topic 20 further trigonometry
TRANSCRIPT
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7/22/2019 Topic 20 Further Trigonometry
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Revision Topic 20: Sine and Cosine Rules and 3D Trigonometry
The objectives of this unit are to:
* use the sine and cosine rules to find the length of sides and angles in triangles;* to use the formula for the area of a triangles;
* to solve problems involving the sine and cosine rules;* to solve problems involving trigonometry in 3 dimensions.
Brief recap of Grade B and C material:
Pytagoras! Teorem:
This theorem, which connects the lengths of the sides in right-angled triangles, states that:
a b c+ =
where cis the length of the hypotenuse !i.e. the side opposite the right-angle" and aand bare the
lengths of the other two sides.
#ote that the hypotenuse is the longest side in a right-angled triangle.
Trigonometry
The following formulae lin$ the sides and angles in right-angled triangles:
sin O
xH
=
cos A
xH
=
tan O
xA
=
where His the length of the ypotenuse;
Ois the length of the side oppositethe angle;
Ais the length of the side ad"acentto the angle.
These formulae are often remembered using the acronym %&'()'T&) or by using mnemonics.
'ere is a commonly used mnemonic:
%illy &ld 'arry (ouldnt )nswer 'is Test &n )lgebra
+hen finding angles, remember that you need to use the %'T $ey.
urther notes, eamples and eamination /uestions relating to 0ythagoras1 theorem and
trigonometry are contained in separate revision boo$lets.
%ometimes you need to calculate lengths and angles in triangles which do not contain any right-
angles. This is when the sine and cosine rules are useful.
c
b
a
H
A
O
x
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#a$elling a triangle
To use the sine and cosine rules, you need to understand the convention for labelling sides and
angles in any triangle.
(onsider a general triangle:
Triangles are named after their vertices - the above triangle is called triangle )2(.
The three angles are commonly referred to as angles ), 2 and (.The length of the sides are given lower case letters:
%ide ais the side opposite angle ). t is sometimes referred to as side 2(.
%ide bis the side opposite angle 2. t is e/uivalently called side )(.
%ide cis the side opposite angle (. t is also $nown as side )2.
) triangle doesn1t have to be labelled using the letters ), 2 and (.
or eample, consider the triangle 04 below:
Sine Rule
The sine rule connects the length of sides and angles in any triangle )2(:
t states that:
sin sin sin
a b c
A B C= = .
)n alternative version of the formula is often used when finding the si5e of angles:
sin sin sinA B C
a b c= =
A B
C
c
ab
A B
C
c
ab
P Q
R
r
pq
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%&le: 'inding te lengt of a side
The diagram shows triangle )2(.
(alculate the length of side )2.
Solution:
To find the length of a side using the sine rule, follow these steps:
%tep 6: 7abel the triangle using the conventions outlined earlier.
%tep : 7oo$ to see whether any additional information can be added to the diagram !for eample,
can you deduce the length of any other angles8"
%tep 3: %ubstitute information from the diagram into the sine rule formula.
%tep 9: elete the unnecessary part of the formula.%tep : 4earrange and then wor$ out the length of the re/uired side.
n our eample, we begin by labelling the sides and by wor$ing out the si5e of the 3 rdangle !using
the fact that the sum of the angles in any triangle is 6.
%ubstituting into the formulasin sin sin
a b c
A B C= = , we get:
63.
sin sin 3 sin ?
a c= =
)s we want to calculate the length cand as the middle part of the formula is completely $nown, we
delete the first part of the formula:
63.
sin 3 sin ?
c=
4earranging this formula !by multiplying by sin?" gives:
63. sin ?
sin3c
=
i.e. c@ 6.? cm !to 6 decimal place".
55o
72o
A B
13.2 cm
C
55o
72o
A B
13.2 cm
C
c
ab
53o
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%&le: 'inding te lengt of an angle
The diagram shows triangle 7A#.
(alculate the si5e of angle 7#A.
Solution:
%tep 6: 7abel the triangle using the conventions outlined earlier.
%tep : %ubstitute information from the diagram into the sine rule formulasin sin sinA B C
a b c= = .
%tep 3: elete the unnecessary part of the formula.
%tep 9: 4earrange and then wor$ out the si5e of the re/uired angle.
n our eample, the labelled triangle loo$s li$e:
The cosine rule formula !adjusted for our lettering" is:
sin sin sinL M N
l m n= =
%ubstituting into this gives:
sin sin639 sin
6?. B.C
L N
l= =
+e want to find angle # and we $now the middle part of the formula completely. +e therefore
delete the first part of the formula, leaving
sin639 sin
6?. B.C
N=
f we multiply by B.C, we get:B.C sin639
sin =.
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%&amination style (uestion
n triangle )2(, angle )2( @ B>, angle )(2 @ 3, 2( @ 6 cm.
+or$ out the length of )2.
%&amination style (uestion:
n triangle )2(, angle 2)( @ 66>, )( @ cm and 2( @ C cm.(alculate the si5e of angle )2(.
A
B C
65o
38o
15 cm
Diagram NOTaccurately ra!"
5 cm
9 cm
115o
A
B C
Diagram NOTaccurately ra!"
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Cosine Rule
The cosine rule also connects the length of sides and angles in any triangle )2(:
t states that: cosa b c bc A= +
D/uivalently, we also have these formulae:
cosb a c ac B= + cosc a b ab C = +
Eou need to be familiar with the structure of these formulae. n particular note that the letter that
appears as the subject of the formula also appears as the angle.
#ote that the cosine rule can be considered as an etension of 0ythagoras1 theorem.
%&le: 'inding te lengt of a side
The diagram shows triangle )2(.
(alculate the length of side )2.
Solution:
To find the length of a side using the cosine rule, follow these steps:
%tep 6: 7abel the triangle using the conventions outlined earlier.
%tep : +rite down the appropriate version of the cosine rule formula and substitute information
from the diagram into it.
%tep 3: +or$ out the length of the re/uired side.
&ur labelled diagram here is:
)s we wish to find the length of c, we need the formula with cas the subject:
cosc a b ab C = +
%ubstitute in: 69.? 6?.C 69.? 6?.C cosBc = +
Typing the right hand side into a calculator !in one long string and pressing @ only at the end" gives:
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%&le: 'inding te lengt of an angle
The diagram shows triangle 2(.
(alculate the length of angle 2(.
Solution:
To find the length of a side using the cosine rule, follow these steps:
%tep 6: 7abel the triangle using the conventions outlined earlier.
%tep : +rite down the appropriate version of the cosine rule formula and substitute information
from the diagram into it.
%tep 3: 4earrange and wor$ out the length of the re/uired angle.
The labelled diagram here loo$s li$e:
+e want to find angleD. +e therefore need to write down a version of the cosine rule formula that
contains angleD.
The subject of the appropriate formula would therefore be d:
cosd b c bc D= +
%ubstituting into this formula gives:
66.9 6=.? 63. 6=.? 63. cos
6C.CB @ 669.9C F 6?9.9 -
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125o
7 cm 12 cm
B
A C
)or*ed %&amination +uestion
n triangle )2(, )2 @ C cm, 2( @ 6 cm and angle )2( @ 66=>.
(alculate the perimeter of the triangle.
Give your answer correct to the nearest cm.
Solution:n order to calculate the perimeter, we need to wor$ out the length of the third side.
7abelling the triangle:
Hsing the cosine rule:
cosb a c ac B= + 6 C 6 C cos66=b = + 3C.
(alculate the length of )(.
11#o
15 cm
9 cm
Diagram NOTaccurately ra!".
A B
C
11#o
15 cm
9 cmA B
C
b a
c
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Past e&amination (uestion ,S%G-:
The diagram shows triangle )2(. )2 @
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Hsing the cosine rule:
cosb a c ac B= + ? 6 ? 6 cos3=b = + 9?.=??b =
b@ B.
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%&amination +uestion ,%de&cel une 2001-
iagram #&T
accurately drawn.
a" (alculate the length of )2. Give your answer in centimetres correct to 3 significant figures.
b" (alculate the si5e of angle )2(. Give your answer correct to 3 significant figures.
%&amination +uestion ,%de&cel ovem$er 14-
n the /uadrilateral )2(, )2 @ B cm, 2( @ ?
cm, ) @ 6 cm, angle )2( @ 6=>, angle
)( @ ?=>.
(alculate the si5e of angle )(. Give your
answer correct to 3 significant figures.
A B
C
8#o
1# cm8 cm
12 cm
7 cm
6 cm
B
C
D
A
7#o
12#o
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5rea of a triangle
The area of a triangle can be found using this alternative formula:
)rea of a triangle @6
sin
ab C
)lternative versions are:
)rea @6
sin
ac B or )rea @6
sin
bc A
This can e/uivalently be thought of as
)rea @ J K product of two sides K sine of the included angle.
%&le:
ind the area of the triangle:
+e can use the above formula to find the area of this triangle as we have two sides and the included
angle !i.e. the angle in between the given sides":
)rea @6
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+e can use the formula for the area of a triangle to find the length of )(:
)rea @6
sin
ab C
%o6
9= B= sin6=
b=
9= 3= sin6=
9= 6 as sin6= @ =.
b
b
=
=%o b @ 3= m.
To find the perimeter, we also need the length )2. +e can use the cosine rule:
cosc a b ab C = + 3= B= 3= B=cos6=c = + ?B6?.BC...c =
%o c@
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%&amination style (uestion
n triangle 04, 0 @ C cm, angle 04 @ B?> and angle 04 @ B=>.
(alculate the area of triangle 04.
Pro$lem style (uestions
uestions are often set involving bearings or angles of elevation.
f a diagram has not been drawn in the /uestion, you will need to begin by s$etching a diagram to
illustrate the situation.
There will usually be several steps re/uired in order to get to the solution.
'
(
)
9 cm
6#o
67o
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)or*ed e&amination (uestion ,%de&cel une 2007-
The diagram shows a vertical tower ( on hori5ontal ground )2(. )2( is a straight line.
The angle of elevation of from ) is .
The angle of elevation of from 2 is 9>.
)2 @ m.
(alculate the height of the tower. Give your answer to 3 significant figures.
Solution:
%tep 6: Hse triangle )2 to find the length 2.
%tep : Hse triangle 2( to find the height (.
%tep 6: rom the original diagram, we can deduce that angle )2 @ 6B> and angle )2 @ B>.
Hsing the sine rule:
sin < sin6B
a b=
sin B=
%o sin < B.??sin B
a m= =
%tep :
Hsing trigonometry for right-angled triangles:
sin9B.??
B.?? sin9
6.?m
CD
CD
CD
=
=
=
%o the tower is 6.? m tall.
A B C
D
54o
28o
25 m
Diagram NOTaccurately ra!".
A B
D
28o
25 m
a
&
126o
B C
D
54o
26.77 m
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%&amination (uestion ,%5B-
) helicopter leaves a heliport ' and its measuring instruments show that it flies 3. $m on a bearing
of 6 to a chec$point (. t then flies 9.? $m on a bearing of =BB> to its base 2.
a" %how that angle '(2 is 66.
b" (alculate the direct distance from the heliport ' to the base.
128o
66o
4.7 *m
3.2 *m
C
B
+
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3D trigonometry
Grade )L )* /uestions often involve you finding distances and angles in 3 dimensional objects.
The $ey to these /uestions is to identify and draw the relevant dimensional triangle.
%&le 1:
)2(DG' is a cuboid with dimensions
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+e now use 0ythagoras1 theorem to find )G:
)G@ )(F (G
)G @ 6=F
)G@ 6
%o )G @ 66. cm
c" ind angle G)(.
The letters mentioned in the name of the angle tell you which triangle to draw !i.e. triangle G)(".
This is the triangle drawn in part !b".
+e can use trigonometry to find angle G)(.
tanN @
6=
i.e N @ B.B>
d" ind the length )M.
* )M is a diagonal length across the cuboid. #et 8 $e te point vertically $elo. 9. +e draw
triangle )ME:
* There is not yet enough information in the diagram to find length )M.
* +e can wor$ out length )E however if we draw out the base )2(:
+e can use 0ythagoras1 theorem to find )E:
)E@ )F E
)E @ BF 9
)E@
%o )E @ ?.666 cm !note that we don1t round too early".
N
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#ow we can find )M from triangle )ME:
)M@ )EF ME
)M @ ?.666F
)M@ ??
%o )E @
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%&le 2:
)2(D is a s/uare-based pyramid. The length )2 is
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* &ne way to find the area of triangle )D2 would be to find the height of this triangle !by splitting
it into right-angled triangles" and then using the formula J bKh. )lternatively, we could find
angle )D2 !for eample using the cosine rule" and then using the formula: area @ J absinC.
f we use the first method, then we must begin by finding the height h of the triangle.
Hsing 0ythagoras1 theorem:
hPF 9P @ CP
hPF 6B @
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%&amination (uestion ,%de&cel ovem$er 200-
The diagram shows a pyramid.
The base, )2(, is a hori5ontal s/uare of side 6= cm.
The verte Q is vertically above the midpoint, A, of the base.
QA @ 6 cm.
(alculate the si5e of angle Q)A.
%&amination (uestion ,%de&cel-
The diagram represents a prism.
)D is a rectangle.
)2( is a s/uare.
D and ( are perpendicular to plane )2(.
)2 @ ) @ B= cm.
)ngle )2D @ C= degrees.)ngle 2)D @ 3= degrees.
(alculate the si5e of the angle that the line D ma$es with the plane )2(.
M
1# cm
1# cmA B
CD
AD
BC
,-
6# cm
6# cm
3#o