WEBVTT

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Hi, I'm Richard.

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Let's look at TAMUA 2021 paper two,
question 18.

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So let's start by reading it through.

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It says a student chooses 2 distinct real
numbers X&

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Y and we have 0 less than X less than Y
less than one.

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The student then attempts to draw a
triangle ABC with AB equals 1,

b9dd2b7a-ce5c-43c0-b8f9-6a7c461bab28-1
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sine of A = X, sine of B = y,
and we are asked which of the following

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is or are correct.

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So statement one for some XY there is
only one exactly 1 possible triangle.

abaafcc6-8569-4212-bccc-31f464874436-0
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Statement 2 for some XY there are exactly
2 possible triangles the student could

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have drawn.

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And statement 3 for some XY.

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There are exactly 3 different possible
triangles the student could have drawn.

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OK, a few things about this question.

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First of all,
it's really important that we realise

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that this means the length of the line AB
equals one.

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It would be very easy to think this means
multiplication of A&B, but it doesn't.

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It means it's telling us about the length
of what we will consider to be the base

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in the triangle.

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So we'll have AB as our base and that has
got length of 1,

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sine of A = X and sine of b = y.

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Well, X&Y are between zero and one.

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We know that A&
B need to be angles because we're going

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to draw a picture of a triangle and A&
B are going to be some of the angles in

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the triangle.

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So I think the first thing to do is think
about what this means for the

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possibilities for big A and big BI.

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Think a good approach to this is to draw
a graph of sine.

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OK,
so this graph shows the sine curve

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between 0 and 180.

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We're only interested in values between 0
and 180 of this problem because we know

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this is a problem about angles and
triangles,

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and angles and triangles can be no more
than 180.

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OK,
so we need to imagine what the

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possibilities are for this angle A and
this angle B given that we have 0 less

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than X less than Y less than one.

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So let's imagine that we mark X&
Y here on the Y axis and notice we have 0

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less than X less than Y less than one.

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So the possibilities for the angle A
given that sine of A is this value X,

b839ab47-3e6d-4ad8-8f0a-45daadf194b2-1
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it means that A is either this angle here
or it is this angle here.

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And if we call this angle alpha,
then this angle would of course be 180

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minus alpha.

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Now let's think about B.

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We have sine of B = y.

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So if we look at our graph,
we draw across them down capital B could

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be this angle or it could be this angle.

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So if we call this value here beta,
then this value here would be 180 minus

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beta.

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So we can see there are two different
possibilities for what the angle big A

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could be, this value and this value,
and similarly two different possibilities

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for what the angle big B could be,
which is this one and this one.

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So actually this will mean that there are
four possible cases to consider overall.

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So case one would be when A equals alpha
and B equals beta.

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So this is when they're the two acute
angles we see alpha and beta.

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We would have a second case which is when
A is alpha but B is the other possibility

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for B which is 180 minus beta.

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OK.

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So these are the four cases we need to
consider.

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We've got A&
B being the two acute tangles,

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alpha and beta.

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We've got A being the acute tangle alpha
and B being the obtuse angle 180 minus

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beta,
A being the obtuse angle 180 minus alpha,

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and B being beta.

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And then the final case is when A&
B are the two possible obtuse angles

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there.

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So let's think about what the picture
might look like in each of these cases.

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So always we will start with our base
which will label as AB.

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And we know we were told this has got
length one in the problem.

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So the angle A needs to be at this point
and the angle B needs to be at this point.

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So in case one,
we would have the angle at A being alpha.

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We can see alpha is the smaller of alpha
and beta because the angle at B is beta

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which will be a larger acute angle.

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But we can see we would get a triangle
with these possibilities.

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So we can see our triangle here with the
points at the top where the angles are as

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in case 1.

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So we would certainly get at least one
triangle.

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So now let's look at case 2.

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So in case 2A is the acute angle alpha.

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So we can draw that as before.

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Let's imagine it like this.

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B has become the obtuse angle 180 minus
beta.

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So if we imagine what we had before with
beta like this as an acute angle but

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larger than alpha,
if the angle at B is now 180 minus beta,

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that means it would look something like
this.

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But you can see from our picture that
this line segment is steeper than this

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line segment.

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And so we can imagine that our two line
segments would actually cross at some

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point and we would actually get a
triangle in this case.

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So, so far we have established that yes,
we would have a triangle where at where

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angles like this and we would also have a
triangle if our angles were like this.

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Let's move on to case 3.

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So in case 3A has become the obtuse angle
118 minus alpha.

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So if we imagine that alpha is like this,
we need to take an angle which is 180

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minus alpha.

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So actually if we reflect this line here,
we can imagine that this would be an

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angle of 180 minus alpha.

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With the alpha appearing here,
the angle at B is beta and beta is larger

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than alpha.

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So what that would mean is we would have
a steeper line here.

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So this would be the angle beta.

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This would be the angle AA is less than
beta.

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So this line segment is shallower than
this line segment.

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And we can see that these two lines would
not meet.

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So we do not get a triangle in this case.

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Our final case is when both of the angles
are obtuse.

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So in the final case,
we have the angle A is 180 minus alpha.

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So if we imagine alpha to be here,
we can imagine a reflection of alpha here,

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which means we would get our angle of 180
minus alpha at A like this,

9b66f2da-bf32-4527-b19d-66849c76e7c4-2
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and at B we've got an angle of 180 minus
beta.

a739130f-109e-41d4-87b3-1c47cd2c5c6c-0
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So if we imagine our angle beta,
which is bigger than alpha here,

a739130f-109e-41d4-87b3-1c47cd2c5c6c-1
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and imagine reflecting that,
so we've got an angle beta over here,

a739130f-109e-41d4-87b3-1c47cd2c5c6c-2
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that would then give us 180 minus beta
here.

2ad1d7cb-b926-47ff-8fde-963dbba4c56d-0
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But you can certainly see that the two
line segments are not going to meet with

2ad1d7cb-b926-47ff-8fde-963dbba4c56d-1
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these two angles becoming obtuse at A&
B.

cc346937-c033-41d2-a80b-dc2406e601f2-0
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And so this will not give us a triangle
either.

884d568c-7a7c-4892-98c0-ea41cb09d936-0
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So it looks like the answer is that for a
choice of X&Y like this,

884d568c-7a7c-4892-98c0-ea41cb09d936-1
00:07:28.109 --> 00:07:32.633
0 less than X less than Y less than one,
there are two possible triangles that the

884d568c-7a7c-4892-98c0-ea41cb09d936-2
00:07:32.633 --> 00:07:33.560
student can draw.

617b8a03-9aa7-4abd-bcdd-c2a4ff3e08c2-0
00:07:33.920 --> 00:07:36.760
And that means we are looking for answer
2 here.