## Wednesday, February 13, 2013

1. In the right-angled triangle MNO, MO = 4.5 metres and angle MON = 32 o.  Calculate the length of MN.

Trigonometry usually begins with a sketch of a right angled triangle and definitions of the hypotenuse and the opposite and adjacent ‘legs’ for a given angle. Next up, of course, are the sine, cosine and tangent functions.

Consequently we begin by identifying the side opposite the 32 o angle, the side adjacent to the 32 o angle and the hypotenuse.   We write down the mnemonic ‘SOH CAH TOA’ to help us recall that which function is which.
As we are given the hypotenuse and are asked for the opposite side, our best choice is the sine function.  (If you choose the cosine function you will calculate the length of segment ON.  You can still find MN by following on with Pythagoras (a2 + b2 = c2) but that option is more time-consuming.   If you go with ‘tan θ’ however, you are at a dead end.
sine 32 o = MN/MO  = MN/4.5
MN = 4.5* sine 32 o = 2.384…
To get an answer with three significant figures, we ought to jot down four figures. The digit on the right (alternatively, the digit for the thousandths place) will be the ‘decider’. As the rule for ‘4’ is to round down, we write MN = 2. 38 metres (to 3 significant figures).
If the teacher wanted to make the question a little bit harder s/he could change the orientation of the triangle.  Or the student could be given the opposite side and required to find the hypotenuse.
2.           GHI is a right triangle;  GH  = 8 cm and HI = 7 cm;  angle H = 90° Calculate the size of angle HIG.
Again this question follows the standard development of trigonometry.  After learning to add, we learn to subtract; after learning to multiply, we learn to divide.  After learning the standard ‘trig’ functions, we learn the inverse functions: arc-sine, arc-cosine and arc-tangent.

Your calculator and your text probably have the notation sin– 1,   cos– 1   and tan – 1. The latter notation emphasizes that these are inverse functions, whereas the somewhat old-fashioned ‘arc’ notation emphasizes that the output is an angle.
tan (θ) = 8/7
tan – 1(tan (θ)) = tan – 1 (8/7)
θ = 48.81°
After giving the required rounding, angle HIG = 48.8°  (to one decimal place).
3. UVWX is a dart.
a) Calculate the length of side UV.     nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn
This question is best solved with the cosine rule, c2 = a2 + b2 – 2ab cos C.  The cosine rule is an extension of Pythagoras.  The last term disappears when C = 90°, since cos 90° = 0.   The cosine rule is (in my opinion) most easily proved with vectors, so for today we will simply take it as it is.  To apply the formula we will re-label triangle UVW as follows:

c2 = a2 + b2 – 2ab cos C
c2 = 3.92 + 5.72 – 2*3.9*5.7 cos 124°
c2 = 3.92 + 5.72 + 24.86… (Note that cos 124° is negative so the last term becomes positive.)
c2 = 72.5617…
Some students, carelessly, write c 2 to 3 figures.  The table below shows that their final answer will be marked wrong and a very ‘easy’ mark has been lost as a result.
 If c 2 is written down as then ‘c’ is calculated as and ‘c’ written down to 3 significant figures as 72.56217… 8.518343 8.52 72.561… 8.518274 8.52 72.5… 8.514693 8.51

b) Calculate the size of angle UXW.
A syllabus certainly has more topics than a quiz (or a test or an exam) can cover.  The examiner is unlikely to ‘waste’ a precious resource by asking the same question twice.  (Of course if your examiner has a question bank from which a computer program samples randomly, all bets are off.  I once had a 25 question multiple choice paper for which ‘oxytocin’ was the correct answer, thrice!)
Yes, you could find the angle by doing another cosine rule calculation.  But the givens are two sides and one angle so it is much easier to use the sine rule, instead.  The formula is shorter, and there is no need to take a square root at the end.
sin x/5.7 = sin 108°/8.2
sin x = 0.661100…
x =   41.4°  (to one decimal place).
Once again, it is advisable to keep many more figures than required, and round off as your final step. The sine rule is proved quite easily from the definition of sine and the formula for the area of a triangle, but that is a story for another day.

4.                    Square-based pyramid PQRST is shown in the diagram below.  All the edges are 3 cm in length.

This is a so-called 3-D problem, but in reality three points (not all on a straight line) define a triangle and a planar (or 2-D) surface.  The skill is in finding the correct triangle.  With 6 points (P, Q, R, S, T and M) we can find 6C3 = 20 different triangles, where (for example) PQR and PQS are different, but PQR and RQP are the same.  So guesswork will take too long.   You need to become proficient at looking at a fairly cluttered diagram and extracting just the bits you need.

a) Calculate the length of PR.

To calculate PR we only need the base PQRS.  In fact, we only need the isosceles right triangle PQR.  It is a right triangle because angle Q is the vertex of a square, hence 90°.  It is isosceles, because PQ and QR are the sides of a square so both are the same length, i.e., 3 cm.

Notice that the sketch on the right is very rough.  There is no attempt made to make angle Q look like a right angle , but it is labelled as a right angle, which is enough.
Since the triangle is isosceles, both angle P and angle R = 45°.  In keeping with the theme (trigonometry) we can calculate PR with sine or cosine.  For example
sin 45° = 3/PR  ß à PR = 3/(sin 45°) = 3/(1/2) = 32 = 4.14 cm (to 3 ‘sig figs’)
Undoubtedly, most students will use Pythagoras instead, with the same result.

b) Calculate the vertical height, MT.

As the length of PR was 32, the length of PM is 1.5*2 or 2.12 cm.   Point M is vertically under T.  As the square is symmetrical, point M is where the diagonals PR and QS intersect, i.e. M is the midpoint of PR.
We could use the inverse cosine function to find angle TPM and with the angle we could use any of sin, cos or tan to solve for h.   Or we could use Pythagoras, base2 + height2 = hypotenuse2 à hypotenuse2  –  base2 = height2.
If we do part c) before part b), we get another isosceles right triangle, which from the symmetry of the situation should not be terribly surprising.  In other words, the height h is the same as the base, 1.5*2 or 2.12 cm.

c) Calculate the angle between PT and the base PQRS.
As already indicated in part b), we have hypotenuse and the side adjacent to the required angle, so this is an inverse cosine problem.

cos (θ) = 2.12/3
cos – 1(cos (θ)) = cos – 1 (0.70666…)
θ = 45.0°  (to 3 sf). Had you calculated with the exact value, ( 1.5*2)/3 = 1/2, you would realise that indeed θ is exactly 45°.

What else could have gone into this trigonometry quiz?  Reading graphs of the sine or cosine function, bearings and scale drawings are the likely candidates.