Lesson Notes By Weeks and Term - Senior Secondary School 2

Cosine and Sine Rule Relating to Triangle

SUBJECT: MATHEMATICS

CLASS:  SS 2

DATE:

TERM: 3rd TERM

REFERENCE BOOKS

• New General Mathematics SSS2 by M.F. Macrae etal.
• Essential Mathematics SSS2 by A.J.S. Oluwasanmi.
• Exam Focus Mathematics.

WEEK TWO

TOPIC: Cosine and Sine Rule Relating to Triangle

Content

-Sine Rule for Acute and Obtuse Angled Triangle.

-Application of Sine Rule to Triangle.

-Cosine Rule for Acute and Obtuse Angled Triangle.

-Application of Cosine Rule.

Sine Rule for Acute and Obtuse Angled Triangle.

Consideration is given to other triangles than a right angled triangle. The angles of any triangle are denoted by capital letters such as; A, B, C, while the sides are represented by small letters; a, b, c, respectively.

A

c               b

B                  a                     C

Acute Triangle: This is a type of triangle in which the angles are less than 900.

Obtuse Triangle: Is a type of triangle in which one of the angles is more than 900 but less than 1800.

Deductive Proof of Sine Rule

The sine rule is the same for acute and obtuse angled triangle.

Given: Any triangle ABC (acute-angled or obtuse-angled triangle).

A

c               b

B                  a                     C

To prove:        a       =  b       =       c

Sin A    SinB         Sin C

Construction: Draw a perpendicular A  to   BC ( produced, if necessary)

Proof:

Sin B  =  h     ……………………………………(1)

c

In  fig. 7.10 a)

.         Sin C =  h     ……………………………………..(2)

b

In   fig.7.10 b)

Sin(1800 - C ) =  h

b

Hence, Sin C = h    [sin(180 – Æ ) = sin Æ]   ……………(2)

b

From (1)   h = c SinB

From(2)    h = b Sin C

Hence, cSinB=bSinC

b   =   c

SinB    SinC

Example

In triangle ABC, A= 380, B = 270, and b = 17cm. find a and c.

Solution;

Using sine rule; Sin A = Sin B

a             b

A

sin 380  = sin 270        38

a              17

a = 17 sin 380

sin 270

a  =  23cm                            C        B

To find  c; Angle C must be known; A + B + C = 1800

C = 1800 – 380 – 270 ,                  C = 1150

Sin B = Sin C

b         c

sin 270 = sin 1150 ,                          c =  17 x sin 1150

17          c                                              sin 270

c= 33.9 approximately, c = 34cm.

NB: In any triangle, the longest side correspond to the largest angle while the shortest side corresponds to the smallest angle.

Evaluation:                                                                                                                                    1. Solve the â completely; A = 390, a = 8.2m and b = 5.6m

2.Calculate the values of angles P and R of â PQR, where q = 14.35cm, p = 7.82cm and Q = 115.60

Deductive Proof of Cosine Rule

The cosine rule is also the same for the acute and obtuse angled triangle.    A

Given: any         ABC

(a)        A

(b)c

bh

c    b

h

B    N

B    a    C    x

a-x                  N      x        C    a+x

a

To prove: c2 = a2 + b2 – 2abCos C

Construction: Draw a perpendicular from A to B(produced if necessary).

Proof: Using the acute triangle;                                                  using the obtuse triangle;

c2 = (a-x)2 + h2                                                                         c2 = (a +x)2 + h2

c2 = a2 -2ax + x2 + h2                                                                c2 =a2+2ax+x2+h2

From â ACN; b2 = x2 + h2, and Cos C = x / b                 From     ACN,x2+h2=b2

x= b Cos C                                                           =a2+2ax+b2

c2 = a2 + b2 – 2ax                                                          From         ACN,x/b=CosACN

x =bCosC                                                             =Cos(1800-C)

=  -Cos C ,x=  -bCos C

c2 = a2 + b2 – 2abCosC                                                          c2 = a2 + b2 +2a(-bCos C)

c2 = a2 + b2 – 2abCos C

Similarly, for other sides and angles. Therefore the cosine rule can be written as thus:

c2 = a2 + b2 – 2abCosC                        OR         Cos C = a2 + b2 – c2

a2= b2 + c2 – 2bcCosA                                                            2ab

b2 = a2 + c2 – 2acCosB                                        Cos A = b2 + c2 – a2

2bc

Cos B = a2 + c2 – b2

2ac

Conditions Necessary for Use: The rule is used for solving acute  and  obtuse angled triangles in which two sides and included angles are given.

Example; Given that A = 1200, b = 7cm, c= 12cm. Solve the triangle completely.

Solution

Using cosine rule;             a2 = b2 + c2 – 2bcCosA

a2= 72 + 122 – 2(7×12)Cos 1200

a2= 49 + 144 – 168 (-0.5)

a2 = 193 +84

a =√277

a = 16.6cm.

To find angle B,     Cos B = a2 + c2 – b2

2ac

Cos B = 16.62 + 122 - 72

2x 16.6 x12

Cos B = 275.56 + 144 – 49

398.4

Cos B = 0.9301,               B = cos-10.9301,          B = 21.50.

To find < C;   < A + < B + 0,

C = 1800 – 1200 – 21.50,          C = 38.50.

Hence, a = 16.6cm, B = 21.50 and C = 38.50.

NB: 1. In any triangle, the longest side corresponds to the largest angle and the shortestside to the smallest angle.

1. It is advisable to always find the smallest angle first , since the angle must be acute.

Evaluation

1.Calculate  the  angles  of  the âs  ABC  whose  sides  are  given  in  centimeters.Give  the  final  answers  to the  nearest  0.10

a=5.2,  b = 6.5cm  ,c = 7.8

General  Evaluation

1.Calculate the smallest angle in the triangle PQR such that p = 7.92m, q= 15.9m and c= 8.44m.

2.Calculate the length of the side opposite the given angle in â XYZ given that x =13.1m, y = 24.2m and Z = 47.80.

Revision Questions

1 Given that sin Æ =5/13 for 0<Æ<900 find

a   sinÆ -cosÆ

b   cos Æ -3

tanÆ

2 If cos 3y=sin 2y find y for 00

Essential Mathematics SSS2, page 180-181, exercise13.2, nos 11-15;exercise 13.4,page 185,nos 1a-1f.

Weekend Assignment

Objectives

Use the information below to answer question 1 – 3. In âABC, a = 7.8m, b= 8.5m and B = 57.70. correct answers to 1 d.p.

1. Find A;         A. 50.90     B. 510       C. 71.40   D. 700
2. Find C;         A. 510       B. 71.40   C. 710       D. 800
3. What is c?    A.10m       B.  12m     C. 9m       D. 9.5m
4. In â ABC, b = 4cm, c= 5cm and A = 1150. Find a to 2 s.f.  A. 7.66cm B 7.6cm  C.8cm D.7.7cm
5. In âPQR, p=1.8cm q = 2.5cm and r = 3.6cm. Calculate P.  A. 27.50   B. 300       C. 320   D.28

Theory

1.A triangle has sides of length 7cm, 8cm, 9cm. Express the cosine of the smallest angle of the triangle as a fraction in its lowest terms.

2.Solve the triangle completely in the âABC such that B = 34.50, c = 2.8cm, â¡â½

a = 5.1cm.