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7/25/2019 Ch6 - Trigonometry - Prelim Maths
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Trigonometry Chapter 6
Trigonometric Ratios and the Calculator
Angles can be given in degrees, minutes and seconds. We usually use angles in
degrees and minutes.
25 32 '43 ' '
We need to know how to round of angles correctly. ust like time, minutes and
seconds are out o! 6", not #"", i.e.$
To %nd unknown sides in right&angled triangles, you will use the !ollowing buttons
on your calculator$
sin cos tan
To calculate angles you will use$
sin1
cos1
tan1
To change a decimal to degrees, minutes and seconds, or vice versa, use the
button$
' '( or )*+
Examples
#. ound of to the nearest
degree$-.
a 23 16 '34 ' '
/.
b 52 43 '14 ' '
0.1. ound of to the nearest
minute$6.
c 23 16 '34 ' '
2.
d 52 43 '14 ' '
3.
4. Change 34 54 ' to a
decimal.
#".
##.Change 76.42 to degrees
and minutes.#-.#/.5ind correct to - decimal
places$#0.
a sin12 38 '
#1.
b tan 56
13
'#6.
age 7 #
6" minutes 8 9999 degree
(60'=1 )
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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#2.5ind in degrees and
minutes$#3.
a sin=0.456
#4.
b cos =( 35 )
age 7 -
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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20.
21. Answers
#. ound of to the nearest degree$--.
a 23 16'34
' '=23
-/.
b 52 43'14
' '=53
-0.-. ound of to the nearest minute$-1.
a 23 16 '34 ' '
-6.
b 52 43 '14 ' '
27.
/. 34 54'=34.9
-3.-4.
0. 76.42 =76 25'12
''
/". 7625 '
/#.1. 5ind correct to - decimal places$/-.
a sin 12 38'=0.21871
//. 0.22
/0.
b tan 56 13'=1.4947
/1. 1.49
/6.
6. 5ind
in degrees and minutes$/2.
a sin=0.456
/3. = sin1
0.456
/4. 27.129
0". 27 7'46 ' '
0#. 27 8'
0-.
b cos=( 35 )
The minutes are less than /"
so you round downto the
The minutes are more than/" so you round upto the
The seconds are more than
/" so you round upto the
The seconds are less than /"
so you round downto the
Type in /0 : ' '( 10 : ' '( enter ;or
8.
: ' '(
Type in 26.0- then : ' '( . Write
what your calculator says and
then round of to the nearest
Type in sin then #- : ' '( /3 : ' '(
then to close the brackets and
Type in tan then 16 : ' '( #/ : ' '(
then to close the brackets and
Type in shi!t ;or -nd sin then
".016 then to close the brackets
and then enter ;or 8. ress : ' '(
and then round appropriately.
Type in shi!t ;or -nd cos then
35 then to close the
brackets and then enter ;or 8.
: ' '(
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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hypotenuse
ad=acent
ad=ace
nt
hypotenuse
>
>
0/. = cos
1( 35 )00. 53.1301
01. 53 7'48.368 ' '
06. 53 8 '
02.
48.
49. Trigonometric Ratios
1".To calculate the length o! a side or the si?e o! an angle in right&angled
triangles, we can use certain ratios.
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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66.
62.
63.
64.
2".
2#.
2-.
!. Reci"rocal #dentities
20.TAB $ # 8 +C $ # 8 CD+C $ # 8 CDT $ # 8
21.
26.TAB 8 +C 8 CD+C 8CDT 8
22.
23.Eere are the / reciprocalratios or reciprocal identities$
79.
80.
81.
82.
83.
84.
8.
8!.
32.
88.
89.
90.
91.
92. Examples
#. Write down the ratios o! sin ,cos , tan , cosec , seccot .
Cosecant cosec = 1
sin =
Secant sec= 1
cos =
Cotangent cot = 1
tan=
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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#2
#1
3
>
#2
#1
3
>
Dpposite ;D
Ad=acent ;A
Eypotenuse ;E
>
4/.40.41.46.42.43.44.
-.
sin=2
5 , %nd the eact ratios o!
cos , tan,cosec,seccot
.
100. Answers
#.#"#.
#"-.
#"/.
#"0.
#"1.sin =
O
H=
8
17
cosec = 1
sin =
1
8
17
=17
8
#"6.cos
=
A
H=
15
17
sec= 1
cos =
1
15
17
=17
15
#"2.tan =
O
A=
8
15
cot = 1tan
= 18
15
= 158
-. sin=
2
5
#"3.#"4.##".###.
##-.##/.##0.##1.##6.
Food idea to label
sides. Dnce you
become !amiliar
they should become
To begin, draw a right angled
triangle with an unknown
angle. Bet use the !act that
sin =O
Hand write in these
sides on the triangle. To %nd
the third side use tha oras(
1 -
sin = OH
= 25
cosec = 1
sin =
1
2
5
=5
2
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>
4" >
;Gsing the angle sum o! a triangle is #3".A
HCa
b
c
##2. c2
=a2
+ b2
##3. 52
=a2
+ 22
##4. 25=a2
+4
#-". 44
#-#. 21=
a
2
#--. a=21
#-/.#-0.
#-1. Co$unction #dentities
#-6. Gse your calculator to evaluate the !ollowing to /decimal places
#-2.
#-3. +in -": 8 cos 2": 8
#-4.
#/". +in 1": 8 cos 0": 8
#/#.
#/-. +in /1: 8 cos 11: 8
#//. What have you noticedI
#/0.
#/1. Why do you think the answers are the sameI
#/6.
#/2. We have established our %rst important result, namely that
138. " the sine o# an an$le is e%ual to the cosine o# it&s
complement "
#/4. This is represented using the mathematical symbols
140. 'in ( ) cos *90 + ( ,
%4%.
#0-.
#0/.
#00.
#01.
#06.
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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#02.(90 ) =sin=sin
#03.(90 )=cos=cos
#04.(90 ) =tan=tan
#1".sec =
1
cos=
1
a
c
=c
asec (90 )= 1
cos (90 )=
1
b
c
=c
b
#1#.
(90 ) = 1
sin (90 )=
1
a
c
=
cosec = 1
sin =
1
b
c
=cosec
#1-.
(90 ) = 1tan (90 )
=1
a
b
=
cot = 1
tan =
1
b
a
=cot
#1/. 5rom these ratios come the
results$
#10.
#11.
#16.
#12.
#13.
#14.
#6".
#6#.
#6-.
sin = (90 )
cos= (90 )
tan = (90 )
cot = (90 )
sec= (90 )
cosec = (90 )
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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#6/.
#60.
1!.
1!!. Examples
#. +how sin 32 =cos58 .
#62.
-. +impli!ysec 65
sec 65 + cosec25
#63.
/. 5ind the value o! x i! tan 27=cot (x+20 ) .
#64.
0. 5ind the value o! p i! sin (2p+15 )=cos (3p20)
170. Answers
#. sin 32=cos (9032 )
#2#. cos68
#2-.
-.sec 65
sec 65 + cosec25 =
sec 65
sec 65 +sec (9025 )
#2/. sec65
sec 65 + sec 65
#20.
sec65
2 sec 65
#21.
1
2
#26.#22.#23.
/. tan 27= cot (x +20)
#24.
#3". tan 27=cot (9027 )
#3#. cot 63
#3-.
#3/. cot (x+20 ) =cot 63
#30.
#31. x +20=63
#36. 2020
#32. x=43
#33.
+ince - o! the / ratios are
sec change the third one to
sec. Ji.e.
+impli!y by collecting like
terms on the bottom.
+impli!y by cancelling
sec65 !rom to and
These should now be the
same because they are
These should now be the
same because they are both
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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#- cm cm
#/.0 m
y m
a b
0. sin (2p+15 )=cos (3p20)
#34.
#4". sin (2p+15 )= cos [ 90 (2p +15 )]
#4#. cos [902p15 ]
#4-. cos (752p )
#4/.
#40. cos (3p20 ) =cos (752p )
#41.
#46. 3p20=752p
#42. +2p+ 2p
#43. 5p20=75
#44. +20+20
-"". 5p=95
-"#. 5 5
-"-. p=19
-"/.
&'4.
&'(.&'). Right-Angled Triangle Pro*lems
-"2. Trigonometry can be used to %nd unknown sides or angles in
triangles.
208. -indin$ a 'ide
-"4.
;other than the right angle you can %nd the length o! any other side in thetriangle using the trigonometric ratios. *ake sure you use the ratios
correctly, as the unknown side can be on the top or the bottom o! the
!raction when you try and solve.
210. Examples
#. 5ind the length o! the unknown side to - decimal places in the !ollowing
triangles$-##.-#-.
-#/.-#0.-#1.-#6.-#2.
He care!ul when taking away.
Always put in brackets and
then multiply out. He care!ul
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-#3.
219.
220.
221. Answers
#. Locate the known angle, decide what sides you have and then which ratio
to use$---.
a < have an opposite side and a hypotenuse so it is sine.
--/. sin=
O
H
--0. sin 23 45
'=
x
12
--1. 1212
--6. 12 sin 23 45'=x
--2. 4.8329 =x
--3. x=4.83
--4.b < have an ad=acent and hypotenuse so it is cosine.
-/". cos=
A
H
-/#. cos67 12'=13.4
y
-/-. y y
-//. y cos67 12'=13.4
-/0. cos67 12' cos67 12
'
-/1. y=34.57925 .
-/6. x=34.58
-/2.
238. -indin$ an An$le
-/4.
%nd the si?e o! any o! the other two angles in the triangle using the
trigonometric ratios. emember to %nd angles we will be using
sin1
,cos1
, tan1
.
240.
241. Examples
Botice how the pronumeral is on the
bottom. He care!ul and don(t mi themround as you will get a diferent answer.
+ome people like to always put the
pronumeral on the top o! the !raction.
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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a b3.- cm
#".2 cm
16 m
2- m
#. 5ind the si?e o! the unknown angle to the nearest minute in the !ollowing
triangles$
-0-.
-0/.
-00.
-01.
24!.
247.
248.
249.
20. Answers
#. Locate the unknown angle, decide what sides you have and then which
ratio to use$-1#.
a < have an ad=acent side and an hypotenuse, so it is cosine.
-1-. cos=
A
H
-1/. cos=
8.2
10.7
-10. = cos
1( 8.210.7 )-11. =39.9722
-16. =39 58'20.271 ' '
-12. =39 58'
-13.-. < have an opposite side and an ad=acent side, so it is tangent.
-14. tan=O
A
-6". tan =
56
72
-6#. = tan
1( 5672 )-6-. =37.8749
-6/. =37
52
'29.941
' '
-60. =39 52'
-61.
emember to close the
emember i! the seconds are
less than /" you round down
to the nearest minute.
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-66.
&).
&)8.
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&)9. A""lications
-2". Trigonometry can have many practical applications, !rom such areas
as building, construction, surveying and navigation.
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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278. earin$s
-24. Hearings are based on directions related to the compass. True
*earingsmeasure angles cloc+,ise!rom orth. Hearings are usually
written as / numbers e.g. 124 ,048 , etc.
-3". Always draw a diagram i! one is not provided.-3#. The bearing o! a point is the number o! degrees in the angle
measured in a clockwise direction !rom the north line to the line =oining the
centre o! the compass with the point.
-3-. A bearing is used to represent the direction o! one point relative to
another point.
-3/.
-30. 5or eample, the bearing o!A !rom Bis "61M. The bearing o! B !romAis -01M.
285.
-36.
287. Examples
#. 5rom a pointA, level with the !oot o! a vertical pole and 25m !rom it, the
angle o! elevation o! the top o! the pole is 40 . Calculate$
a The height o! the pole ;to the nearest metre,b The angle o! elevation !romAo! a point hal!way up the
pole ;to the nearest minute.-33.
-. An observer in a lighthouse 100 m above sea level is watching a ship
sailing towards the lighthouse. The angle o! depression o! the ship !rom
the observer is 15 .
a Eow !ar is the ship !rom the lighthouseI Correct to the
nearest metre.
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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b +ome time later, the angle o! depression is measured as
25 . Eow !ar has the ship travelledI Correct to the
nearest metre.-34.
/. Bick cycles 15km due north, then 12km due east and %nally 20km
due south. What are his distance and bearing !rom his original positionI
-4".
291.
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A
292. Answers
#. )raw a diagram$-4/.-40.-41.-46.-42.
-43.-44./"".
/"#. tan 40 =
h
25tan =
10.5
25
/"-. 2525 =tan1( 10.525)
/"/. 20.97749=h 22.782
/"0. h=21m 22 47 '
/"1./"6./"2.-. )raw a diagram$/"3./"4./#"./##./#-./#/.
/#0./#1.
/#6.
/#2. tan 15=
100
x
tan 25=100
y
/#3. x x y y
/#4. x tan 15 =100y tan 25=100
/-". tan 15 tan 15 tan 25 tan 25
/-#. x=
100
tan 15 y=
100
tan 25
/--. 373.2050 =214.450
/-/. 373m 214m
/-0.
DistanceTravelle=373214
/-1.159m
b
a
21 mh
10.5 m
25m25 m
A
Bow that you
have !ound the
height you can
use it in part b
to help solve,
but remember
to halve it !or
'hal!way up theole(.
ba25
Gse alternate
100 m
25
yx
)on(t !orget to
answer the @uestion.
-#0 m is how !ar it is
still !rom the
lighthouse, we want
to know how !ar it
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32!.
327.
328.
329.
330.
/.
331.
332.
333.
334.
33.
33!.
337.
!!8.
12m
Gsing =ust the triangle the
height is 5 m , i.e.15m
tan =5
12
= tan1( 512 )
22.6198
'
20 m
5m
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#
#
--
#
!!9. /act Ratios
/0". We can %nd the eact values o! certain angles, that being
30 ,45 60 . o! # we can %nd the hypotenuse using ythagoras(s Theorem$
/0-.
/0/.
/00.
/01.
346.
347.
/03. Gsing the above triangle we get the eact ratios$
/04.
/1".
351.
352.
353.
/10. We can %nd the ratios !or 30 and 60 using an e@uilateral
triangle and then splitting it in hal!. 5or this we use sides o! - so that when
we split in hal! we can have a side o! #. +imilar to above, we can %nd the
height ;a short side by using ythagoras( Theorem$
/11.
/16.
/12.
/13.
/14.
360.
/6#. Gsing the above triangle we get the eact ratios$
/6-.
c2
=a2
+ b2
12
+12
1+1
4
sin 45= 1
2
cos 45= 1
2
c2
=a2
+ b2
22
=a2
+12
4= a2
+ 1
sin 60=3
2
cos60 =1
2
sin 30 =1
2
cos30 =3
2
1
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/6/.
/60.
365.
366.
3!7. Examples#. 5ind the eact value o! $
a sin45 +cos 45
b 3 sec 60
c sin2
30 +cos2
30
/63.
-. 5ind the eact value o! x $
/64./2"./2#./2-./2/.
374. Answers
#.
a sin 45+cos 45=
1
2+ 1
2
/21. 1+1
2
/26.
2
2
/22. 22
2
/23. 2
/24./3".
b 3 sec 60 =3
1
cos 60
/3#. 3
1
1
2
/3-. 3 1
2
1
/3/. 6
/30.
Nou can
rationalise the
denominator by
multiplying top
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5irst Ouadrant+econd Ouadrant
hird Ouadrant5ourth Ouadrant
c sin230 +cos
230 =( 12 )
2
+(32)2
/31.1
4+3
4
/36. 1
/32.-.
/33.
/34. Alsotan 30=
1
3
/4".
x
12=
1
3
/4#. 12 12
/4-. x= 12
3
/4/.
!94. Angles o$ 0agnitude
/41. The angles in a right&angled triangle are always acute ;ecept the
right angle o! course, but we may need to know angles greater than
90 , such as with bearings. Also negatives angles are used in diferent
situations such as engineering and science. Angles are measured around a
circle starting !rom the positive direction o! thex&ais.
/46.
/42.
/43.
/44.
0"".
0"#.
0"-.
0"/.
404.
0"1. When %nding the value o! angles greater than 90 we always
begin at 0 and move in an anti&clockwise direction. The angle is
always taken of thex&ais ;i.e. the hori?ontal line. The sign o! the ratio
Nou can rationalise thedenominator and get
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S
A
TC
S
A
T C
can be determined by the !ollowing acronym$ A+TC ;All +tations To
Central.
0"6.
0"2.
0"3.
0"4.
0#".
0##.
0#-.
0#/.
0#0.
0#1.
0#6.
0#2.
0#3.
0#4.
420. Examples
#. 5ind the eact value o!$
a sin135
b tan 300
c cos (150 )
d sin 690
421. Answers
#. 5or the answers to be eact you know you are going to use angles o!
30 , 45 60 .
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S
A
T C
S
A
T C
0-2.
0-3.
b tan 300 =tan60
0-4. 3
0/".
0/#.
0/-.
0//.
0/0.
0/1.
c cos (150 )=cos30
0/6.
0/2.
0/3.
0/4.
00".
00#.
d
00-.
1
2
00/.
000.
001.
006.
002.
003.
449.
The angle it makes with the
x&ais is 60 and only cos
is positive in this @uadrant, so
tan has to be negative
tan 300 =tan 60
Hecause it is a negative
angle you =ust go backwards
;i.e. clockwise. The angle it
makes with thex&ais is
30 and only tan is
positive in this @uadrant, so
To get 690 go around the
number plane once and then
an etra 330 ;
690360 =330 . The
angle it makes with thex&ais
is 30 and onl cos is
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S
A
T C
S
A
T C
4('. Trigonometric 1uations
451. A trigonometric e@uation involves %nding an angle whose
trigonometric ratio is given. There is usually more than one solution to an
e@uation. Also be care!ul to see what the conditions o! the e@uation are
e.g. 0 ! !360180 ! !180 etc. This eercise is the opposite to the
last eercise, we are working backwards. Dnce again it is a good idea todraw diagrams !or these @uestions.
42. Examples
#. +olve !or 0 ! !360 $
a sin=
1
2
b tan =3
c 2cos2=1
01/.
-. +olves
2=
1
2 !or180 ! !180 .
44. Answers
#.
a sin=
1
2
4.
4!.
47.
458.=30 ,18030
014. 30 ,150
4!0.
b tan =3
06#.
06-.
06/.
060.
sin 30 =1
2 and it is positive.
+in is positive in the %rst and
second @uadrants. We make -
angles that are 30 !rom thex&
ais in these @uadrants.
tan 60=3 , but it is negative.
Tan is negative in the second and!ourth @uadrants. We make -
angles that are 60 !rom thex&
ais in these @uadrants.
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A
T C
465. =18060 ,36060
066. 120 , 300
062.
063.
064.
02".
c 2cos2=1
02#. 22
02-.
02/.
020.021.026.022.023.024.03".03#.03-.03/.030.031.
486. 2=60 ,36060 ,360 +60 ,72060
487. 60 ,300 ,420 ,660
033. =30 , 150 , 210 , 330
034.04".
04#.04-.04/.040.041.046.
-. +olves
2=
1
2 !or18 0 ! ! 180 .
042. sin =" 12043.
" 1
2
Can(t get an eact value with # using cos, so
divide by - and then we can use Q which has an
cos60 =1
2 , and it is positive.
Cos is positive in the %rst and
!ourth @uadrants. We make -
The conditions !or this @uestion are
0 ! !360 , but we have 2 , so
multiplying everything by - we get the
condition 0 ! 2 ! 720 , so we are going
)ividing everything by - so
we et instead o! 2 .
Botice how the 0 answers are
within the original conditions
o! the @uestion$
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499.
00.
01.
02.
03.
504. =180 +45 ,0 45 ,45 ,180 45
1"1. 135 ,45 , 45 , 135
(').
To get rid o! the s@uared we s@uare
root. Hut remember when we s@uare
root in an e uation we alwa s et " .
sin 45= 1
2 , and it is positive and
negative. +o we make 0 angles that45 &
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('. Trigonometric 2ra"hs
1"3.
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1#/.
1#0. The height o! the graph ranges !rom to #. The graph is
symmetrical and continues with the same shape a!ter 360 .
1#1.
1#6.
1#2. y=tanx
1#3.
519. There are asymptotes at x=90 270 . The graph continues with
the same curve beyond 360 .
20.
#
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1-#. y=cosec x
1--.
1-/. This graph is the inverse o! y=sinx . Hecause you can(t divide by
?ero, we have asymptotes at x=180360 .
1-0.
525.
1-6. y=secx
1-2.
528. This graph is the inverse o! y=cosx . Hecause you can(t divide by
?ero, we have asymptotes at x=90 270 .
#
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1-4. y=cotx
1/".
531. This graph is the inverse o! y=tanx . Hecause you can(t divide by
?ero, we have asymptotes at x=180360 .
532. 5rom all these graphs you can deduce what the values o!
90 ,180 ,270360 .
(!!.
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(!4. Trigonometric #dentities
1/1. There are a number o! trigonometric identities that can be used to
simpli!y @uestions.
manipulate to get the identities that are related to it.
1/6.
1/2.
1/3. cot =
1
tan
1/4.
1
sin
cos
10".
cos
sin
10#.10-.10/.100.101.106.102.103.104.11".
11#.
11-. Hy dividing by sin2
:
11/.sin
2
sin2+cos
2
sin2=
1
sin2
110. 1+cot2
= cosec2
111. Hy dividing by cos2
:
116.sin
2
cos2
+
cos2
cos2
=
1
cos2
112. tan2
+1= sec2
113. we get the - identities$
114.
16".
RR tan =
sin
cos
cot=cos
sin
RR sin
2+cos
2=1
1+cot2=cosec
2
tan2
+1= sec2
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16#.
16-.
!3. Examples
#. rove the !ollowing identities$160.
a1sin
2
sin
2+cos
2
=cos2
161.
b tanA sinA +cosA =sec A
166.
c sin2
tan +cos2
cot +2sin cos =tan +cot
!7. Answers
#. rove the !ollowing identities$163.
a1sin
2
sin2
+cos2
=cos
2
164.12".
12#. #H$= 1sin
2
sin2+cos
2
12-. cos
2
1
12/. cos2
120. %H$
121.
b tanA sinA +cosA =sec A
126.122. #H$=tanA sinA +cosA
123.
sinA
cosA sinA+ cosA
124. sin
2A
cosA +
cosA
1
13". sin
2A
cosA +
cosA
1( cosAcosA)
13#. sin 2A
cosA +
cos2A
cosA
13-.
+tart with the side that you think you
can manipulate. Gsually it will be the
Le!t Eand +ide ;LE+. Food idea to
sin2
+cos2
=1
sin2
sin2
2 = 2
Another good
idea is to write to
the side o! your
working what
Dnce it is e@ual
to the ight
Eand +ide ;E+
you %nish by
tanA =sinA
cosA
Beed the same
denominator to add
!ractions, so
multiply top and
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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13/.
1
cosA
130. sec A
131. %H$
136.
c132.
133. #H$=sin2
tan +cos2
cot + 2sin cos
134. sin
2
sin
cos + cos
2
cos
sin+ 2sin cos
14". (1cos2 ) sin
cos +( 1sin2 ) cos
sin +2sin cos
14#.
1sin
cos cos
2
sin
cos+1
cos
sinsin
2
cos
sin + 2sin cos
14-.
sin
cos sin cos +
cos
sin sin cos + 2sin cos
14/.
sin
cos +
cos
sin
140. tan +cot
141. %H$
146.
142.
143.
99.
sin2
+cos2=1
sin2
sin2
5rom the %rst to the second
line o! working < used the
identities$
tan =sin
cos cot =
cos
sin
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A
H Ca
bc
#-.#
#3./
)''. on-Right-Angled Triangle
Results
6"#. A non&right&angled triangle is named so that its angles and opposite
sides have the same pronumeral ;i.e. angle A is opposite side a, etc.
6"-.
6"/.
6"0.
)'(.
6"6.
!07. he 'ine ule
6"3. 5or %nding a side$
6"4.
6#".
6##. 5or an angle we =ust Sip it upside down$
6#-.
6#/.
6#0.
6#1. Nou can use this rule in non&right&angled triangles when you have -
pairs o! opposite sides and angles.
!1!. Examples
#. 5ind the length o! x in the !ollowing to # decimal place$
6#2.
6#3.
6#4.
6-".
-. 5ind the si?e o! to the nearest minute$
6-#.6--.6-/.
6-0.6-1.6-6.6-2.
a
sinA=
b
sin&=
c
sin
sinA
a =
sin&
b =
sin
c
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#-.#
#3./
!28. Answers
#. Gsing the sine rule$
6-4.a
sinA=
b
sin&=
c
sin
6/". xsin59 34 '= 12.1
sin 3212'
6/#. sin 5934 ' sin 5934 '
6/-.x=
12.1
sin32 12' sin 59 34
'
6//. 19.5783
6/0. 19.6 units
6/1.-.
6/6.sinA
a =
sin&
b =
sin
c
6/2. sin27.4= sin 28 8
'
18.3
6/3. 27.4 27.4
6/4. sin =sin 28 8
'
18.327.4
60". = sin1(sin 28 8
'
18.327.4)
60#. 44.910..
60-. 44 54'37.164
' '
60/. 44 55 '
!44.
Hetter to type this in your
calculator so you are not
rounding of until the last
7/25/2019 Ch6 - Trigonometry - Prelim Maths
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A
H Ca
bc
/
2
#-
#"
6
/
2
!4. he osine ule
606.
602.
603.
604.
61".
61#. 5or a side$ 5or an angle$
61-.
61/.
610. Gse this when you have - sides and the angle in between. Gse
this rule when you have all / sides.
!. Examples
#. 5ind the length o! y in the !ollowing correct to - decimal places$
616.612.613.614.66".66#.
66-.66/.
-. 5ind the si?e o! ( in the !ollowing correct to the nearest minute$
660.
661.
666.
662.
663.
!!9. Answers
#. c2
=a2
+ b2
2 ab cos
62". y2
=32
+72
2 3 7 cos34
62#. 23.180421
62-. y=23.180421..
62/. 4.81460
620. 4.8 units
621.
cos=a
2+b
2c
2
2abc2=a
2+b
22abcos
)on(t round of until the %nalemember to s@uare root to
get rid o! the s@uared.
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#-
#"
6
626.622.623.
-. cos=a
2+b
2c
2
2ab
624. cos(=10
2+12
26
2
210 12
63".
208
240
63#. (=cos
1( 208240 )63-. 29.9264
63/. 29 55'35.166
''
630. 29 56 '
631.636.632.
!88.
emember to %nd
=ust the angle use
Nou can but this
straight in your
calculator butremember to put (=cos1
[(102+12262)
(21012 )]
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C
/" km
-" km
!89. 'ine and osine ro5lems
64". This eercise looks at using the sine and cosine rules in real li!e
situations.
!91. Examples
#. A, H and C are three towns such that H is -" km !rom A on a bearing o!330 and C is /" km !rom A on a bearing o! 204 . 5ind the distance
!rom H to C to the nearest kilometre.
-. An aircra!t Sies !rom a point A to a point H 0"" km on a course o! 040 .
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2#". &A=180 60 43 51'
2##. 769 '
2#-.
a&
sin76 9'
= 500
sin60
2#/. sin 76 9'
sin 76 9'
2#0. &=
500
sin 60 sin 76 9
'
2#1. 560.5642434
2#6. 561km ;nearest km
b &earin)=040 +76 9'
2#2. 116 9 '
718.
Couldn(t use sine rule or cosine rule
yet to %nd the distance, but i! we
!ound a missing angle, i.e. HAC, wecould use the sine rule to %nd the
distance between A and H. To %nd
HAC we had to use the sine rule to
%nd HCA and then the angle sum o!
a triangle. To %nd the bearing < add
the an le HCA and the bearin o!
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A
H Ca
bc
A
H C
%9. Area2-". We can %nd the area o! a triangle with a right angle using the
!ormulaA=
1
2bh
. We can also %nd the area o! a non&right angled
triangle using a similar !ormula$
2-#.
2--.
2-/.
2-0.
2-1. emember when we are %nding area we use mm2
, cm2
, m2
etc, or
*nits2
i! the units are not stated.
72!. Examples
#.
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2/2. 58 45 '
2/3.
2/4. A=
1
2ab sin
20".
1
280 100 sin 58 45
'
20#. 3419.647
20-. 3420m2
;nearest m2
20/. nd o! Trigonometry
Chapter 6
To %nd the area you need an
angle so use the cosine rule
to %nd any angle in the
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