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Explainer: Here's the problem with the Leaving Cert honours maths paper

Warning: the following article contains trigonometry.

AROUND 15,000 STUDENTS were expected to take the Higher Level papers in the Maths Leaving Cert – perhaps hoping to get 25 valuable bonus points for the CAO race later this summer.

All of them were expected to take Question 8, worth 30 marks – the one which has caused minor outrage around the country because of a printing blunder.

The error meant one question had multiple correct answers, while at the same time also being technically impossible to solve.

Here’s how.

The question

A port P is directly east of a port H. To sail from H to P, a ship first sails 80 km, in the direction shown in the diagram, to the point R before turning through the angle of 124° and sailing 110 km directly to P.

(i) Find the distance from R to HP.

In other words: let’s pretend you’re the captain of a ship which is currently docked at Port H. You’ll shortly be setting off to Port P – but you’re not going to be travelling directly there: you’ll need to take a detour, going off to one side to visit a dock R in between.

You’re told that if you drew a line between H to R, and then from R to P, the angle at R would be 124°. You’re also told that it’s 80 kilometres from H to R, and 110 kilometres from R to P.

You’re also told – incorrectly - that if you drew straight lines between H and P, and H and R, the angle between them would be 36°. (Look very closely – you’ll see that the width of the angle isn’t explicitly described in the text, and the figure is only written into the diagram.)

You’re asked to work out the distance between R and the straight line which joins H and P.

Alarm bells

From the off, the inclusion of four pieces of information is the sort of thing that should warrant some alarm bells, according to Eamonn Toland of, who has put together a walk-through video showing how the question was botched.

Ordinarily for a trigonometry problem you’ll be given three pieces of data – with the length of one side and the width of the two angles at either end, or the length of two lines and the width of the angle in between them.

Because the three angles in a triangle will always add up to 180°, and because you can use basic trigonometry to work out the length of the other sides, you should only ever be given three pieces of information.

Any more than that – particularly at higher level – and there ought to be a suspicion that maybe the pre-provided information is too good to be true.

First steps

Remember that the three angles of a triangle would always add to up to 180° – so, having been given the width of two angles, you can very easily work out the width of the third by adding the first two together, and subtracting your answer from 180°.

In this case, you can add 124° and 36° to get 160° – meaning the other angle would, by definition, be 20°.

So - as you can see in this screen grab from Eamonn's video - you can simply draw a line between the angle R and the line HP, which then splits your large, shapeless ('scalene') triangle into two right-angled triangles. (For the sake of explanation, let's mark a dot at the spot on HP where you draw the line, and call it 'A'.)

This makes it trivial to figure out the length of the line between R and A - because you can use basic trigonometry to work it out.

You only need to decide which triangle you're going to use - whether it's the one between points H, A and R, or the one between P, A and R.

If you use use HAR, you can work out the length of the line AR with the 'sine formula' (the sine of the angle is equal to the length of the opposite side, divided by the length of the hypotenuse) based on the angle at H - and you'll get an answer around 47km.

But! If you use triangle PAR, and work it out using the sine formula based on the angle at P, you'll get an answer of around 38km.

Each is a legitimate way of approaching the question - and because of the data the student has been given, each is a perfectly legitimate answer.

As Eamonn explains, a strong student might have found this question to be so basic that it was almost routine - so they might have double-checked their answer by working out the length of the line between A and R with both methods, and then become flustered by getting two perfectly valid results.

"It's a bit ironic that the more you know about maths, the more problems that this triangle will cause you," Eamonn succintly explains.

The problem is...

...that it's actually impossible to draw a triangle where two lines are as long as the paper says, and where the two angles are as wide as the paper says.

If you follow all of those instructions and try to draw a triangle - as Eamonn did - you get this:

If you're not told the width of the angle at H, it's much simpler to construct a triangle - you only need to draw a straight line between H and P:

As it happens, you can then apply some routine trigonometry to discover that the angle at H is not 36° at all - it's actually 33°, meaning the angle at P is 23° and not the 20° that students were (implicitly) told.

This, of course, actually changes the whole complexion of the question - because if the question was actually put as originally intended, and the student decided to answer it by splitting the triangle into two, they would have found the correct answer to be around 43 kilometres.

But what if you read the question properly?

Eamonn points out that the description given in the written part of the question didn't actually reflect the triangle drawn out beside it. The image above is his drawing of what a triangle would actually look like if it followed the written instruction.

The description talks about a ship turning at an angle of 124° - when in fact this would mean a turn far sharper than a right angle, leading to a triangle which can't possibly correspond to the instructions given.

It's more a case of sloppy wording on the part of the question setter rather than any critical error - but a thorough student who read every question diligently may well have been thrown by the disconnect between the words and the image, and become even more flustered as a result.

...and going on...

The question above is Part (a) of the botched question. This is Part (b):

The point T is directly east of the point R. | HT | = 110 km and | TP | = 80 km. Find | RT |.

In layman's terms: it turns out we also have to visit a dock at another location, T - which, as it happens, is exactly 110 km from H and exactly 80 km from P. Figure out the distance between R and T.

Most students will have noticed that the triangle between H, P and T is basically a mirror image of the original one between H, P and R - the distance between the two end points are still the same.

Here's what happens if you copy the triangle HPR (which we've shaded in pink) and paste it onto HPT:

Many students are likely to have treated this shape as a trapezoid with four corners - H, R, T and P - and borrowed some of the stuff they had learned from the first question.

This, crucially, would include the distance between R and the line HP - which, as we've seen, could be either 43km, 38km or 47km.

Having drawn a four-cornered shape with two parallel lines, they could then split the shape into three - consisting of two right-angled triangles with a rectangle in between:

Based on the information given to them in the first question, students should have been able to figure out that the distance between H and P was 170 kilometres. (Again, this is fairly basic trigonometry for an honours Leaving Cert paper.)

A Leaving Cert honours Maths student should see that the two triangles are identical - one is simply the mirror image of the other - and therefore the distance between R and T is the same as the length between H and P, minus the length of the two triangles (i.e. between H and A).

By now, students will know the length of two sides and the width of two angles in the triangle HAR - but if they use the wrong ingredients, plugging in the answers they got from the first question, they're going to get the wrong result again.

If the angle at H was 33°, as it should have been, the length of HA is 67km - meaning the distance between R and T is 36 kilometres - but if they were using the mistaken 36° figure, they will have found the length of HA to be 65km and then discovered RT to be 40km.

So what happens now?

The marking scheme for the question will have to reflect the fact that there were three possible answers for part (a) - and give full marks to anyone who may have gotten the wrong answer, but used a legitimate method to get there.

Similarly, anyone who used their legitimate answer for part (a) and plugged it into part (b) will also be accommodated.

Where things aren't so clear is what happens to the students who may have done enough work to get full marks - but who might have been flustered or flummoxed by the error, and spent extra time doing the question.

Q8 in total was worth 30 marks out of a possible 300 - so it was worth 10 per cent of the entire paper, and therefore would have been worth 15 minutes of a student's time. Indeed, the question would be worth 5 per cent of a pupil's total Maths grade - and five valuable CAO points.

It's easy to see how a diligent pupil checking their work could have spent far longer than 15 minutes poring over their answer, desperately trying to find an error that did not exist.

How that will be reflected is up to the chief examiner.

Read: Marking of Leaving Cert Maths papers ‘will reflect time lost by students’

Plus: Further errors in Leaving Cert maths exams highlighted

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