Download Visual Fractions (TI-Nspire) file to accompany this article (0.34Mb )

The diagram shows a pair of parallel lines
and a pair of perpendicular lines.
Find the size of angle v when angle u is 20˚.
u
v
Maths Medicine
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GEO
U
© Dexter Graphics 2012
www.mathsmed.co.uk
The Problem
This file offers several ways of solving this problem.
Try to solve the problem before going through the file.
Try to find several methods, then compare them to the methods in this file.
Can you add to the methods in the file?
Page 1
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u
u
v
v
Method 1
This is a standard method; it is very direct and elegant. The new parallel line splits the given right-angle into
two angles. These angles are equal to u and v [using the ‘alternate angles’ or ‘Z-angles’ property of parallel
lines and a transversal].
Thus u + v = 90˚, or v = 90˚ – u.
So v = 70˚ when u = 20˚.
Page 2
Note: As is common in geometry, the symbols u and v are being used in two
ways simultaneously: as names for angles and as numbers of degrees.
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u
v
v
Method 2
When one is stuck, a good strategy is to ‘doodle’.
Here, one of the given lines has been extended to produce a triangle.
It turns out that we know its angles: u [given], v [alternate angles] and
90˚ [angles on a straight line].
Thus u + v + 90˚ = 180˚ [interior angle sum of triangle], so v = 90˚ – u.
Page 3
In a sense, this is less direct than Method 1.
If we think of geometry as a logically ordered
system, then ‘angle sum of triangle’ is based
on angle properties of parallel lines. For
the above triangle, one standard angle-sum
proof involves drawing the line constructed in
Method 1!
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u
u
v
Method 2b
Here is an alternative doodle.
A different line has been extended to produce a triangle.
Again we know its angles.
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u
e
90˚–e
v
Method 3
Another doodle.
Here we have (somehow) hit upon drawing a completely new line, which again produces a triangle.
We know one of its angles is 90˚ [given] so the sum of the other two is 90˚. We don’t know their specific
values but we can write them as, say, e and 90˚– e.
Thus u + e + 90˚– e + v = 180˚ [‘supplementary angles’ property], so u + v = 90˚, so v = 90˚ – u.
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u
45˚
45˚
v
Method 3b
Using the dummy variable e is quite sophisiticated and abstract. We can assign a specific value to e - in this
case 45˚, by turning our triangle into an isosceles triangle by stretching one of the green lines.
This step is perfectly legitimate here, because we have not changed the size of the given angles or v. But
it is also sophisticated - and might be seen as dangerous, as it appears to lend credence to the practice of
assigning properties to a diagram on the basis of how it looks rather than on the basis of what is given.
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u
v
u
Method 4
This is very much like Method 1, in that we have again constructed a parallel line, this time parallel to a
green line.
This produces a new angle adjacent to the give angle v, of size u [corresponding angles].
Altogether the adjacent angles u and v are equal to the given right-angle [alternate angles].
Thus u + v = 90˚, so v = 90˚ – u.
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u
v
u
Method 4r
We can achieve the same thing (as Method 4) by rotating both of the parallel red lines through an angle u.
This produces an angle u adjacent to the give angle v.
Again, the adjacent angles u and v are equal to the given right-angle [alternate angles].
Thus u + v = 90˚, so v = 90˚ – u.
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u
u
v
Method 5r
This again involves a rotation, this time through 90˚.
The original two green lines now form a single straight line (actually, ‘line segment’).
So we have formed a triangle with angles u and v; the third angle is 90˚ because one
of the red lines (which was parallel to the other red line), has turned through 90˚.
Thus u + v +90˚ = 180˚, so v = 90˚ – u.
Page 9
This is rather neat, but
if we see geometry as a
logically ordered system
it can also be regarded as
rather convoluted compared
to Method 1, say. This is
because we use parallel line
properties to prove the angle
sum of a triangle.
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u
w
v
PS
The problem can be made more general by replacing the given right-angle by an angle w.
What is the relationship between the angles u, v and w ?
Can we still apply the previous methods?
Page 10
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