Teaching & learning
Additional mathematics Form 4
CHAPTER 10
NAME:\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202
FORM :\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202
Date received : \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 Date completed \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 . Marks of the Topical Test : \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026 .. Prepared by : Additional Mathematics Department Sek Men Sains Muzaffar Syah Melaka For Internal Circulations Only
a)
a
sin A
=
b
sinB
=
c
b)
sinC
c) Area of triangle
1 =ab sinC 2
a = b +c 2
2
2
- 2bc cosA
Students will be able to: 1.
Understand and use the concept of sine rule to solve problems.
1.1 1.2 1.3 1.4
Verify sine rule. Use sine rule to find unknown sides or angles of a triangle. Find the unknown sides and angles of a triangle involving ambiguous case. Solve problems involving the sine rule
To solve the triangle means we must find three interior angle and three sides of any triangle
A Sine Rule
Label the triangle with alphabets A,B,C and a,b,c . For any triangle ABC, whether it is an acute-angled or obtus angled triangle, sine Rule state that : a
sin A
=
b SinB
=
c
sinC
It is correct when the label of the triangle are correct. So label on the triangle is vary importance to verify sine rule.
Homework Text Book Exercise 10.1.1 page 242 Example 1 . Solve each of the following triangle
a)
b)
2
Exercise 1 : Solve the triangles below 0 0 0 2. . Ans \u2220MKL =38 ,MK=17.663.cm, Ans \u2220ABC=49.05 , PQR =42 ,PQ=10.49 cm, a) Ans \u2220 ML=11.04 cm
PR=7.38
\u2220ACB =65.950, AB = 6.05 cm
L
R
12 cm
K
C
100
0
72
0
0
66 P
10 cm
5 cm
42
0
Q
A
M
65
0
6 cm B
Homework Text Book Exercise 10.1.2 page 243 1.3 Finding the unknown sides and angles of a triangle involving ambiguous case. Ambiguous case (Ambiguous means having more than one value ) a) When two sides and an acute angle which is not an included angle are given with the acute angle opposite the shorter of two given sides, then ambiguity arises. b) Ambiguity means it is possible to form two triangle as shown below A situation like this will not happen if a) the angle given is an obtuse angle b) the side that is opposite the given angle is longer than the second side.
Example 2 : a) { Answer C =57.7 or 123.3 BC 12.8 cm or 23.5cm o a) ABC is a triangle with \u2220 B = 25 , AB = 20 cm and AC = 10 cm. Solve the triangle. o
o
3
Exerice 2: For each of the following triangle , state if ambiguity arises. Hence, solve the triangles o a) ∆ ABC : ∠ B = 30 , c = 8cm and b = 6cm o c) ∆ ABC : ∠ B= 64 22' a = 13.3 cm and b = 12 cm
o
∠ C=41 49' , 138
b)
∠X
∆
XYZ :
o
= 53
∠
o
o
11' ∠ A=108 11' 11o49' a =11.4 cm , 2.46 ∠ A cm = 87 o
Y = 69 10' , x = 8.2cm and y = 9.5 cm
47'
∠
o
Z = 57
3' z = 8.53
d)
∠P
∆
o
o
o
o
47', 92 13'∠ C = 27 51' 23 25' c = 6.22 , 5.29 o
PQR : ∠ R = 28 , p = 8 cm and r = 11 cm
o
= 19 58'
∠Q
o
= 132 2' q = 17.4 cm
4
Homework Text Book Exercise 10.1.3 page 244
1.4
Solving problems involving the sine rule
Example 3
L
Example 4
9.5 cm P
S
R
o
PSR is a straight line as shown in the diagram above o Given that SR = 6 cm , SQ = 7 cm∠ ,PQS = 23 and ∠SRQ
=
0
65 , Calculate
a) ∠ SQR
35
7 cm
b) the length of PS
J
K
The diagram shows a triangle JKL, a) Draw and label another triangle JLK ’ such that JK ‘ = 7cm JL = 9.5 cm and∠ LJKremains fixed at 35 o . b) Calculate the obtuse angle ∠ of JKL c) Find ∠ KL K ′
Homework Text Book Exercise 10.1.4 page 245 Students will be able to:
2.
Understand and use the concept of cosine rule to solve problems.
2.1 Verify cosine rule. 2.2 Use cosine rule to find unknown sides or angles of a triangle. 2.3 Solve problems involving cosine rule. 2.4 Solve problems involving sine and cosine rules. 2.1
Verify cosine rule
Cosine Rule 2
2
2
a = b + c - 2bc CosA 2 2 2 b = a + c - 2ac C osB 2 2 2 c = a + b - 2ab CosC
The cosine rule can be used to solve any triangle when a) two sides and the included angle are given, b) three sides are given
5
Example xample 5: Write the cosine ule rule r forfor each each of triangles of triangles below below a) b) c) C 8 cm
70
B
0
S
12 cm
t
β
a A
p
b
β
c
d)
q
r e
β
Homework Text Book Exercise 10.2.1 page 247
2.2
Using cosine rule to find unknown sides or angles of a triangle
Example 4 o a) Given ∆ ABC such that ∠ A = 50.5 , c = 4.5 cm and b) Find all the angle of triangle PQR. Given that p = 4 cm, b = 6 . 5 cm .Find the valueo f a [ Ans a = 5.029 cm ] q = 6 cm, and or = 8 cm. o o (Ans ∠ R = 103 29' , ∠ P = 28 58' ∠Q =46 33' )
Exercise 4
a) Given that ∆ ABC, b = 5 cm, c = 4 cm and∠ A = 66o .b) Given that ∆ ABC , a = 12.5 cm, b = 20.5 cm and o Find the valueof a [Answer 4.97 ] ∠ C = 124 25' . Find the value of c and the remaining angle of the triangle. o o [ Answer c = 29.43 cm ∠ 20 31', ∠ C = 41 25' ] , A
Homework Text Book Exercise 10.2.2 page 248 2.3 Solve problems involving cosine rule
6
2.4 Solve problems involving sine andne cosi rules Example 5 From the dia gram given , find∠ PQS and∠ RSQ o o [ Answer 67.49 and 17.7 ]
Exercise 6
a) In the diagram below, ABC is a straig ht line. Find (a) length of BD (b) length of CD [Ans 3.538 cm5 8.3 cm ] D 4 cm A
Example 6 ∠ ACD From the dia gram given , find i) ii) length of AB [ Answer .4 42 , 12.76 cm ]
b)
A 15 cm
7 D
5 B
6 cm
C
L
C
12 cm
B
BCD is a straig ht linen ithe above diagram. Given that BC 0 ∠ CAD = 12 cm, AC = 15 cm, = 29o and∠ABC = 68 . a) Find ∠ BAC, b) the length of CD o [ answer 47.88 , 12.64 cm ]
Homework Text Book Exercise 10.2.4 page 250
7
Students will be able to: 3. 3.1 3.2
Understand and us e the formula for areas of triangles to solve problems.
1
Find the area of triangles using the formula ab sinC or its equivalen t.
2
Solve problems involving three-dimensional objects.
Area of Triangles
Area of
∆ABC
=
1
ab sin C 2 1 = ac sin B 2 1 = bc sin A 2
Find the area of the following triangles 2 a) Answer 50.07 cm
b) Answer [ 57.22 cm ]
c) Answer [40.13 cm ]
d) Answer [ 56.82 cm ]
2
2
2
Homework Text Book Exercise 10.3.1 page 254
3.2 Solve problems involving three-dimensiona l objects a)Calculate∠ BGE[ Answer 70.17 o ] b) Calculate i) Length of PS ∠ ii)PSQ o [Answer 16.76 cm , 83.41 ]
8
Exercise 7 o 2 c) [Answer i) 25.8 ii) 15.82 cm ] ABE and DCF are equilateral triangle BCFE and ADFE are rectangle. Calculate i) ∠ ACE ii) Area of∆ AEB
d) [ 70.2
o
ii ) 34.66 cm
2
] Base on the diagram given on the left Calculate i) ∠PUQ ii) area of∆ PUQ
Homework Text Book Exercise 10.3.2 page 255
9
.SPM 2003 SPM 2004 A diagram shows a tent VABC in the shape of a The diagram shows a quadrilateral ABCD such that pyramid with triangle ABC as the horizontal base. V is∠ ABC is acute angle . the vertex of the tent and the angle between the 0 inclined plane VBC and the base is 50 . D
5.2 cm
9.8 cm
C
12.3 cm
9.5 cm
A
B
(a) Calculate (i) ∠ ABC (ii) ∠ ADC 2 (iii) the area, in cm , of quadrilateral ABCD. Given that VB = VC = 2.2 m and AB = AC = 2.6 m, calculate (a) the length of BC if the area of the base is 3 m [3 marks] (b) the length of AV if the angle between AV and the 0 base is 25 [3 marks] (c) the area of triangle VAB [4 marks] 2
[8 marks]
(b) A triangle A’B’C’ has the same measurements as those given for triangle ABC, that is,A’C’ = 12.3 cm,C’B’ = 9.5 cm and ∠ B ' A'C ' = 40.5 , but which is different in shape to triangle ABC. (i) Sketch the triangle A’B’C’, ∠ A' B' C' (ii)State the size of [2 marks] o
AnswerSPM 2004 Answer (i) ∠ABC = 57.23 o
Answer: (a) Length BC = 2.700 cm (b) Length AV = 3.149 cm 2 (c) Area = 2.829 cm
(ii) ∠ADC = 73.93
o
(iii) Area of ABCD = 80.96 cm
2
(b) (i)
C'
A
B
B '
(ii) ∠A' B'C ' = 180 − 57.23 = 122.77 o
o
o
10
