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.