## Angles or Angels?

When I marked my year 11 books the other day I noticed that quite a few had been working that morning on “Angels in triangles’. This peturbed me a little, surely by Year 11 they should know the difference and be able to spell each one.

To counteract this massive literacy issue I played a game of “Angles and Angels”. I spoke to them first about the difference, then about the spelling and then did a show me activity where I showed them various pictures and they had to show me on their whiteboards if it was an angle or an “Angel”. I was impressed that they even got the picture of Kurt Angle, although none of them recognised David Boreanas…..

The activity led to a discussing with a couple of them as to why it was important to discuss these things in maths lessons. Stemming from the inevitable question “why we learning about this? It’s maths not English.”

I explained my opinion that we may be learning maths, but that literacy is important in all subjects. As a maths teacher I educate these students and literacy has to be a big party of that, as I hope numeracy is a big party of those subjects that deal with numbers but aren’t maths. I also expressed the importance of maths specific vocabulary, such as ‘angles’ and how it’s not necessarily going to be covered in English.

It is these sorts of things that we need to be thinking about, literacy wise, to ensure our students are in the best position when they leave.

## A lovely angle puzzle

I’ve written before about the app “Brilliant“, which is well worth getting, and I also follow their Facebook page which provides me with a regular stream questions. Occasionally I have to think about how to tackle them, and they’re excellent. More often, a question comes up that I look at and think would be awesome to use in a lesson.

Earlier this week this question popped up:

What a lovely question that combines algebra and angle reasoning! I can’t wait to teach this next time, and I am planning on using this as a starter with my y11 class after the break.

The initial question looks simple, it appears you sum the angles and set it equal to 360 degrees, this is what I expect my class to do. If you do this you get:

*7x + 2y + 6z – 20 = 360*

*7x + 2y + 6z = 380 (1)*

I anticipate some will try to give up at this point, but hopefully the resilience I’ve been trying to build will kick in and they’ll see they need more equations. If any need a hint I will tell them to consider vertically opposite angles. They should then get:

*2x – 20 = 2y + 2z (2)*

*And*

*3x = 2x + 4z (3)*

I’m hoping they will now see that 3 equations and 3 unknowns is enough to solve. There are obviously a number of ways to go from here. I would rearrange equation 3 to get:

*x = 4z (4)*

Subbing into 2 we get:

*8z – 20 = 2y + 2z*

*6z = 2y + 20 (5)*

Subbing into 1

*28z + 2y + 6z = 380*

*34z = 380 – 2y (6)*

Add equation (5) to (6)

*40z = 400*

*z = 10 (7)*

Then equation 4 gives:

*x = 40*

And equation 2 gives:

*60 = 2y + 20*

*40 = 2y*

*y = 20.*

From here you can find the solution x + y + z = 40 + 20 + 10 = 70.

A lovely puzzle that combines a few areas and needs some resilience and perseverance to complete. I enjoyed working through it and I’m looking forward to testing it out on some students.

*Cross-posted to Betterqs here.*

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October 21, 2015
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## Parallelograms

Parallelograms, you know, the weird quadrilaterals that look like a sheared rectangle. These:

I’ve never rally thought that deeply about them, to be honest. They have some uses in angle reasoning lessons, and we need to be able to find their area in the GCSE, but I’ve not thought too deeply about them recently at all.

When teaching how to find the area I normally do this:

It’s a fine method, and easy to show that it works by showing that you can cut the end off, stick.it in the other end and get a rectangle which is clearly of the same area.

But last week I marked a mock exam in which one of my year 11s had done this:

I love this method, it’s much, much nicer than the other. I couldn’t wait to question him. When I did he said that he “couldn’t remember” how to do it, but knee how to find the area of a non right angled triangle so split it into two of them which were congruent using SAS.

I asked him what would happen if you split the parallelogram across the other diagonal. He thought about it for a while, and eventually told me it would be fine because of “how the sine curve is” and because, “the angles add up to 180”.

I was impressed by his reasoning. He has clearly understood this method and generalised the area of a parallelogram in a way I’d never considered. I would have phrased is slightly differently though:

*The area of a parallelogram is equal to the product of two adjecent sides multiple by the sine of one of the angles. (Either will so as Sin x = Sin (180 – x) )*

## More triangles!

Earlier this evening, Jane Moreton (@PGCE_Maths) tweeted this:

I looked at it for a moment and started pondering. In this case comes was clearly always going to be 45. I wondered whether the others would change and then realised they wouldn’t. I decided to draw it out and play around with some angle reasoning.

When I drew it out something else occured to me, there would be 3 right angled triangles, all with the same opposite side (from the given angle) and differing adjacent sides that were multiples of each other. It occurred that the sidelengths didn’t matter, and I could reason it out with tangent ratios instead (and everyone knows trigonometry is more fun).

*Tan (a) = 1/3
Tan (b) = 1/2
Tan (c) = 1*

I realise that as the angles will always be the same I could evaluate each one and show that a+b = c. But a) I didn’t have a calculator on me and b) that’s no fun!

Instead, I used the addition formula for Tan(a+b):

*Tan (a+b) = (Tan (a) + Tan (b))/(1-Tan(a)Tan(b))*

Using our values for Tan (a) and Tan (b) we get a numerator of 1/3 + 1/2 which equals 5/6. We get a denominator of 1-(1/3)(1/2) which also equals 5/6. So we get Tan (a+b) = 1, Tan (c) = 1 too so Tan (a+b) = Tan (c). As a,b and c are all in right angled triangles they all fall between 0 and 90 degrees, so a + b must equal c which equals 45 degrees.

A nice little mental workout. I will show some of my classes tomorrow and next week.

Here’s my back of an envelope workings:

## Tangrams and Angle Reasoning

Yesterday I came across this tweet from @learningmaths

I thought; “what a lovely little activity”. I’ve been a fan of tangram puzzles for a while, but this task was an entirely new one on me, and I love it.

The first thing that stuck me was the angles. Even without the lengths, this is a fantastic problem, which uses a lot of angle reasoning. When you add lengths it involves all sorts of other maths which can be tackled in many different ways! I intend to trial this as starters for my year 11 (angles only),12 and 13 (angles and lengths) classes, and I hope to build it into future teaching of angle rules and reasoning.

I am also tempted to dig out my own tangram set and investigate what other shapes I can create which will produce equally interesting problems!

I think there is potential to use this in area and enlargement problems, “if the triangle has area x, what sf have I enlarged by?”, “if I alter the hexagon to have side length 5cm, what’s the area?” I’m sure there are many more uses too. I will post again if and when I think of them!

If you have any further ideas I’d live to hear them!

Here is a larger version of the picture:

## Angle problem

Today has been quite a geometric based day for me. I spent a couple of hours solving non-RAT trigonometry problems with year 10 and then a while with year 11 looking at various algebra angle problems. Then I went on Twitter and saw this from Ed Southall (@solvemymaths):

A couple of nice parallel lines questions that I might grow at y11 tomorrow.

Both are fairly straight forward to solve. I looked at the first one, imagines a third parallel line through the join if x and saw x must be the sum of 40 and 60 hence 100.

The second I saw an alternate angle to the 50 in the top triangle and used angle sum of a triangle is 180 to spot that x is a right angle. I glanced down at the responses and saw the vast majority had the same answers as me. That would probably have been the end of it but then I noticed this response:

The same thought process for the first one, but a significantly different approach to the second.

It made me wonder what approach others would take, and which approach my students would take. I wondered if the first problem had led this respondent into this solutions the second, and if so why it hasn’t had the same effect as me.

I don’t know if either approach is better, I just thought the differences were interesting. I’d love to hear your thoughts on it and how you would approach it.

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