Add Math Solution Of Triangle

Teaching & learning

Additional mathematics Form 4

CHAPTER 10

NAME:…………………………………………………. FORM :…………………………………………………

Date received : ……………………………… Date completed …………………………. Marks of the Topical Test : …………………………….. Prepared by : Additional Mathematics Department Sek Men Sains Muzaffar Syah Melaka For Internal Circulations Only

a)

a b c = = sin A sin B sin C

c) Area of triangle

=

b)

1 absin C 2

a2 = b2 + c2 - 2bc cosA

Students will be able to: 1.

Understand and use the concept of sine rule to solve problems.

1.1 Verify sine rule. 1.2 Use sine rule to find unknown sides or angles of a triangle. 1.3 Find the unknown sides and angles of a triangle involving ambiguous case. 1.4 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 obtuseangled triangle, sine Rule state that : a b c = = sin A SinB sin C

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

a) Ans

∠ PQR =420 ,PQ=10.49 cm,

PR=7.38

2. . Ans ∠MKL =380,MK=17.66 cm, ML=11.04 cm L

R

12 cm

3. Ans

∠ACB =65.950,

K

100 10 cm

P

5 cm

6 cm

0

42

0

66

AB = 6.05 cm C

0

720

∠ABC=49.050,

Q M

A

650

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 =57.7o or 123.3o BC 12.8 cm or 23.5cm a) ABC is a triangle with ∠ B = 25o , AB = 20 cm and AC = 10 cm. Solve the triangle.

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Exerice 2: For each of the following triangle , state if ambiguity arises. Hence, solve the triangles a) ∆ ABC : ∠ B = 30o , c = 8cm and b = 6cm c) ∆ ABC : ∠ B= 64o 22' a = 13.3 cm and b = 12 cm

∠ C=41o49' , 138o 11' ∠ A=108o11' 11o49' a =11.4 cm , 2.46 cm

∠ A = 87o 47', 92o13' ∠ C = 27o51' 23o25' c = 6.22 , 5.29

b) ∆ XYZ :

d) ∆ PQR : ∠ R = 28o , p = 8 cm and r = 11 cm

∠ Y = 69o 10' , x = 8.2cm and y = 9.5 cm

∠ X = 53o 47' ∠ Z = 57o 3' z = 8.53

∠ P = 19o58' ∠ Q = 132o2' q = 17.4 cm

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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

35o

R

PSR is a straight line as shown in the diagram above Given that SR = 6 cm , SQ = 7 cm , ∠PQS = 23o and

∠SRQ = 65 0 , Calculate a) ∠SQR

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 ∠LJK remains 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 a 2 = b2 + c2 - 2bc CosA b2 = a2 + c2 - 2ac CosB c2 = a2 + b2 - 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

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Example xample 5: Write the cosine rrule ule for each of triangles below a) b) c) C 8 cm

70

0

S

β

b

p

12 cm a

B

d)

t

β

q

r

β c

e

A

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 a) Given ∆ABC such that ∠ A = 50.5o , c = 4.5 cm and b = 6 . 5 cm . Find the value of a [ Ans a = 5.029 cm ]

Exercise 4

a) Given that ∆ABC , b = 5 cm, c = 4 cm and ∠ A = 66o . Find the value of a [ Answer 4.97 ]

b) Find all the angle of triangle PQR. Given that p = 4 cm, q = 6 cm, and r = 8 cm. (Ans ∠ R = 103o 29' , ∠ P = 28o 58' ∠ Q =46o 33' )

b) Given that ∆ABC , a = 12.5 cm, b = 20.5 cm and ∠ C = 124o 25' . Find the value of c and the remaining angle of the triangle. [ Answer c = 29.43 cm, ∠ A 20o 31', ∠ C = 41o 25' ]

Homework Text Book Exercise 10.2.2 page 248

2.3 Solve problems involving cosine rule

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2.4 Solve problems involving sine and cosine rules Example 5 From the diagram given , find ∠ PQS and ∠ RSQ [ Answer 67.49o and 17.7o ]

Example 6 From the diagram given , find i) ∠ ACD ii) length of AB [ Answer 42.4 , 12.76 cm ]

Exercise 6 a) In the diagram below, ABC is a straight line. Find (a) length of BD (b) length of CD [Ans 3.538 cm 8.35 cm ] D

b)

A

15 cm 4 cm

7 D

A

5 B L

6 cm

C

C

12 cm

B

BCD is a straight line in the above diagram. Given that BC 0 = 12 cm, AC = 15 cm, ∠CAD = 29o and ∠ABC = 68 . a) Find ∠BAC , b) the length of CD [ answer 47.88o , 12.64 cm ]

Homework Text Book Exercise 10.2.4 page 250

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Students will be able to: 3. 3.1

Understand and use the formula for areas of triangles to solve problems. 1 Find the area of triangles using the formula ab sin C or its equivalent.

2

3.2 Solve problems involving three-dimensional objects. Area of Triangles

1 ab sin C 2 1 = ac sin B 2 1 = bc sin A 2

Area of ∆ABC =

Find the area of the following triangles a) Answer 50.07 cm2

b) Answer [ 57.22 cm2 ]

c) Answer [40.13 cm2 ]

d) Answer [ 56.82 cm2 ]

Homework Text Book Exercise 10.3.1 page 254

3.2 Solve problems involving three-dimensional objects b) Calculate i) Length of PS ii) ∠PSQ a)Calculate ∠BGE [ Answer 70.17 o ] [Answer 16.76 cm , 83.41 o ]

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Exercise 7 c) [Answer i) 25.8o ii) 15.82 cm2 ] 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

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.SPM 2003 A diagram shows a tent VABC in the shape of a pyramid with triangle ABC as the horizontal base. V is the vertex of the tent and the angle between the inclined plane VBC and the base is 500.

SPM 2004 The diagram shows a quadrilateral ABCD such that ∠ABC is acute angle . D

5.2 cm 9.8 cm C

12.3 cm A

9.5 cm

B

(a) Calculate (i) ∠ABC (ii) ∠ADC (iii) the area, in cm2, 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 m2 [3 marks] (b) the length of AV if the angle between AV and the base is 250 [3 marks] (c) the area of triangle VAB [4 marks]

[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’, (ii)State the size of ∠A' B ' C ' [2 marks] o

Answer SPM 2004 Answer (i) ∠ABC = 57.23

o

Answer: (a) Length BC = 2.700 cm (b) Length AV = 3.149 cm (c) Area = 2.829 cm2

(ii)

∠ADC = 73.93o

(iii) Area of ABCD = 80.96 cm2

(b) (i)

C'

A

B

B '

(ii) ∠A' B ' C ' = 180 − 57.23 = 122.77 o

o

o

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