side length is 3 and our adjacent side length is 4. 

I have also uploaded a worksheet under the worksheets tab. This worksheet contains a wide variety of problems and will be used for a large chunk of the Trigonometry unit. For today, just look at the very first question. This is very similar to the examples I have just gone through. 

As always, if you have any questions or concerns feel free to comment on this post or email me. 

Revision Lesson

Today we went through the revision sheet (which can be found in the worksheets tab). We focused on question 1, 4, 6, and 8 from the Factorising section and then questions 1, 5, 7, 10, and 11 from the Expanding section. These are the main types of questions you will be faced with on the test. I have included worked examples of all of these questions also in the worksheets tab.

When you are working through the revision sheet, ignore questions 11, 12, 13, and 17 from the factorising section as well as questions 16 and 17 from the expanding section.

Again, a reminder that we do have a test on Wednesday the 10th June on Algebra. You are allowed an A5 information sheet (single sided). I suggest you include an example of each of the questions highlighted in the first paragraph on your information sheet. You are also allowed a calculator, however this may not be a calculator on your phone.

Use the posts below to help you with your revision. They have good explanations for all the topics we have covered in this unit. They also have useful links to websites that may help as well.

As always, if you have any questions feel free to comment on any of the posts on this site.

Finishing Off

Wednesday's lesson was spent finishing off factorising quadratics as well as factorising using the differences of two squares method.
We also got a revision sheet, which will be quite helpful as practice for the test. Friday's lesson will be dedicated to looking through this sheet, so I suggest looking through this sheet before Friday so if there are any problems that you have issues with, we can work through those as a class on Friday. This revision sheet can be found in the worksheets section of this page.

A reminder that we do have a test on Wednesday 10th June. It will be on all areas of algebra that we have covered this term (i.e. expanding and factorising). You are able to bring in an A5 single sided information sheet, which can contain any information you think will be helpful. You are also allowed calculators, however you may not use a calculator on your phone.

I have also put the answers for the expanding assignment up in the assignments tab of this page.

As always, feel free to comment on this post or any others if you do have any questions or comments.

Recap lesson

So today's lesson was just a recap on factorising quadratics as well as factorising using the difference of two squares method. If you have any issues with either of these areas, look at my posts below. If you are still having troubles, put up a comment or email or see me.

We also discussed when we would do our algebra test. We came up with Wednesday 10th of June. This gives us this Wednesday's lesson to finish up on factorising, then Friday's lesson to do some revision. We do have a public holiday on Monday, so we won't have class then.
For the test, you will be allowed to bring in a calculator (not one on your phone!), as well as an A5 single sided information sheet which you can include any information you want on. You will have to make this information sheet in your own time but it will be very helpful, so I strongly suggest spending some time on this.

As always, if you have any questions or concerns regarding the test or anything we have covered in this algebra unit you can comment on this post or shoot me an email.