Tso math sine cosine

Trigonometry
S4 Credit

Sine Rule Finding a length
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Sine Rule Finding an Angle
Cosine Rule Finding a Length
Cosine Rule Finding an Angle
Area of ANY Triangle
Mixed Problems

23 Mar 2013 Created by Mr. Lafferty Maths Dept.

Starter Questions
S4 Credit

1. Multiply out the brackets and simplify

5(y - 5) - 7(5 - y)

2. True or false the gradient of the line is 5
3
y = 5x -
4

3. Factorise x2 - 100

Sine Rule
S4 Credit

Learning Intention Success Criteria

1. To show how to use the 1. Know how to use the sine
sine rule to solve REAL rule to solve REAL LIFE
LIFE problems involving problems involving lengths.
finding the length of a
side of a triangle .


Sine Rule
S4 Credit Works for any Triangle

The Sine Rule can be used with ANY triangle
as long as we have been given enough information.

B
a b c a
= =
SinA SinB SinC c C
b
A
23 Mar 2013 Created by Mr Lafferty Maths Dept

The Sine Rule

Deriving the rule Consider a general triangle ABC.

C CP
SinB = ⇒ CP = aSinB
a
CP
a b also SinA = ⇒ CP = bSinA
b
⇒ aSinB = bSinA
P aSinB
B A ⇒ =b
c SinA
Draw CP perpendicular to BA a b
⇒ =
SinA SinB

This can be extended to

a b c SinA SinB SinC
= = or equivalently = =
SinA SinB SinC a b c

Calculating Sides
Using The Sine Rule
S4 Credit

Example 1 : Find the length of a in this triangle.
B

10m a
34o
41 o
C
A
Match up corresponding sides and angles:

a 10 a b c
= = =
sin 41o
sin 34o sin Ao sin B sin C

Rearrange and solve for a.
10sin 41o 10 × 0.656
a= a= = 11.74m
sin 34o 0.559

Calculating Sides
S4 Credit
Using The Sine Rule
Example 2 : Find the length of d in this triangle.
D
10m

133 o

37o E
C
Match up corresponding sides and angles: d
d 10 c d e
o
= = =
sin133 sin 37 o sin C o sin D sin E
Rearrange and solve for d.
10sin133o 10 × 0.731
d= d= = 12.14m
sin 37o 0.602

What goes in the Box ?
S4 Credit

Find the unknown side in each of the triangles below:

12cm
(1) (2) b
47o
32 o
a
72o
93o 16mm

A = 6.7cm
B = 21.8mm


Sine Rule
S4 Credit

Now try MIA Ex 2.1
Ch12 (page 247)


Starter Questions
S4 Credit

1. True or false 9x - 36 = 9(x + 6)(x - 6)

2. Find the gradient and the y - intercept
3 1
for the line with equation y = - x +
4 5

3. Solve the equation tanx - 1 = 0

Sine Rule
S4 Credit


1. To show how to use the 1. Know how to use the sine
sine rule to solve problems rule to solve problems
involving finding an angle involving angles.
of a triangle .


Calculating Angles
Using The Sine Rule
S4 Credit
B
Example 1 : 45m
38m
Find the angle A o

23o C
A
Match up corresponding sides and angles:
45 38 a b c
= = =
sin Ao sin 23o sin A sin B sin C
Rearrange and solve for sin Ao
45sin 23o
sin A =
o
= 0.463 Use sin-1 0.463 to find Ao
38

Ao = sin −1 0.463 = 27.6o

Calculating Angles
S4 Credit
Using The Sine Rule
75m
Example 2 : X Z

Find the angle Xo

143o 38m

Y
Match up corresponding sides and angles: x y z
38 75 = =
= sin X sin Y sin Z
sin X o
sin143o
Rearrange and solve for sin Xo
38sin143o
sin X o = = 0.305 Use sin-1 0.305 to find Xo
75
−1
X = sin 0.305 = 17.8
o o

What Goes In The Box ?
S4 Credit

Calculate the unknown angle in the following:

(1) 8.9m (2)
100o
12.9cm Bo
Ao
14.5m

14o
A = 37.2
o o

14.7cm

Bo = 16o

Sine Rule
S4 Credit

Now try MIA Ex3.1
Ch12 (page 249)


Starter Questions
S4 Credit

1. Find the gradient of the line that passes

through the points ( 1,1) and (9,9).

2. Find the gradient and the y - intercept
for the line with equation y = 1 - x

3. Factorise x2 - 64

Cosine Rule
S4 Credit


1. To show when to use the 1. Know when to use the cosine
cosine rule to solve rule to solve problems.
problems involving finding
the length of a side of a 2. Solve problems that involve
triangle . finding the length of a side.


Cosine Rule

The Cosine Rule can be used with ANY triangle
as long as we have been given enough information.

a =b +c - 2bc cos A
2 2 2
B
a
c C
b
A

The Cosine Rule

The Cosine Rule generalises Pythagoras’ Theorem and 1

takes care of the 3 possible cases for Angle A.
A
Deriving the rule Consider a general triangle ABC. We
require a in terms of b, c and A.
B a2 = b2 + c2
BP = a – (b – x)
2 2 2

Also: BP2 = c2 – x2 2

c a
⇒ a2 – (b – x)2 = c2 – x2
⇒ a2 – (b2 – 2bx + x2) = c2 – x2 A

P ⇒ a2 – b2 + 2bx – x2 = c2 – x2
A x b b-x C a2 > b2 + c2
b ⇒ a2 = b2 + c2 – 2bx*
3
⇒ a = b + c – 2bcCosA
2 2 2
Draw BP perpendicular to AC
*Since Cos A = x/c ⇒ x = cCosA
A
When A = 90o, CosA = 0 and reduces to a2 = b2 + c2 1 Pythagoras

When A > 90o, CosA is negative, ⇒ a2 > b2 + c2 Pythagoras + a bit
2
a2 < b2 + c2
When A < 90o, CosA is positive, ⇒ a2 > b2 + c2 3 Pythagoras - a bit

The Cosine Rule

The Cosine rule can be used to find:
1. An unknown side when two sides of the triangle and the included
angle are given (SAS).
2. An unknown angle when 3 sides are given (SSS).
B
Finding an unknown side.

a2 = b2 + c2 – 2bcCosA c a

Applying the same method as
A b C
earlier to the other sides
produce similar formulae for b2 = a2 + c2 – 2acCosB
b and c. namely:
c2 = a2 + b2 – 2abCosC

Cosine Rule

How to determine when to use the Cosine Rule.

Two questions
1. Do you know ALL the lengths.
OR SAS
2. Do you know 2 sides and the angle in between.

If YES to any of the questions then Cosine Rule

Otherwise use the Sine Rule

Using The Cosine Rule

Example 1 : Find the unknown side in the triangle below:
L

5m
43o

12m Identify sides a,b,c and angle Ao
a= L b= 5 c = 12 Ao = 43o
Write down the Cosine Rule.

a2 = 52 + 122 - 2 x 5 x 12 cos 43o Substitute values to find a .
2

a2 = 25 + 144 - (120 x 0.731 )
a2 = 81.28 Square root to find “a”.
a = L = 9.02m

17.5 m
Example 2 : 137 o
12.2 m
Find the length of side M.

M
a = M b = 12.2 C = 17.5 Ao = 137o Identify the sides and angle.

Write down Cosine Rule

a2 = 12.22 + 17.52 – ( 2 x 12.2 x 17.5 x cos 137o )
a2 = 148.84 + 306.25 – ( 427 x – 0.731 )
Notice the two negative signs.
a2 = 455.09 + 312.137

a2 = 767.227

a = M = 27.7m

S4 Credit

Find the length of the unknown side in the triangles:
43cm

(1)
78o

31cm L = 47.5cm
L

(2)
5.2m M

38o M =5.05m

8m

Cosine Rule
S4 Credit

Now try MIA Ex4.1
Ch12 (page 254)


Starter Questions
S4 Credit

1. If lines have the same gradient

What is special about them.

2. Factorise x2 + 4x - 12
54o
3. Explain why the missing angles
are 90 o and 36o

Cosine Rule
S4 Credit


1. To show when to use the 1. Know when to use the cosine
cosine rule to solve REAL rule to solve REAL LIFE
LIFE problems involving problems.
finding an angle of a
triangle . 2. Solve REAL LIFE problems
that involve finding an angle
of a triangle.


Finding Angles

Consider the Cosine Rule again:
We are going to change the subject of the formula to cos A o

b2 + c2 – 2bc cos Ao = a2 Turn the formula around:

-2bc cos Ao = a2 – b2 – c2 Take b2 and c2 across.

a2 − b2 − c2 Divide by – 2 bc.
cos Ao =
−2bc
Divide top and bottom by -1
b +c −a
2 2 2
cos A =
o
You now have a formula for
2bc
finding an angle if you know all
three sides of the triangle.

Finding Angles

Example 1 : Calculate the
unknown angle Ao .

b2 + c 2 − a 2
cos Ao =
2bc Write down the formula for cos Ao

Ao = ? a = 11 b = 9 c = 16 Label and identify Ao and a , b and c.

92 + 16 2 − 112
cos A =
o

2 × 9 × 16
Substitute values into the formula.
Cos Ao = 0.75 Calculate cos Ao .

Ao = 41.4o Use cos-1 0.75 to find Ao

Finding Angles

Example 2: Find the unknown
Angle yo in the triangle:

b2 + c 2 − a 2
cos Ao = Write down the formula.
2bc
A o = yo a = 26 b = 15 c = 13 Identify the sides and angle.

152 + 132 − 262 Find the value of cosAo
cos Ao =
2 ×15 ×13
The negative tells you
cosA = - 0.723
o
the angle is obtuse.
A o = yo = 136.3o

S4 Credit

Calculate the unknown angles in the triangles below:

(1) (2)
5m Ao 7m 12.7cm
Bo
7.9cm 8.3cm
10m

Ao =111.8o Bo = 37.3o

Cosine Rule
S4 Credit

Now try MIA Ex 5.1
Ch12 (page 256)


Starter Questions
S4 Credit

1. True or false

2( x + 3) − (4 − x) = 3 x − 2

2. Find the equaton of the line passing
through the points ( 3,2) and (10, 9) .

3. Solve the equation sin x - 0.5 = 0

S4 Credit


1. To explain how to use the 1. Know the formula for the
Area formula for ANY area of any triangle.
triangle.
2. Use formula to find area of
any triangle given two length
and angle in between.


Labelling Triangles
S4 Credit

In Mathematics we have a convention for labelling triangles.

B

a

c C

b
A Small letters a, b, c refer to distances
Capital letters A, B, C refer to angles

Labelling Triangles
S4 Credit

Have a go at labelling the following triangle.

E

d

f F

e
D

General Formula for
S4 Credit
Co
Consider the triangle below:
b a
h

Ao Bo
c
Area = ½ x base x height
What does the sine of Ao equal
1
A = ×c×h h
2 sin A =
o

b
1
A = × c × b sin Ao Change the subject to h.
2
h = b sinAo
1
A = bc sin Ao Substitute into the area formula
2

Key feature
S4 Credit To find the area
you need to knowing
The area sides and the angle be found
2 of ANY triangle can
by the following formula.
in between (SAS)

B 1
Area = bc sin A
a 2
Another version
c C 1
Area = ac sin B
2
Another version
b
A 1
Area = ab sin C
23 Mar 2013 Created by Mr Lafferty Maths Dept 2

S4 Credit

Example : Find the area of the triangle.

The version we use is
B
1
20cm Area = ab sin C
2

c 30o C
1
Area = × 20 × 25 × sin 30o
2
25cm
A Area = 10 × 25 × 0.5 = 125cm 2

S4 Credit

Example : Find the area of the triangle.

The version we use is
E
1
10cm Area= df sin E
60o 2

8cm F
1
Area = × 8 ×10 × sin 60o
2

D Area = 40 × 0.866 = 34.64cm 2

Key feature

Remember
S4 Credit
(SAS)
Calculate the areas of the triangles below:
(1)

12.6cm
A = 36.9cm2
23o
15cm
(2)

5.7m
A = 16.7m2
71 o
6.2m

S4 Credit

Now try MIA Ex6.1
Ch12 (page 258)


Starter Questions
S4 Credit

1. A washing machine is reduced by 10%

in a sale. It's sale price is £360.
What was the original price.

2. Factorise x - 7x + 12
2

3. Find the missing angles. 61o

Mixed problems
S4 Credit


1. To use our knowledge 1. Be able to recognise the
gained so far to solve correct trigonometric
various trigonometry formula to use to solve a
problems. problem involving triangles.


Exam Type Questions

The angle of elevation of the
Angle TDA = 180 – 35 = 145o
top of a building measured
Angle DTA = 180 – 170 = 10o from point A is 25o. At point
T D which is 15m closer to the
building, the angle of
elevation is 35o Calculate the
height of the building.
10o
36.5

35o 145o 25o
B D A
15 m
t d a
= =
sin T sin D sin A SOH CAH TOA
TD 15 TB
= Sin 35o =
Sin 25o Sin 10o 36.5
15Sin 25o ⇒ TB = 36.5Sin 35o = 20.9 m
TD = = 36.5 m
Sin 10

Exam Type Questions

A fishing boat leaves a harbour (H) and travels due East for 40 miles to a
marker buoy (B). At B the boat turns left and sails for 24 miles to a
lighthouse (L). It then returns to harbour, a distance of 57 miles.
(a) Make a sketch of the journey.
(b) Find the bearing of the lighthouse from the harbour. (nearest degree)

572 + 402 − 24 2
CosA =
2x 57x 40
µ
A = 20.4o
L
∴ Bearing = 90 − 20.4 = 070o

57 miles
24 miles
H
A 40 miles
B

Exam Type Questions
The angle of elevation of the top of a column measured from point A, is 20o.
The angle of elevation of the top of the statue is 25o. Find the height of the
statue when the measurements are taken 50 m from its base
T
Angle BCA =180 – 110 = 70 Angle ACT = 180 – 70 = 110 Angle ATC = 180 – 115 = 65
o o o

65o

t d a TC 53.21 110o
= = = C
sin T sin D sin A Sin 5o Sin 65o 70o

53.21 Sin 5 m
⇒ TC = = 5.1 m (1dp ) 21
Sin 65o 53.

5o
25o
SOH CAH TOA
20o
A 50 m B
50 50
Cos 20o = ⇒ AC = = 53.21 m (2dp )
AC Cos 20o

Exam Type Questions
An AWACS aircraft takes off from RAF P
Not to Scale
Waddington (W) on a navigation
exercise. It flies 530 miles North to
a point (P) as shown, It then turns
left and flies to a point (Q), 670
miles away. Finally it flies back to
base, a distance of 520 miles.
670 miles
Find the bearing of Q from point P.

530 miles
b 2 + c 2 − a2
CosA =
2bc
Q

5302 + 6702 − 5202
CosP =
2x 530x 670 520 miles
µ W
P = 48.7o
∴ Bearing = 180 + 48.7 = 229o

Mixed Problems
S4 Credit

Now try MIA
Ex 7.1 & 7.2
Ch12 (page 262)


Tso math sine cosine

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