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Trigonometry
Outcomes
Must: Use label sides of triangles correctly
Should: Be able to do calculations involving trig functions
Could: Use trig ratios to find missing lengths in triangles.
Thursday 9 August 2012
Using all 3 trig ratios to find missing lengths
Right-angled trianglesA right-angled triangle contains a right angle.
The longest side opposite the right angle is called the hypotenuse.
Thursday 9 August 2012
The opposite and adjacent sidesThe two shorter sides of a right-angled triangle are named with respect to one of the acute angles.
The side opposite the marked angle is called the opposite side.
The side between the marked angle and the right angle is called the adjacent side.
x
Label the sides
Calculate the following ratiosUse your calculator to find the following to 3 significant figures.
1) sin 79° = 0.982 2) cos 28° = 0.883
3) tan 65° = 2.14 4) cos 11° = 0.982
5) sin 34° = 0.559 6) tan 84° = 9.51
7) tan 49° = 1.15 8) sin 62° = 0.883
9) tan 6° = 0.105 10) cos = 0.55956°
θ
OPPOSITE
HY
PO
TE
NU
SE
A D J A C E N T
The three trigonometric ratios
Sin θ =Opposite
Hypotenuse S O H
Cos θ =Adjacent
Hypotenuse C A H
Tan θ =OppositeAdjacent T O A
Remember: S O H C A H T O A
The sine ratio
The ratio of the length of the opposite sidethe length of the hypotenuse
is the sine ratio.
The value of the sine ratio depends on the size of the angles in the triangle.
θ
OPPOSITE
HY
PO
TE
NU
SE
We say:
sin θ =opposite
hypotenuse
Using sine to find missing lengths
65°
x cm
11 cm
opposite
hyp
S O H C A H TO A Sin = opposite
hypotenuse θ
Sin = opp
hyp θ
Sin 65 = x
11
Sin 65 x 11 = x
9.97 (2dp) = x
32°
x cm
6 cm 72°
x cm
12 cm 54°
x cm
15 cm
(1) (2) (3)
Sin 32 = x 6
Sin 32 x 6 = x
3.18cm (2dp) = x
Sin 72 = x 12
Sin 72 x 12 = x
11.41cm (2dp) = x
Sin 54 = x 15
Sin 54 x 15 = x
12.14cm (2dp) = x
Using sine to find missing lengths
47°
8 cm
x cm
opposite
hyp
S O H C A H TO A Sin = opposite
hypotenuse θ
Sin = opp
hyp θ
Sin 47 = 8
x
x = 8
Sin 47
x = 10.94 cm (2dp)
32°
6 cm
x cm 72°
12 cm
x cm 54°
3 cm
x cm
(1) (2) (3) hyp
opp
Sin 32 = 6 x
x = 6Sin 32
x = 11.32 cm(2dp)
Sin 72 = 12 x
x = 12Sin 72
x = 12.62 cm(2dp)
Sin 54 = 3 x
x = 3Sin 54
x = 3.71 cm(2dp)
Using cosine to find missing lengths
S O H C A H TO A cos = adjacent
hypotenuse θ
53°
x cm
10 cm
hyp
adj
cos = adj
hyp θ
cos 53 = x 10
cos 53 x 10 = x
6.02 (2dp) = x66°
12 cm
x cm
hyp
adj
cos = adj
hyp θ
cos 66 = 12 x
x = 12cos 66
x = 29.50 cm(2dp)
(1)
31°
x cm
9 cm
(2)
49°
22 cm
x cm
hyp
adj
cos = adj hyp θ
cos 31 = x 9
cos 31 x 9 = x
7.71 (2dp) = x
cos = adj hyp θ
cos 49 = 22 x
x = 22cos 49
x = 33.53 cm(2dp)
Using tangent to find missing lengths
S O H C A H TO A tan = oppositeadjacent
θ
71°
x cm
10 cmadj
opp
tan = opp
adj θ
tan 71 = x 10
tan 71 x 10 = x
29.04cm (2dp) = x
318cm xcm
adj
opp
tan = opp adj θ
tan 31 = 8 x
x = 8tan 31
x = 13.31 cm(2dp)
31xcm
7cm
48
xcm
43cm(1) (2)
4.21cm (2dp) = x x = 38.72 cm(2dp)
Using all 3 trig ratios to find missing lengths
S O H C A H TO A
477cm
xcm hyp
opp
sin = opp
hyp θ
sin 47 = 7 x
x = 7sin 47
x = 9.57 cm(2dp)
33°
x cm
opp
adj
7cm
tan = opp adj θ
tan 33 = 7 x
x = 7tan 33
x = 10.78 cm(2dp)
53°
x cm
10 cm
hyp
adj
cos = adj
hyp θ
cos 53 = x 10
cos 53 x 10 = x
6.02cm (2dp) = x
Finding Angles using Trig
S O H C A H TO A
θ
5 cm
8 cm
adj
opp
tan = opp
adj θ
tan = 8 5
θ
= 8 5
θ57.99 (2dp)
tan-1
θ10cm
6cm
hyp
opp
sin = opp
hyp θ
sin = 6 10
θ
= 6 10
θ sin-1
θ = 36.87 (2dp)
Trigonometry 2Objective: Use trig to find missing lengths
and angles in right angled triangles for worded questions. Grade A
Outcomes
Must: Use trig ratios to find missing lengths and angles in triangles.
Should: Use trig to answer worded problems.
Could: Use trig to answer more difficult worded problems.
Thursday 9 August 2012
Finding side lengthsA 5 m long ladder is resting against a wall. It makes an angle of 70° with the ground.
5 m
70°x
What is the distance between the base of the ladder and the wall?
We are given the hypotenuse and we want to find the length of the side adjacent to the angle, so we use:
cos θ =adjacent
hypotenuse
cos 70° =x5
x = 5 × cos 70°= 1.71 m (to 2d.p.)
Area of a triangle (using ½ ab Sin C)
Outcomes
Must: Understand when you can use this formula for triangle area.
Should: Be able to find the area of a triangle using ½ ab sin C.
Could: Answer exam questions
Thursday 9 August 2012
Objective Find the area of a triangle using Area = ½ ab sin C
A
B C4cm
5cm
Triangle Area : When to use ½ ab Sin C ?• When you have not been given a perpendicular height (straight height).
• When you have been given two lengths and an angle between them.
35
Area of triangle ABC = ab sin C12
A = ½ x 4 x 5 Sin 35A = ……….cm2
4cm6cm
40
A =½ ab Sin C
A = ½ x 4 x 6 Sin 35A = ……….cm2
6cm
9cm23
A =½ ab Sin CA = ½ x 6 x 9 Sin 35A = ……….cm2
50
4cm
7cm
A = ½ ab Sin CA = ½ x 7 x 4 Sin 50A = ……….cm2
Find the area of the triangles
6cm(1)
35
4cm
7cm
A = ½ ab Sin CA = ½ x 7 x 4 Sin 35A = ……….cm2
6cm(2)
453cm
7cm
A = ½ ab Sin CA = ½ x 5 x 3 Sin 45A = ……….cm2
5cm
(3)
Sine Rule
Outcomes
Must: Use sine rule to find lengths in triangles.
Should: Use sine rule to find angles in triangles.
Could: Answer mixed questions
Thursday 9 August 2012
Objective Use sine rule to find angles and lengths in triangles. (not right angles triangles)
Sine Rule
• Find the sides and angles of a triangle whether it’s a right angle or not.
C
A B
b
c
a
asin A
=b
sin B=
csin C
6cm y cm4025
Examples: Find y using sine rule
A
a
B
b
aSin A
=b
Sin B
3Sin 25
=y
Sin 40
3Sin 25
x Sin 40 = y
4.56 cm (2dp) = y
C
A B
b
c
a
Sine Rule aSin A
=b
Sin B=
cSin C
Exercise: Find lengths y using sine rule
6cm
y cm34
15 A
a
C
c
ySin 15
=6
Sin 34
y =6
Sin 34 x Sin 15
y = 2.8 cm (1dp)
y
4.2cm 41
22 B
bC
c
4.2Sin 22
=y
Sin 41
4.2Sin 22
x Sin 41 = y
7.4 cm (1dp) = y
(1) (2)
y
8.9cm
6236 B
bC
c
(3)
Y = 5.9 cm
C
A B
b
c
a
asin A
=b
sin B=
csin C
or
sin A sin B sin Ca
=b
=c
Sine Rule
43
x
3.5cm
2.3cm
C
cB
b
Sin Bb
=Sin C
cSin x2.3
=Sin 43
3.5
Sin x =Sin 43
3.5 x 2.3
Sin x = 0.44817….. x = 0.44817….. sin-1
x = 26.6 (1dp)
Use this when finding a length
Use this when finding an angle
C
A B
b
c
a
Sine Rule
Sin Aa
= Sin Bb
= Sin Cc
Exercise: Find angle y using sin rule
4.3cm
2.9 cm63
y A
a
C
c
(1)
Sin Aa
=Sin C
cSin y2.9
=Sin 63
4.3
Sin y =Sin 63
4.3 x 2.9
Sin y = 0.600911….. y = 0.600911….. sin-1
y = 36.9 (1dp)
8.4cm
7.3 cm y
53 B
b
C
c
(2)
Sin Bb
=Sin Y
c
Sin 537.3
=Sin y
8.4Sin 537.3
x 8.4 = Sin y
0.9080…= Sin y
0.9080… sin-1 = y
65.2 (1dp) = y
Sine Rule Practice
Higher GCSE for AQA (Oxford) Book
Page 387 Exercise 4r Q1 – Q19
Cosine Rule
Outcomes
Must: Use cosine rule to find lengths in triangles.
Should: Use cosine rule to find angles in triangles.
Could: Answer mixed questions
Thursday 9 August 2012
Objective Use cosine rule to find angles and lengths in triangles. (not right angles triangles)
The cosine ruleA
B C
c
a
b
a2 = b2 + c2 – 2bc cos Aa2 = b2 + c2 – 2bc cos A
or
cos A = b2 + c2 – a2
2bc
Use when finding a length of a side.
Use when finding an Angle.
B
C A7 cm
4 cm
48°
a
Example 1Find a
When you are given two lengths and an angle between them use;
a2 = b2 + c2 – 2bc cos Aa2 = 72 + 42 - 2 x 7 x 4 cos 48a2 = 27.52868…..a = 5.25 (2dp)
B
C A7 cm
4 cm
48°
a
Example 1Find a
When you are given two lengths and an angle between them use;
a2 = b2 + c2 – 2bc cos Aa2 = 72 + 42 - 2 x 7 x 4 cos 48a2 = 27.52868…..a = 5.25 (2dp)
Exercise Find the length marked x
2cm
5cm
52x
(1)
98x 9cm
5cm
(2)
a2 = 52 + 22 - 2 x 5 x 2 cos 52a2 = 16.68677…..a = 4.08 (2dp)
a2 = 52 + 92 - 2 x 5 x 9 cos 98a2 = 118.52557…..a = 10.89 (2dp)
The cosine ruleA
B C
c
a
b
a2 = b2 + c2 – 2bc cos Aa2 = b2 + c2 – 2bc cos A
or
cos A = b2 + c2 – a2
2bc
Use when finding a length of a side.
Use when finding an Angle.
(given 2 lengths and an angle)
(given 3 lengths)
Example 1Find angle A
4 cm
8 cm6 cm
A
B
C
You are given 3 sides and asked for an angle.
cos A = b2 + c2 - a2
2bca
b
ccos A = 42 + 62 - 82
2 x 4 x 6cos A = - 0.25
A = - 0.25 cos-1 A = 104. 5 (1dp)
Exercise Find the length marked x
2cm
5cm
x4cm
(1)x
11cm9cm
5cm
(2)
a
b
c
A
cos A = 22 + 52 - 42
2 x 2 x 5cos A = 0.65
A = 0.65 cos-1 A = 49.5 (1dp)
a
A
b
c
cos A = 92 + 52 - 112
2 x 9 x 5cos A = - 0.1666….
A = - 0.166.. cos-1 A = 99. 6 (1dp)
Example 1Find angle A
4 cm
8 cm6 cm
A
B
C
You are given 3 sides and asked for an angle.
cos A = b2 + c2 - a2
2bca
b
c cos A = 42 + 62 - 82
2 x 4 x 6cos A = - 0.25
A = - 0.25 cos-1
A= 104. 5 (1dp)