Lesson Notes By Weeks and Term - Senior Secondary School 2
Cosine and Sine Rule Relating to Triangle
Cosine and Sine Rule Relating to Triangle
SUBJECT: MATHEMATICS
CLASS: SS 2
DATE:
TERM: 3rd TERM
REFERENCE BOOKS
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 + 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. Evaluation 1.Calculate the angles of the âs 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 0 Reading Assignment 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 âABC, a = 7.8m, b= 8.5m and B = 57.70. correct answers to 1 d.p. 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 âABC such that B = 34.50, c = 2.8cm, â¡â½ a = 5.1cm.
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