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 Pearson Malaysia Sdn Bhd Form 4 Chapter 9: Trigonometry II.

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Presentation on theme: " Pearson Malaysia Sdn Bhd Form 4 Chapter 9: Trigonometry II."— Presentation transcript:

1  Pearson Malaysia Sdn Bhd Form 4 Chapter 9: Trigonometry II

2  Pearson Malaysia Sdn Bhd Quadrants and angles in the unit circle 0°0° 90° 180° 270° 360° Quadrant I Quadrant II Quadrant III Quadrant IV Cartesian plane can be divided into four parts called quadrants. Quadrants are named in the anticlockwise direction. y x

3  Pearson Malaysia Sdn Bhd Quadrants and angles in the unit circle O Angle  is measured by rotating the line OP in the anticlockwise direction from the positive x -axis at the origin, O. y x P 

4  Pearson Malaysia Sdn Bhd Verify sin  = y -coordinate in quadrant I of the unit circle O sin  = = = y y x  P ( x, y ) 1 y x Q  sin  = y -coordinate

5  Pearson Malaysia Sdn Bhd Verify cos  = x -coordinate in quadrant I of the unit circle O cos  = = = x P y x  ( x, y ) 1 y x Q  cos  = x -coordinate

6  Pearson Malaysia Sdn Bhd Verify tan  = in quadrant I of the unit circle O tan  = = P y x  ( x, y ) 1 y x Q  tan  =

7  Pearson Malaysia Sdn Bhd y x III III IV A ll s in t an c os Determine whether the value is positive or negative Quadrant I = All positive Quadrant II = sin  positive Quadrant III = tan  positive Quadrant IV = cos  positive

8  Pearson Malaysia Sdn Bhd Example 1: sin 213° y x Determine whether the value is positive or negative The angle 213° lies in quadrant III. Therefore, the value of sin 213° is negative. 213° O Sin  is positive in quadrant II. Not quadrant II

9  Pearson Malaysia Sdn Bhd Example 2: cos 321° y x Determine whether the value is positive or negative The angle 321° lies in quadrant IV. Therefore, the value of cos 321° is positive. 321° O Cos  is positive in quadrant IV. It is quadrant IV.

10  Pearson Malaysia Sdn Bhd Example 3: tan 123° y x Determine whether the value is positive or negative The angle 123° lies in quadrant II. Therefore, the value of tan 123° is negative. 123° O Tan  is positive in quadrant III. Not quadrant III

11  Pearson Malaysia Sdn Bhd Example 4: sin 32° y x Determine whether the value is positive or negative The angle 32° lies in quadrant I. Therefore, the value of sin 32° is positive. 32° O It is quadrant I. All positive in quadrant I.

12  Pearson Malaysia Sdn Bhd Determine the values of sine, cosine and tangent for special angles 45° 1 1  sin 45° =cos 45° = tan 45° = 1

13  Pearson Malaysia Sdn Bhd Determine the values of sine, cosine and tangent for special angles 30° 60° 1 2 sin 30° =cos 30° =tan 30° =

14  Pearson Malaysia Sdn Bhd Determine the values of sine, cosine and tangent for special angles 30° 60° 1 2 cos 60° =sin 60° =tan 60° =

15  Pearson Malaysia Sdn Bhd Determine the values of sine, cosine and tangent for special angles y x O (1, 0) (0, 1) (–1, 0) (0, –1)  0°0°90°180°270°360° sin  010–10 cos  10–101 tan  000

16  Pearson Malaysia Sdn Bhd Summary: Determine the values of sine, cosine and tangent for special angles  0°0°30°45°60°90°180°270°360° sin  010–10 cos  10–101 tan  0100

17  Pearson Malaysia Sdn Bhd Determine the values of sine, cosine and tangent for special angles Question 1: Calculate the values of the following: 7 sin 90° + 4 cos 180 ° Solution: 7 sin 90° + 4 cos 180 ° = 7 × (1) + 4 × (–1) = 7 – 4 = 3

18  Pearson Malaysia Sdn Bhd Values of angles in quadrant II  y x O  between x -axis and line O P  = corresponding angle in quadrant I P

19  Pearson Malaysia Sdn Bhd Values of angles in quadrant II y x  O  P    where  = 180° –  sin  = + sin  cos  = – cos  tan  = – tan   =  – 90° X

20  Pearson Malaysia Sdn Bhd Values of angles in quadrant III y x  O  P    where  =  – 180° sin  = – sin  cos  = – cos  tan  = + tan   = 270° –  X

21  Pearson Malaysia Sdn Bhd Values of angles in quadrant IV y x  O  P    where  = 360° –  sin  = – sin  cos  = + cos  tan  = – tan   =  – 270° X

22  Pearson Malaysia Sdn Bhd Solution: = – sin 51° Finding the value of an angle Question 1: Find the value of sin 231°. 231° y x O  P  quadrant III sin 231°  negative sin 231° = – sin (231° – 180°) = – 0.7771

23  Pearson Malaysia Sdn Bhd Solution: = cos 56° 43' Finding the value of an angle Question 2: Find the value of cos 303° 17‘. 303° 17'  quadrant IV cos 303° 17'  positive cos 303° 17' = cos (360° – 303° 17') = 0.5488 303° 17' x y O  P

24  Pearson Malaysia Sdn Bhd Solution: = – tan 62° 47' Finding the value of an angle 117° 13'  quadrant II tan 117° 13'  negative tan 117° 13' = – tan (180° – 117° 13') = – 1.945 Question 3: Find the value of tan 117° 13'. 117° 13' x y O  P

25  Pearson Malaysia Sdn Bhd Solution: Finding angles between 0° and 360° 0.9511  positive Therefore, the acute angle is in quadrant I or II. Question 1: For sin x = 0.9511 where 0° ≤ x ≤ 360°, find the value of x. x O 72° P Quadrant I: x = Quadrant II: x = 72° x P x y and x y Corresponding acute angle, x = 72° 72° 180° – 72° = 108°

26  Pearson Malaysia Sdn Bhd Solution: Finding angles between 0° and 360° – 1.746  negative Therefore, the acute angle is in quadrant II or IV. Question 2: For tan x = – 1.746 where 0° ≤ x ≤ 360°, find the value of x. 60° 12' O P x x y Quadrant IV: x = Quadrant II: x = 60° 12' P x and x y Corresponding acute angle, x = 60° 12' 180° – 60° 12' = 119° 48' 360° – 60° 12' = 299° 48'

27  Pearson Malaysia Sdn Bhd Solution: Finding angles between 0° and 360° 0.5  positive Therefore, the acute angle is in quadrant I or IV. Question 3: For cos x = 0.5 where 0° ≤ x ≤ 360°, find the value of x. 60° O P x x y Quadrant IV: x = Quadrant I: x = 60° P x and x y Corresponding acute angle, x = 60° 60° 360° – 60° = 300°

28  Pearson Malaysia Sdn Bhd Solve problems involving sine, cosine and tangent Question:In the diagram below, HMS and JHN are straight lines. H is the midpoint of JN. Given that HM = 12 cm, MN = 13 cm and FJ = 4 cm, calculate: (a)the length of HN, (b)the value of cos x °, (c)the value of tan y °. N HM S JF y°y° x°x°

29  Pearson Malaysia Sdn Bhd Solution: Solve problems involving sine, cosine and tangent N HM S JF y°y° x°x° HN 2 = (a) = 169 – 144 = 25  HN = 5 cm Pythagoras’ theorem 13 2 – 12 2 12 cm 13 cm 4 cm

30  Pearson Malaysia Sdn Bhd Solve problems involving sine, cosine and tangent N HM S JF y°y° x°x° Solution: x° x° = 180° –  HMN (b) cos x ° = HMS is a straight line = – cos  HMN

31  Pearson Malaysia Sdn Bhd Solve problems involving sine, cosine and tangent N HM S JF y°y° x°x° Solution: y° y° = 180° –  FHJ (c) tan y ° = JHN is a straight line = – tan  FHJ

32  Pearson Malaysia Sdn Bhd


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