sine, cosine and area rules - brettonwood high school
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Grade 11
CAPS
Mathematics
Series
Sine, Cosine
and Area Rules
Apply the Sine and the Cosine rules to solve
problems in 2-dimensions. Unit 4
In this topic we will:
Calculate the area of a triangle given an angle and the two
adjacent sides. Unit 1
Apply the Sine Rule for triangles to calculate an unknown
side or an unknown angle of a given triangle. Unit 2
Outcomes for this Topic
Apply the Cosine Rule for triangles to calculate an unknown
side or an unknown angle of a given triangle. Unit 3
Grade 11
CAPS
Mathematics
Series
Unit 1
The Area Rule
sin opposite
hypotenuse
oppositehypotenuse
Trigonometric Ratios
In a , the 3 trigonometric ratios for an angle
are defined as follows:
right angled triangle
cos adjacent
hypotenuse
adjacent
tan opposite
adjacent
Some basic definitions – a reminder
Consider a non-right angled triangle ABC.
, and are the sides opposite angles , and respectively.
( This is the conventional way of labelling a triangle ).
a b c A B C
A B
C
b a
c N
h
1Area of base height
2
12
A re a c h --- (1)
In ,A C N sin Ah
bs in b A h
Substituting for h in (1)
12
A re a c s in b A12
s inA re a b c A
The area formula of a triangle
Draw the perpendicular, , from to . h C BA
Any angle can be used as such in
area formula, so
12
sin Bca12
sinab C 12
sinbc AArea = = =
90
A similar argument gives the same formula for the area
if is obtuse i.e. B B
•
•
The formula always uses
2 sides and the
angle formed by those sides (Included )
AB
C
b
a
c N
h
90
Different forms of the area formula
c
baa
C
BA
c
baa
C
BA
c
baa
C
BA
Three possible approaches to find
the area of a triangle
12
Area sin ab C
Any angle can be used in the formula, so
12
Area sin bc A
12
Area sin Bca
r
B
A
C
r
21sin
2 Area r
Find the area of in circle .ABC CA useful application of the area formula occurs when we
have a triangle formed by a radius and a chord of a c ircle.
The area of a triangle – Example 1
1
Area sin2
CA CB C
But CA CB r
Find the area of PQR.
We know PQ and RQ so use the included angle Q
The area of a triangle – Example 2
Solution: We must use the angle formed by the
2 sides with the given lengths.
64
1Area of sin
2PQR QP QR Q
218 7 sin 64 cm
2
225,2 cm
T u to r ia l 1 : A re a o f a T r ia n g leFind the areas of the triangles shown in the diagrams.
Give your answers accurate to 2 decimal digits
PAUSE
• Do Tutorial 1
• Then View Solutions
radius 6 cm
120AOB
1. 2.
radius 6 cm
120AOB
212. Area sin
2AOB r O
21
6 cm sin1202
215,59 cm
Given:
T u to r i a l 1 : P r o b le m 1 : A r e a o f a T r i a n g le : S o lu t io n
40
308 cm
10 cm
11. Area sin
2XYZ XY YZ Y
180 40 30 110Y
1sin
2z x Y
218 10 sin 110 cm
2
237,59 cm
T u to r i a l 1 : P r o b le m 2 : A r e a o f a T r i a n g le : S o lu t io n
Grade 11
CAPS
Mathematics
Series
Unit 2
The Sine Rule
One way to find unknown sides and angles in
is by using the :
non - right angled
triang Sine Rs ulele
T h e S in e R u le f o r T r ia n g le s
In ACN, sinh
Ab
sinh b A
Suppose is a scalene triangleABC
In , sinh
BCN Ba
sinh a B sin sinb A a B
sin sinor
A B
a b
ab
c N
Drop CN AB
h
C
A B
b a
c
A
B C
b
a
c
h
Now sin sin
sin sin
h c B b C
B C
b c
can be turned so that is the base.
We then get
ABC BC
T h e C o m p le te S in e R u le f o r T r i a n g le s
sin sin sinSo
A B C
a b c
W h e n d o w e u s e t h e S in e R u le ?The sine rule can be used in a triangle when:
Two angles and a side are given•
Two sides and the non-included angle are given•
• To calculate second side
•
To calculate second angle
sin sinSolution: Use
A B
a b
sinsin
a BA
b10 sin 62
sin12
A
1 8 0 6 2 4 7 , 4 7 0 , 6 CT h u s
A p p li c a t io n o f t h e S in e R u le - E x a m p le 1
ABC A CIn , find the size of angles and .
47,4 A is opposite the shorter of the 2 given sides.
62 must be an acute angle.
(Only one possibility as can be seen from sketch)
A
A A
sin 5 sin 48sin
4
p QP
q
2
1
68, 3
180 68, 3 111, 7
or
P
P
A p p li c a t io n o f t h e S in e R u le - E x a m p le 2
is opposite the longer of the 2 given sides.
48 can be an acute or obtuse angle.
( Two possibilities as can be seen from sketch below)
P
P P
1P
2Psin sin Use Solution :
Q P
q p
5, 4 48 .
.
PQR
QR PR Q
P
In it is given that:
and
Determine
13 sin 55
sin 29
z
2 2 , 0 z
sin sin
z y
Z Y
In , find the length .XYZ XY
A p p li c a t io n o f t h e S in e R u le - E x a m p le 3
As the unknown is a side, we use the sine rule in
its reciprocal form. The unknown side is then at the top.
Solution :
sin
sin
y Zz
Y
2. In , find and the area of PQR QR PQR
1. In , find .
(Correct to two decimal places)
ABC B
PAUSE
• Do Tutorial 2
• Then View Solutions
T u to r i a l 2 : S in e R u le
T u t o r i a l 2 : P r o b le m 1 : S i n e r u le : S o lu t i o nFind .
(2 decimal places)
B
10sin 35sin
7B
1 55,02B
2or 180 55,02 124,98B
Given:
35 acute or obtuseB B
sin sin sin 35 sin
7 10
A B B
a b
Two Possible Answers!!
T u to r i a l 2 : P r o b le m 1 : W h y tw o s o lu t io n s ?
1
2
Obtained: 55,02
or 124,98
B
B
Given:
35 acute or obtuseB B
1 (Obtuse)B 2 (Acute)B
2. Find and the area of .QR PQR
67
sin 64 sin 80
QR
T u t o r i a l 2 : P r o b le m 2 : S i n e r u le : S o lu t i o n
Given:
80R
67sin 64
sin 80QR
61,15 cm
1Area of sin 36
2PQR QP QR
167 61,15 sin 36
2
21 204,09 cm
Grade 11
CAPS
Mathematics
Series
Unit 3
The Cosine Rule
2 2 2
2 2 2
2 cos
or
2 sin
b a c ac B
c a b ab C
c
baa
C
BA
The Cosine Rule for is given by:ABC
T h e C o s in e R u le f o r T r i a n g le s
2 2 2 2 cosa b c bc A
We use this form to find the third side when
two sides and included angle are given.
Symmetry also implies that:
P r o o f o f t h e C o s in e R u le
b a
In :CAD
2 2 2cos and x
A b x hb
In :BCD
22 2 2 2 22a h c x h c cx x
x c x
h
2 22 2 2cos2a c xb x c b A
2 2 2 cosb c bc A Proofs for symmetrical results are similar.
A s e c o n d f o r m o f t h e C o s in e R u le 2 2 2 2 cos a b c bc AKnow:
2 2 22 cos bc A b c a 2 2 2
cos2
b c aA
bc
We use this form to find any angle of
a triangle when we know all 3 sides.
7PR
Q
6
p
120
Find in the p PQR
A p p li c a t io n s o f t h e C o s in e R u le - E x a m p l e 1
Apply the Cosine Rule
2 2 2 2 cosp q r qr P
2 2 27 6 2 7 6 cos120p 127 11,3 1 decimal accuracyp
6
Y
Z8
4
X
2 2 28 6 4cos
2(8)(6)
X
Find in the X XYZ
Solution: Use the Cosine Rule
A p p li c a t io n s o f t h e C o s in e R u le - E x a m p l e 2
29,0 ( 1 dec )X
2 2 2
cos2
y z xX
yz
2 2 2 2 c o s c b a b a C
A
BC
c
30
Sine rule:sin sin
B C
b c
Find side and in the given . c B ABC
15b
19a
Cosine rule:
A p p li c a t io n s o f t h e C o s in e R u le E x a m p le 3
2 2 215 19 2(15)(19)cos30 c
9,61 c ( 2 decimal places )
15 sin 30sin
9,61
B
51,3 B ( 1 dec. )
T u to r i a l 3 : C o s in e R u le
2. Find all the angles in , giving your
answers to one decimal place accuracy.
XYZ
1. Given with 6 cm; 4 cm and
60 . Find correct to 2 decimal digits.
ABC AB BC
ABC AC
PAUSE
• Do Tutorial 3
• Then View Solutions
T u t o r i a l 3 : P r o b le m 1 : C o s i n e R u le : S o lu t i o n
2 2 2 2 cosAC BC AB BC AB ABC
Given:Find (2 dec accuracy):AC
2 24 6 2 4 6 cos60
28
28 5, 29 cmAC
T u t o r i a l 3 : P r o b le m 2 : C o s i n e R u le : S o lu t i o n
sin sin sin 4sin 48.2Now sin
7
X Y y XY
x y x
Given:
Determine all angle measures of XYZ.
2 2 2
cos2
y z xX
yz
2 2 24 9 7Hence cos
2 4 9X
48,2X
25, 2Y
Then 180 106,6Z X Y
CAPS
Mathematics
Series
Unit 4
Basic Applications:
Problems in 2-D
42 65
Problems in 2 dimensions: Example 1 1. Points and are in the same horizontal plane as ,
the foot of a vertical tower . 42 ; 65
and 25 . Calculate .
A B C
PC B PAC
AB m PC
P
CAB25 m
65 42 23BPA
25
sin 42 sin 23
AP
25sin 4242,81 m
sin 23AP
sin 65PC
AP
sin 65 42,81sin 65PC AP 38,8 m
23
Sine rule:
is unknown and needed to
find ( PAC is known)
AP
PC
2. In the figure represents a proposed tunnel.
and are visible from a point .
The three points are in the same plane.
QR
Q R P
Q R
P
Given:
100 m; 60 m
and 110
QP PR
QPR
Calculate the length of tunnel.
100 60110
2 2 2100 60 2 100 60 cos110QR
133 mQR
Problems in 2 dimensions: Example 2
T u to r i a l 4 : P a r t 1 : P r o b le m s i n 2 D
1. From the ends of a bridge , 101 metres long,
the angles of depression of a point on the ground
directly under the bridge is 42,2 and 70,1 .
Find the height, , of the bridge und
AB
P
h
er this point.
42,2 70,1A B101 m
P
h PAUSE
• Do Tutorial 4 Part 1
• Then View Solutions
T u to r i a l 4 : P a r t 1 : S u g g e s t e d S o lu t io n
42,2 70,1A B101 m
P
h
Question: Find h
180 42,2 70,1 67,7APB
101
sin 70,1 sin 67,7
AP
101 m sin 70,1
sin 67,7AP
102,65 mAP But sin 42,2102,65
h
102,65 sin 42,2 68,95 mh
T u to r i a l 4 : P a r t 2 : P r o b le m s i n 2 -D2. is a wall of a room, being the
line of the ceiling. is a picture rail,
with being directly below .
= 2 metres; and
2cos(a) Prove that
sin( )
(b)
ABCD AD
EF
E A
AE ACB x ECB y
xEC
x y
Find the length and height
of the wall if 33
and 20 .
x
y
2 m
xy
PAUSE
• Do Tutorial 4 Part 2
• Then View Solutions
T u to r i a l 4 : P a r t 2 ( a ) S o lu t io n
2 m
xy
Now and ACE x y CAD x
x
Hence, 90CAE x
From :AEC
2
sin 90 sin
EC
x x y
2sin 90
sin
xEC
x y
2cos
sin
x
x y
2.(a) Prove that
2cos
sin( )
xEC
x y
2 m
xy
Know:
2cos
sin
33 and 20
xEC
x y
x y
2cos337,46 m
sin13EC
cos cosBC
y BC EC yEC
Length of room 7,46 m cos20 7,01 mBC
Tutorial 4: Part 2: Length & Height of Room
Height of room:
2 sin 2 2 7,46 sin33 6,06 mEB EC y
End of the Topic Slides on
Sine, Cosine and Area Rules
REMEMBER!
•Consult text-books and past exam papers and
memos for additional examples.
•Attempt as many as possible other similar examples
on your own.
•Compare your methods with those that were
discussed in these Topic slides.
•Repeat this procedure until you are confident.
•Do not forget: Practice makes perfect!