Trigonometry Solving Triangles ADJ OPP HYP Two old angels Skipped over heaven Carrying a harp...
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Transcript of Trigonometry Solving Triangles ADJ OPP HYP Two old angels Skipped over heaven Carrying a harp...
Trigonometry
Solving Triangles
ADJ
OPPHYP
opposite
adjacent
opposite
hypotenuse
adjacent
hypotenusecos
sin
tan Two old angels
Skipped over heaven
Carrying a harp
Solving Triangles
Trigonometric ratios in surd form
1
1
1
2 30º
45º
45º
60º
Angles are given in radians
π radians = 180ºπ
3= 60º
π
2= 90º
π
4= 45º
π
6= 30º
3
1
3 3
2
3
1tan 60 º =
sin 60 º =
1
2cos 60 º =
tan 30 º =
1
2 sin 30 º =
3
2 cos 30 º =
2
tan 45 º = 1
1
2 sin 45 º =
1
2cos 45 º =
Page 9 of tables
c
a b
Cosine Rule
C
A B
a2 = b2 + c2 – 2bccosA
b a
c
b2 = a2 + c2 – 2accosB
c2 = a2 + b2 – 2abcosC
Page 9 of tables
By Pythagoras’ Theorem
a2 = (c – x)2 + h2
a2 = c2 – 2cx + x2 + h2
a2 = b2 + c2 – 2cx
Cosine Rulec
a b
a2 = b2 + c2 – 2bccosA
c
b a
x c – x
b2 = x2 + h2
cosx
Ab
cosx b A
A
h
The Cosine Rule can be used to find a third side of a triangle if you have the other two
sides and the angle between them.
89o13.8
6·2
w
147o
8 11
m
Cosine Rule
6
10
65o
l
Included angle
Find the unknown side in the triangle below:
l5 m
12 m
43oIdentify sides a,b,c and angle Ao
a = l b = 5 c = 12 A = 43º
Write down the Cosine Rule
Substitute values and find a2
a2 = 52 + 122 – 2 5 12 cos 43o
a2 = 81·28
Take square root of both sides
a = 9·02 m
Examples
a2 = b2 + c2 – 2bccosA
a2 = 25 + 144 – 120(0·731)
137o17·5 cm
12·2 cm
a2 = 12·22 + 17·52 – ( 2 12·2 17·5 cos 137o )
a2 = 148·84 + 306·25 – ( 427 – 0·731 )
a2 = 455·09 + 312·137
a2 = 767·227 a = 27·7 cm
Find the unknown side in the triangle below:
Examples
a2 = b2 + c2 – 2bccosA
a = ?
b = 12·2
c = 17·5
A = 137º
20o
x 6
62 = 102 + x2 – (2 10 x cos 20o)
36 = 100 + x2 – 20x( 0·9397)
0 = x2 – 18·79x + 64
Find the two possible values for the unknown side.
Examples
10
a = 6
b = 10
c = x
A = 20ºa2 = b2 + c2 – 2bccosA
2 4
2
b b acx
a
20o
x 6
0 = x2 – 18·79x + 64
Find the two possible values for the unknown side.
Examples
10
2( 18·79) ( 18·79) 4(1)(64)
2(1)x
18·79 353·2 256
2x
2 4
2
b b acx
a
18·79 9·86
2
14·3 or 4·5x
a = 6
b = 10
c = x
A = 20º
a 1
b –18·79
c 64
Find the length of the unknown side in the triangles below:
(1)78o
43 cm
31 cmL
(2)
8 m
5·2 m
38o
M
(3) 110o
6·3 cm
8·7 cm
GL = 47.5cm
M = 5·05 m
G = 12.4cm
c
a b
Sine Rule
C
A Bba
c
Page 9 of tables
sin sin sin
a b c
A B C
____________
5
a
cbd
7
50º
82º
(i) Find dc , correct to the nearest cm.
Multiply both sides by sin 50º
dc sin 48º
In the triangle abc, d is a point on [bc].bd = 5 cm, ac = 7cm, bca = 82º and cad = 50º .
48º
=sin 50º
7
= 7·215…= 7 cm
Angle sum of ∆ is 180º
7
= =c
sin C
asin A
bsin B
_____ _____ _____
Sine Rule
__
(ii) Find ab , correct to the nearest cm.
In the triangle abc, d is a point on [bc].bd = 5 cm, ac = 7cm, bca = 82º and cad = 50º .
5 d
50º
7
+ = 1212
Cosine Rule:
a2 = b2 + c2 – 2bc cosA
ab 2 = 122 + 72 – 2(12)(7)cos82º
= 144 + 49 – 168cos82º
= 169·6189…
= 13·02…
= 13 cm
ab =
169·6189
a
cb
7
82º
c
a b
Area of triangle
C
A Bba
c
Page 6 of tables
1Area sin
2ab C
Must be the included angle
Calculate the area of the triangle shown.
Give your answer correct to one decimal place.
Area of triangle = absin C12
Area = (3)(4) sin 55 12
= 4·9149… 4
cm
3 cm
C must be the included angle= 4·9 cm2
55º
Find the area of triangle abc, correct to the nearest whole number.
Area of triangle = absin C12
Area = (14)(18·4) sin 70 12
= 121·0324… 18·4
14
c
a b44º
66º
C must be the included angle
|abc | = 180 – 44 – 66 = 70
70º
= 121units2