Trigonometry Quiz: Answers
1. In the rightangled 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 (a^{2} + b^{2} =
c^{2}) but that option is more timeconsuming. 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: arcsine, arccosine
and arctangent.
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 oldfashioned ‘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, c^{2} = a^{2} + b^{2} – 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 relabel triangle UVW as follows:
c^{2} = a^{2} + b^{2}
– 2ab cos C
c^{2} = 3.9^{2} +
5.7^{2} – 2*3.9*5.7 cos 124°
c^{2} = 3.9^{2} +
5.7^{2} + 24.86… (Note that cos 124° is negative so the last term
becomes positive.)
c^{2} = 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. Squarebased pyramid PQRST is shown in the diagram below. All the edges are 3 cm in length.
This is a socalled 3D problem, but in reality
three points (not all on a straight line) define a triangle and a planar (or
2D) surface. The skill is in finding
the correct triangle. With 6 points (P,
Q, R, S, T and M) we can find _{6}C_{3} = 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) = 3√2 = 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 3√2, 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, base^{2} + height^{2} = hypotenuse^{2} à hypotenuse^{2 }– base^{2} = height^{2}.
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.
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