## Thursday, June 13, 2013

### Leaving Certificate Mathematics 2013: Paper 2, Question 8 “Solution”

So, the State Examination Commission in Ireland made a giant cock-up. At issue is a particular mathematics question, so — as someone not entirely ignorant of basic mathematics — I thought I'd explain the problem.

 Leaving Cert Mathematics 2013, Paper 2, Question 8

The problem is that a triangle is completely specified by three values — either angles or lengths — provided that at least one of them is a length: one length and two angles, two lengths and one angle, or three lengths. Given one of these, you can compute the missing values: two lengths and one angle, one length and two angles, or three angles, respectively.

The question text specifies two lengths: $$|HR|=80~\text{km}$$ and $$|RP|=~110\text{km}$$ and one angle: $$\angle{r}=124^\circ$$ (or $$\angle{HRP}=124^\circ$$ if you prefer the more verbose notation). The diagram shows one angle consistent with the text, $$r=124^\circ$$, and — in the English version of the exam — another angle, $$h=36^\circ$$, which is inconsistent with the values given in the text.

If we take the text as correct, and ignore the diagram, let's see what happens — the problem now is: find $$\angle h$$ given (dropping the units and angle symbols for convenience): $|HR|=80 \\ |RP|=110 \\ r=124$
We know that the sum of the internal angles of any triangle is $$180^\circ$$, so: $r+h+p=180 \Rightarrow h+p = 180-r = 56$
If we drop a line vertically from the apex, $$R$$, to a point, $$X$$, on the base, we then have two right-angle triangles “back-to-back”, i.e. sharing the line-segment $$RX$$.

From the right-angle triangle, $$HRX$$, on the left, we have: $|RX|=|HR|\sin(h)$ and from the right-angle triangle on the right, $$PRX$$, we have: $|RX|=|RP|\sin(p)$ Combining these two: $|HR|\sin(h)=|RP|\sin(p)$ But we know that $$h+p=56$$ or $$p=56-h$$, so $|HR|\sin(h) = |RP|\sin(56-h)\qquad\qquad(1)$ Now, a basic trigonometric relation (listed in the “log tables” that every candidate gets in the exam) is $\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$ Applying this to the RHS of (1), we get $|HR|\sin(h)=|RP|(\sin(56)\cos(h)-\cos(56)\sin(h))$ Rearranging: $(|HR|+|RP|\cos(56))\sin(h) = |RP|\sin(56)\cos(h)$ or $\tan(h) = \frac{\sin(h)}{\cos(h)} = \frac{|RP|\sin(56)}{|HR|+|RP|\cos(56)}$ or $h = \tan^{-1}\left( \frac{|RP|\sin(56)}{|HR|+|RP|\cos(56)}\right)$ Substituting in the values we know:  $h = \tan^{-1}\left( \frac{110\times 0.8290}{80+110\times0.5592}\right) = \tan^{-1}(0.6444) = 32.80^\circ$ Therefore $32.80\neq 36 \Rightarrow \text{The SEC are morons}$

Now, the question that is actually asked is “Find the distance from R to HP”. There are at least two further problems:

• the distance from $$R$$ to the line-segment $$HP$$ is not uniquely defined (we must assume that the perpendicular distance, i.e. $$|RX|$$ is intended); and
• even under the simplifying assumption that the Earth is a sphere, latitude makes a significant difference to the answer (consider point $$R$$ being at the North pole vs. line-segment $$HP$$ lying on the equator), and non-Euclidean geometry is not on the syllabus.
If one uses the given value for $$h$$, the problem becomes utterly trivial: $|RX|=|HR|\sin(36^\circ)$ If we don't ignore the diagram, we have two angles and two lengths and must decide to discard either one of the angles or one of the lengths (in the foregoing, we discarded $$h$$ from the diagram and kept the two lengths and angle $$r$$ from the text), so there are actually four choices for how to proceed.

All in all, a gargantuan cock-up and staggering incompetence from the SEC. Imagine that nobody there spotted any of this!