Topic 20 Further Trigonometry

7/22/2019 Topic 20 Further Trigonometry

1/22

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


2/22

#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


3/22

%&ample: '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


4/22

%&ample: 'inding te lengt of an angle

The diagram shows triangle 7A#.

(alculate the si5e of angle 7#A.

Solution:


%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 =.


5/22

%&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!"


6/22

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.

%&ample: '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 : +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:


7/22

%&ample: '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 : +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 -


8/22

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


9/22

Past e&amination (uestion ,S%G-:

The diagram shows triangle )2(. )2 @


10/22

Hsing the cosine rule:

cosb a c ac B= + ? 6 ? 6 cos3=b = + 9?.=??b =

b@ B.


11/22

%&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


12/22

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.

%&ample:

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


13/22

+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@


14/22

%&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


15/22

)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


16/22

%&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

+


17/22

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.

%&ample 1:

)2(DG' is a cuboid with dimensions


18/22

+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


19/22

#ow we can find )M from triangle )ME:

)M@ )EF ME

)M @ ?.666F

)M@ ??

%o )E @


20/22

%&ample 2:

)2(D is a s/uare-based pyramid. The length )2 is


21/22

* &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 @


22/22

%&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

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Topic 20 Further Trigonometry

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