1. Wednesday 8 August 2012
Trigonometry
Using all 3 trig ratios to find missing
lengths
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.
2. Wednesday 8 August 2012
Right-angled triangles
A right-angled triangle contains a right angle.
The longest side opposite
the right angle is called the
hypotenuse.
3. The opposite and adjacent sides
The 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 x
marked angle and the
right angle is called
the adjacent side.
5. Calculate the following ratios
Use 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 56° = 0.559
6. The three trigonometric ratios
Opposite
O
H
Sin θ =
Hypotenuse SOH
Y
P P
P O
O T
Adjacent
S
I
E
N Cos θ =
Hypotenuse CAH
T U
E S
E
θ Opposite
Tan θ =
Adjacent TOA
ADJACENT
Remember: S O H C A H T O A
7. The sine ratio
the length of the opposite side
The ratio of is the sine ratio.
the length of the hypotenuse
The value of the sine ratio depends on the size of the angles
in the triangle.
O H
P
Y
P
We say:
P O
T
O E opposite
S
N
U sin θ =
I S
E
hypotenuse
T
E θ
8. Using sine to find missing lengths
S O H C A H TO A Sin θ =
opposite
hypotenuse
opp
hyp Sin θ =
hyp
11 cm 65° Sin 65 x 11 = x
x
Sin 65 = 9.97 (2dp) = x
11
x cm
opposite
x x x
Sin 32 = Sin 72 = Sin 54 =
6 12 15
Sin 32 x 6 = x Sin 72 x 12 = x Sin 54 x 15 = x
3.18cm (2dp) = x 11.41cm (2dp) = x 12.14cm (2dp) = x
6 cm 32° 12 cm 72° 15 cm 54°
(1) (2) (3)
x cm x cm
9. Using sine to find missing lengths
S O H C A H TO A Sin θ =
opposite
hypotenuse
opp
hyp Sin θ =
hyp
x cm 47° 8
8 x =
Sin 47 = Sin 47
x
8 cm opposite x = 10.94 cm (2dp)
6
Sin 32 = x 12 3
Sin 72 = x Sin 54 = x
x = 6 12
Sin 32 x = x = 3
Sin 72 Sin 54
x = 11.32 cm(2dp)
x = 12.62 cm(2dp) x = 3.71 cm(2dp)
x cm 32° x cm 72° x cm 54°
hyp
(1) (2) (3)
6 cm opp 12 cm 3 cm
10. Using cosine to find missing lengths
adjacent
cos θ =
S O H C A H TO A hypotenuse
adj
cos θ = cos θ =
adj
hyp hyp
hyp hyp
cos 53 = x 12
10 cm 10 x cm cos 66 = x
53° cos 53 x 10 = x x = 12
66° cos 66
x cm 6.02 (2dp) = x 12 cm
adj x = 29.50 cm(2dp)
adj
adj
cos θ = hyp cos θ
adj
= hyp
hyp (2)
(1) cos 31 = x 22
9 cm 9 x cm cos 49 = x
31° cos 31 x 9 = x 49° x = 22
cos 49
x cm 7.71 (2dp) = x 22 cm
x = 33.53 cm(2dp)
adj
11. Using tangent to find missing lengths
opposite
S O H C A H TO A tan θ =
adjacent
opp
opp
tan θ 8cm xcm
= adj
adj 31
71°
10 cm
adj tan 71 = x
10 opp
tan θ = adj
opp x cm tan 71 x 10 = x
8
29.04cm (2dp)= x tan 31 = x
x = 8
tan 31
7c
m (2) 43 x = 13.31 cm(2dp)
(1) c m
m
31
xc
4.21cm (2dp) = x
48 x = 38.72 cm(2dp)
xc
m
12. Using all 3 trig ratios to find missing lengths
S O H C A H TO A hyp
10 cm
opp opp
53°
7cm 7cm x cm
47 adj
33°
adj
xcm hyp x cm adj cos θ =
hyp
opp opp cos 53 = x
sin θ = tan θ = adj 10
hyp
7 7 cos 53 x 10 = x
sin 47 = x tan 33 = x
6.02cm (2dp) = x
x = 7 x = 7
sin 47 tan 33
x = 9.57 cm(2dp) x = 10.78 cm(2dp)
13. Finding Angles using Trig
S O H C A H TO A 6cm
opp
θ
8 cm opp hyp 10cm
opp
5 cm
tan θ =
opp sin θ =
adj hyp
θ adj
6
tan θ =
8 sin θ =
10
5
6
θ = sin-1
θ = 8 tan-1
5
10
57.99 (2dp) θ = 36.87 (2dp)
14.
15. Wednesday 8 August 2012
Trigonometry 2
Objective: 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.
16. Finding side lengths
A 5 m long ladder is resting against a wall. It makes an
angle of 70° with the ground.
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
5m the angle, so we use:
adjacent
cos θ =
hypotenuse
70° x
x cos 70° =
5
x = 5 × cos 70°
= 1.71 m (to 2d.p.)
17.
18. Wednesday 8 August 2012
Area of a triangle
(using ½ ab Sin C)
Outcomes
Must: Understand when you
Objective
can use this formula for
Find the area of a triangle area.
triangle using
Area = ½ ab sin C
Should: Be able to find the
area of a triangle using
½ ab sin C.
Could: Answer exam questions
19. Triangle Area : When to use ½ ab Sin C ?
• When you have not been given a perpendicular height
(straight height). 4cm 40
6cm
• When you have been given two lengths and an
angle between them.
A
A = ½ ab Sin C
5cm A = ½ x 4 x 6 Sin 35
35
A = ……….cm2
B C
4cm
1 6cm
Area of triangle ABC = ab sin C
2 23
9cm
A = ½ x 4 x 5 Sin 35 A = ½ ab Sin C
A = ……….cm2 A = ½ x 6 x 9 Sin 35
A = ……….cm2
20. Find the area of the triangles
(1) (2)
6cm 4cm 6cm 4cm
50 35
(3)
7cm 7cm
5cm 45
A = ½ ab Sin C A = ½ ab Sin C 3cm
A= ½ x 7 x 4 Sin 50 A= ½ x 7 x 4 Sin 35
A= ……….cm2 A= ……….cm2
7cm
A = ½ ab Sin C
A= ½ x 5 x 3 Sin 45
A= ……….cm2
21.
22. Wednesday 8 August 2012
Sine Rule
Outcomes
Objective
Use sine rule to find
angles and lengths in Must: Use sine rule to find
triangles. (not right lengths in triangles.
angles triangles)
Should: Use sine rule to find
angles in triangles.
Could: Answer mixed
questions
23. Sine Rule Examples: Find y using
sine rule
• Find the sides and angles of a triangle
whether it’s a right angle or not.
b 6cm
C y cm
a
25 40
A B
b a a b
=
Sin A Sin B
A B y
c 3
=
Sin 25 Sin 40
a b c
= = 3
sin A sin B sin C x Sin 40 = y
Sin 25
4.56 cm (2dp) = y
24. Sine Rule a b c
C = =
Sin A Sin B Sin C
b a
A c B (3) C
Exercise: Find lengths y using sine rule b
8.9cm
(2) C
(1) C
b
34 4.2cm 41 36 62
B
y cm y c
a
15 22
B
A y c
6cm c
Y = 5.9 cm
y 6 4.2 y
= =
Sin 15 Sin 34 Sin 22 Sin 41
6 4.2
y = x Sin 15 x Sin 41 = y
Sin 34 Sin 22
y = 2.8 cm (1dp) 7.4 cm (1dp) = y
25. Sine Rule
C
b
C
2.3cm 43
b a x
3.5cm B
A B c
c
Sin B Sin C
Use this when finding a length =
b c
a b c Sin 43
= = Sin x
sin A sin B sin C =
2.3 3.5
Sin 43
or Sin x = x 2.3
3.5
Use this when finding an angle
sin A sin B sin C Sin x = 0.44817…..
= = x = 0.44817…..sin-1
a b c
x = 26.6 (1dp)
26. Exercise: Find angle y using sin rule
Sine Rule
C (2) C
(1)
63 y
C 2.9 cm 7.3 cm
b
a
b a y 53
A B
A c B 4.3cm c 8.4cm c
Sin A
= Sin B = Sin C Sin A
=
Sin C Sin B
=
Sin Y
a b c a c b c
Sin y Sin 63 Sin 53 Sin y
= =
2.9 4.3 7.3 8.4
Sin 63 Sin 53
Sin y = x 2.9 x 8.4 = Sin y
4.3 7.3
Sin y = 0.600911….. 0.9080… Sin y
=
y = 0.600911….. -1
sin 0.9080… sin-1 = y
y = 36.9 (1dp) 65.2 (1dp) = y
31. Wednesday 8 August 2012
Cosine Rule
Outcomes
Objective
Use cosine rule to find
angles and lengths in Must: Use cosine rule to find
triangles. (not right lengths in triangles.
angles triangles)
Should: Use cosine rule to
find angles in triangles.
Could: Answer mixed
questions
32. The cosine rule Use when
finding a length
a2 = b2 + c2 – 2bc cos A of a side.
A
or
c b Use when
cos A = b2 + c2 – a2 finding an
B C 2bc Angle.
a
When you are given two lengths and an angle
Example 1 between them use;
Find a B
a2 = b2 + c2 – 2bc cos A
a 4 cm a2 = 72 + 42 - 2 x 7 x 4 cos 48
a2 = 27.52868…..
C 48° A a = 5.25 (2dp)
7 cm
33. When you are given two lengths and an angle
Example 1 between them use;
Find a B
a2 = b2 + c2 – 2bc cos A
a 4 cm a2 = 72 + 42 - 2 x 7 x 4 cos 48
a2 = 27.52868…..
C 48° A a = 5.25 (2dp)
7 cm
Exercise Find the length marked x
(1) (2)
2cm
x 9cm
52
98
x
5cm 5cm
a2 = 52 + 92 - 2 x 5 x 9 cos 98
a2 = 52 + 22 - 2 x 5 x 2 cos 52
a2 = 118.52557…..
a2 = 16.68677…..
a = 10.89 (2dp)
a = 4.08 (2dp)
34. The cosine rule Use when
finding a length
a2 = b2 + c2 – 2bc cos A of a side.
A (given 2 lengths
or and an angle)
c b
cos A = b2 + c2 – a2 Use when
B C 2bc finding an
a Angle.
(given 3 lengths)
Example 1 You are given 3 sides and asked for an angle.
Find angle A
B
cos A = b + c - a
2 2 2
a 2bc
8 cm c
6 cm cos A = 4 + 6 - 8
2 2 2
2x4x6
C 4 cm A cos A = - 0.25
b
A = - 0.25 cos-1 A = 104. 5 (1dp)
35. Example 1 You are given 3 sides and asked for an angle.
Find angle A
cos A = b + c - a
2 2 2
B
2bc
a
8 cm c cos A = 4 + 6 - 8
2 2 2
6 cm
2x4x6
A= 104. 5 (1dp)
C A cos A = - 0.25
b
4 cm
A = - 0.25 cos-1
Exercise Find the length marked x
(2) a
11cm b
(1) b 9cm
2cm
x
A x a A
4cm c 5cm
c 5cm cos A = 9 + 5 - 11
2 2 2
2x9x5
cos A = 2 + 5 - 4
2 2 2
2x2x5 cos A = - 0.1666….
cos A = 0.65 A = - 0.166..cos-1 A = 99. 6 (1dp)
A = 0.65 cos-1 A = 49.5 (1dp)
Notes de l'éditeur
Review the definition of a right-angled triangle. Tell pupils that in any triangle the angle opposite the longest side will always be the largest angle and vice-versa. Ask pupils to explain why no other angle in a right-angled triangle can be larger than or equal to the right angle. Ask pupils to tell you the sum of the two smaller angles in a right-angled triangle. Recall that the sum of the angles in a triangle is always equal to 180 °. Recall, also, that two angles that add up to 90° are called complementary angles. Conclude that the two smaller angles in a right-angled triangle are complementary angles.
We don ’t don’t need to know the actual size of the marked angle to label the two shorter sides as shown. It could be labelled using a letter symbol such as x as shown here. Point out that if we labelled the sides with respect to the other acute angle their name would be reversed.
Use this activity to practice labelling the sides of a right-angled triangle with respect to angle θ . Ask volunteers to come to the board to complete the activity.
This exercise practices the use of the sin, cos and tan keys on the calculator. It also practices rounding to a given number of significant figures. Pupils should notice that the sine of a given angle is equal to the cosine of the complement of that angle. They can use this fact to answer question 10 using the answer to question 5.
Stress to pupils that they must learn these three trigonometric ratios. Pupils can remember these using SOHCAHTOA or they may wish to make up their own mnemonics using these letters.
The sine ratio depends on the size of the opposite angle. We say that the sine of the angle is equal to the length of the opposite side divided by the length of the hypotenuse. Sin is mathematical shorthand for sine. It is still pronounced as ‘sine’.
This ratio can also be demonstrated using the similar right-angled activity on slide 7.
This ratio can also be demonstrated using the similar right-angled activity on slide 7.
This ratio can also be demonstrated using the similar right-angled activity on slide 7.
This ratio can also be demonstrated using the similar right-angled activity on slide 7.
We can use the first form of the formula to find side lengths and the second form of the equation to find angles.
We can use the first form of the formula to find side lengths and the second form of the equation to find angles.
