Archive for the ‘SSM’ Category

Is one solution more elegant?

May 14, 2016 13 comments

Earlier this week I wrote this post on mathematical elegance and whether or not it should have marks awarded to it in A level examinations, then bizarrely the next day in my GCSE class I came across a question that could be answered many ways. In fact it was answered in a few ways by my own students.

Here’s the question – it’s from the November Edexcel Non-calculator higher paper:


I like this question, and am going to look at the two ways students attempted it and a third way I think I would have gone for. Before you read in I’d love it if you have a think about how you would go about it and let me know.

Method 1

Before I go into this method I should state that the students weren’t working through the paper, they were completing some booklets I’d made based on questions taken from towards the end of recent exam papers q’s I wanted them to get some practice working on the harder stuff but still be coming at the quite cold (ie not “here’s a booklet on sine and cosine rule,  here’s one on vectors,” etc). As these books were mixed the students had calculators and this student hadn’t noticed it was marked up as a non calculator question.

He handed me his worked and asked to check he’d got it right.  I looked, first he’d used the equation to find points A (3,0) and D (0,6) by subbing 0 in for y and x respectively. He then used right angled triangle trigonometry to work out the angle OAD, then worked out OAP from 90 – OAD and used trig again to work out OP to be 1.5, thus getting the correct answer of 7.5. I didn’t think about the question too much and I didn’t notice that it was marked as non-calculator either. I just followed his working, saw that it was all correct and all followed itself fine and told him he’d got the correct answer.

Method 2

Literally 2 minutes later another student handed me her working for the same question and asked if it was right, I looked and it was full of algebra. As I looked I had the trigonometry based solution in my head so starter to say “No” but then saw she had the right answer so said “Hang on, maybe”.

I read the question fully then looked at her working. She had recognised D as the y intercept of the equation so written (0,6) for that point then had found A by subbing y=0 in to get (3,0). Next she had used the fact that the product of two perpendicular gradients is -1 to work put the gradient of the line through P and A is 1/2.

She then used y = x/2 + c and point A (3,0) to calculate c to be -1/2, which she recognised as the Y intercept, hence finding 5he point P (0,-1.5) it then followed that the answer was 7.5.

A lovely neat solution I thought, and it got me thinking as to which way was more elegant, and if marks for style would be awarded differently. I also thought about which way I would do it.

Method 3

I’m fairly sure that if I was looking at this for the first time I would have initially thought “Trigonometry”, then realised that I can essential bypass the trigonometry bit using similar triangles. As the axes are perpendicular and PAD is a right angle we can deduce that ODA = OAP and OPA = OAD. This gives us two similar triangles.

Using the equation as in both methods above we get the lengths OD = 6 and OA = 3. The length OD in triangle OAD corresponds to the OA in OAP, and OD on OAD corresponds to OP, this means that OP must be half of OA (as OA is half of OD) and is as such 1.5. Thus the length PD is 7.5.

Method 4

This question had me intrigued, so i considered other avenues and came up with Pythagoras’s Theorem.

Obviously AD^2 = 6^2 + 3^2 = 45 (from the top triangle). Then AP^2 = 3^2 + x^2 (where x = OP). And PD = 6 + x so we get:

(6 + x)^2 = 45 + 9 + x^2

x^2 + 12x + 36 = 54 + x^2

12x = 18

x = 1.5

Leading to a final answer of 7.5 again.

Another nice solution. I don’t know which I like best, to be honest. When I looked at the rest of the class’s work it appears that Pythagoras’s Theorem was the method that was most popular, followed by trigonometry then similar triangles. No other student had used the perpendicular gradients method.

I thought it might be interesting to check the mark scheme:


All three methods were there (obviously the trig method was missed due to it being a non calculator paper). I wondered if the ordering of the mark scheme suggested the preference of the exam board, and which solution they find more elegant. I love all the solutions, and although I think similar triangles is the way I’d go at it if OD not seen it, I think I prefer the perpendicular gradients method.

Did you consider this? Which way would you do the question? Which way would your students? Do you tuink one is more elegant? Do you think that matters? I’d love to know, and you can tell me in the comments or via social media!

Cross-posted to Betterqs here.

Exact Trigonometric Ratios

November 8, 2015 4 comments

This morning I read this interesting little post from Andy Lyons (@mrlyonsmaths) which looked at teaching the exact Trigonometric Ratios for certain given angles (namely 0, 30, 60, 90 and 180 degrees). The post gave a nice little info graphic linked to the unit circle to show what was going on and then focused on methods yo remember the ratios.

While reading it I thought about how I introduce these exact Trigonometric Ratios. I first like to know that my students have a thorough and in depth understanding of right angled triangles and the trigonometry involved with them (including Pythagoras’s Theorem). I feel this is imperative to learning mathematics, the Triangle is an extremely important shape in mathematics and to fully understand triangles you must first fully understand the right angled triangle. The rest follows from that.

Once these are understood then you can move on to the trigonometric graphs, showing how these can be generated from right angled triangles within the unit circle, as shown in the info graphic on Andy’s post. Once the graphs are understood then the coordinates f the x and y intercepts and the turning points give us nice exact values for angles of 0, 90 and 180 degrees. This leaves us with 30, 60 and 45 to worry about.

At this point I introduced 2 special right angled triangles. First up is the right angled isosceles triangle with unit lengths of the short sides. This obviously gives us a right angled triangle that has two 45 degree angles (as the angle sum of a triangle is 180) and a hypotenuse of rt2 (via Pythagoras’s Theorem).


Using our definitions of trigonometric ratios (ie sin x =opp/hyp, cos x = adj/hyp and tan x = opp/adj) we can clearly see that tan 45 = 1 and that sin 45 = cos 45 = 1/rt2. This aids the understanding more than just giving the values and allows students a method of working these values out easily if stuck.

The second triangle is an equilateral triangle of side length 2 cut in half. This gives us a right angled triangle with hypotenuse 2, short side lengths 1 and rt3 (again obtained through Pythagoras’s Theorem) and angles 30, 60 and 90.


Again we can use our definitions of trigonometric ratios to conclude that sin 30 = cos 60 = 1/2, sin 60 = cos 30 = rt3/2, tan 30 = 1/rt3 and tan 60 = rt3.

This is again good for deeper understanding and for seeing why sin x = cos 90 – x, and cos x = sin 90 -x. This can lead to a nice discussion around complementary angles and that the word cosine means “sine of the complementary angle”. This triangle is also a good demonstration that tan x = cot 90 – x, when you come to higher level trig.


October 21, 2015 Leave a comment

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, 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) )

A surprising find

April 18, 2015 3 comments

The other day I my timehop showed me this lovely little post from last year. It includes “Heron’s Formula” for calculating the area of a triangle, as I read it I remembered thinking it was a little strange that not many people had heard of it before.

Today I was looking through a number of textbooks trying to find a decent set of questions on area, perimeter and volume for my year nines as I wanted to consolidate their learning at the start then move onto surface area. I’m not a fan of textbook misuse- ie “copy the example and try the questions” but I do sometimes use them for exercises as we have a very limited printing budget and some of them have superb exercises. For a fuller picture view of textbooks, read this.

I was looking in one of my favourite textbooks:


And I happened across this:


There it is! Plain as day! Heron’s Formula! In a KS3 textbook!

I was disappointed that its function was described and its name wasn’t and there was no mention of why this worked. It basically reduces the question down from a geometry one to a purely algebraic substitution task and I would question the appropriateness of including it in an exercise on area, but still, I was incredibly exciting to find it there!

Are you a fan of Heron’s Formula? Had you even heard of it? Do you have a favourite textbook? I’d love to hear your views.

Quadrilateral Puzzle

January 10, 2015 Leave a comment

Yesterday the fantastic Ed Southall (@solvemymaths) tweeted this brilliant puzzle:


It looked fun, so I thought I’d give it a try. First I sketched it out and gave all the vertices labels. A strategy I advise my students to take.


I then considered triangle wzg, as it was the triangle I knew most about:


My first thought was to find the length of the hypotenuse using Pythagoras’s Theorem. This was something that I didn’t use in the end, but I had yet to really formulate at strategy, and as I tell my students, you can never have too much information. I then thought I’d find the angles, but realised that it is this defaulting to trigonometry that often leads me to overcomplicate matters, so I thought I’d leave that til later, (plus I don’t know tan 2 or tan 0.5 off the top of my head.)

I considered the area of the triangle, then sketched the next triangle I knew stuff about fyz. It was here I saw my strategy.


The right angles meant I could use congruent triangles.


I could work out the area of the triangle yxk, which is congruent to fyz, as half the parallelogram area is 32, which is made up of this triangle and one which has area 8.

Thus the other leg of the right angled triangle must be 6rt2, and so a=12rt2 (as y is it’s midpoint!)

From there it was a question of Pythagoras’s Theorem to find b.


A fantastic little puzzle, one that I enjoyed solving, and one which should be accessible to higher GCSE learners. If you haven’t already do check out Ed’s website.

Area of a semi-circle puzzle

December 6, 2014 3 comments

If you have read this blog before, you may have noticed I enjoy a good mathematical puzzle ever now and then. This week I had a bit of time to pass and I thought I’d have a crack at this one from Ed Southall (@solvemymaths):


Well almost, I didn’t have my phone and I misremebered it slightly, so actually had a good at this:


As you can see, it’s the same puzzle but scaled by a factor if 1/2.

The first thins I did was label the 4 points. I knew ABD was a right angled triangle, as it is a semi-circle. We are told that ABC and BCD are right angled in the question. The calculate the area we need the radius, which should be easy enough to calculate through Pythagoras’s Theorem and Right Angled Triangle Trigonometry.

I used Pythagoras’s Theorem to calculate AB.


I could then calculate the sine of angle ABC, which I called x. I know that ABC and CBD (which I called y) are complementary, and as such Sin x = Cos y,  so Cos y = 1/rt5


This meant I could easily work out the length BD using the fact that Cos y = adjacent/hypotenuse.


This left me with the two short sides of the right angled triangle ABD, so the hypotenuse (the diameter of the semi-circle) was easy to calculate:


From this the radius, then the area, follow easily.



When I came to type this up I realised I’d solved the wrong problem, so for completeness sake, to solve the original problem I can multiply by 4 (as I need to enlarge so lengths gave increased SF 2, and I’m dealing with area) giving a final answer of: 112.5pi (or 225pi/2).

I enjoyed this puzzle, and did it without a calculator. I think had I used a calculator, it may have lost some of its appeal. I would love to see questions like this appear on the non-calculator GCSE paper. The skills/knowledge needed are circle theorems, Pythagoras’s Theorem, Right Angled Triangle Trigonometry, Surds, Area of a Semi-circle and basic number skills. All of which should be within the grasp of a decent GCSE student. There are, of course, many other solutions, some of which are explored here.

More triangles!

June 5, 2014 5 comments

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:


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