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.

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

*any*of this!

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