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MATHEMATICS-I, , ANGLE AND ITS MEASUREMENT, LECTURE NO.– 1 to 5, , 1. Complex, Numbers, BY- -Shermale, Sir, , N. M.V. Jr. COLLEGE PUNE-2

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Let's Study, , , , , , , Directed angle., Angles of different measurements, Units of measure of an angle, Length of an arc of a circle., Area of a sector of a circle.

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DIRECTED, ANGLES, STRAIGHT, ANGLE, , ANGLE IN A, QUADRANT, , Measures of, angle, , POSITIVE, ANGLES, , NEGATIVE, ANGLE, , ZERO, ANGLE, , ONE, ROTATION, ANGLE, , RIGHT, ANGLE, , Standard, Angles, , QUADRANTAL, ANGLE., , CO-TERMINAL, ANGLES

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DIRECTED ANGLES: , , Definition:-, , , , The ordered pair of rays (𝑂𝐴,𝑂𝐵) together with the rotation of, , , , the ray 𝑂𝐴 to the position of the ray 𝑂𝐵 is called the directed angle ∡AOB., , , , Where, , , , 1) the ray OA is called the initial arm, , , , 2) the ray OB is called the terminal arm., 3) O is called the vertex, 4) It is denoted by ∡𝐴𝑂𝐵, 5) Measure of directed angle may be positive or negative, , , , 

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, , Measures of angles:-, , , , The amount of rotation from the initial ray, , , , 𝑂𝐴 to the terminal ray 𝑂𝐵 gives the measure, of angle ∡AOB., , , , It is denoted by m∡𝐴𝑂𝐵 = 𝑥, , 

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POSITIVE ANGLES, If the rotation of the initial ray is anticlockwise,  then the measure of directed angle is,  considered as positive., , , , , 𝜃, , m∡𝑨𝑶𝑩 = 𝜽

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NEGATIVE ANGLE, If the rotation of the initial ray is clockwise,  then the measure of directed angle is, , , , , , , considered as negative., , −𝜽, , m∡𝑩𝑶𝑨 = −𝜽

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 ZERO ANGLE, , , If amount of rotation of the ray OA about vertex O is zero then the, , , , directed angle ∡𝐴𝑂𝐵 is called the zero angle., , , , In this case initial ray OA and terminal OB coincide. m∡𝐴𝑂𝐵 = 0°, , STRAIGHT ANGLE:After the rotation, if the initial ray OA and the terminal ray OB are in opposite, directions then directed angle ∡𝐴𝑂𝐵 is called as, , straight angle, , m∡𝐴𝑂𝐵 = 180° or −180°

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One rotation angle: After one complete rotation if the initial ray OA coincides with,  the terminal ray OB then directed angle AOB is called as one rotation,  angle m∡ AOB = 360°. Or −360°, , , Right angle:One fourth of one rotation angle is called as one right angle, it is also, half of a straight angle. m∡ AOB = 90°, B, A, O

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Angles in Standard position :, A directed angle with its vertex at origin O and,  the initial ray along the positive X-axis, is called,  angle in standard position., , , ∡𝑋𝑂P, ∡𝑋𝑂Q, ∡𝑋𝑂R, ∡𝑋𝑂Y, ∡𝑋𝑂Y’

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ANGLE IN A QUADRANT, A directed angle in standard position,  whose terminal ray lies in a quadrant is,  called Angle in a Quadrant, , , ∡𝑋𝑂P,  ∡𝑋𝑂Q,  ∡𝑋𝑂R, 

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Co-terminal angles:, Directed angles of different amount of,  rotation having the same positions of,,  initial rays and terminal rays are called,  Co-terminal angles, , , , , 𝐦∡𝑨𝑶𝑩 = 𝟑𝟎𝟎, , , , 𝐦∡𝑨𝑶𝑩 = −𝟑𝟑𝟎𝟎, , , , 𝐦∡𝑨𝑶𝑩 = 𝟑𝟗𝟎𝟎, , , , If the two directed angle are co-terminal angle then, difference between measures of two directed angles is an, integral multiple of 3600

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, , Measures of angles:-, , , , The amount of rotation from the initial ray 𝑂𝐴 to, the terminal ray 𝑂𝐵 gives the measure of angle, AOB., , , , It is denoted by m∡𝐴𝑂𝐵 = 𝑥, , , , , , It is measured in two systems., 1) Sexagesimal system (Degree measure), 2) Circular system (Radian measure)

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Sexagesimal System (Degree Measure):, , In this system, the unit of measurement of, angle is a degree., ,  One rotation angle is divided into 360 equal parts, the, measure of each part is called as one degree angle., , , One degree angle is divided into 60 equal parts, the, measure of each part is called as one minute angle., , , , One minute angle is divided into 60 equal parts, the, measure of each part is called as one minute angle.

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, , ∴, , 1 𝑡ℎ, 360, , ∴, , 1 𝑡ℎ, 60, , ∴, , 1 𝑡ℎ, 60, , , , , , , , , part of one complete rotation is called one degree, and it is denoted by 1°., , part of one degree is called one minute, and it is denoted by 1′., part of one minute is called one second, and it is denoted by 1′′., , , , , 1′ = 60′′ and 1′′ =, , , , 𝑚 ∡ one rotation angle = 3600, , , , 𝑚 ∡ straight angle = 1800, , , , 𝑚 ∡ right angle = 900, , = 60′, , and 1′, , 1 0, 60, 1 ′, 60, , 10, , =

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Circular System (Radian Measure):, , , , , , , , , , , , , , In this system, the unit of measurement of an angle is a radian., one radian is the measure of an angle subtended at the Centre of, a circle by an arc whose length is equal to the radius of the circle., Let r be the radius of a circle with Centre O., Let A and B be two points on circle, such that the l(arc AB )=𝑟., Then the measure of the central, angle AOB is defined as 1 radian., It is denoted by 1𝑐 ., i.e. 𝑚∡AOB=1𝑐

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Relation between angle and time in a clock., (R =rotation.), Minute Hand, 1𝑅 = 360, , 0, , 1𝑅 = 60 min, , 60 min = 3600, 1 min = 60, , Hour Hand, 1𝑅 = 3600, , 1𝑅 = 12 Hrs, 12 Hrs = 3600, , 1 Hr = 300, 1 Hr = 60 𝑚𝑖𝑛, 60 min = 300, 10, 1 min =, 2

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Exercise 1.1, Q.1 A) Determine which of the following, pairs of angles are co-terminal.,  i) 2100 , −1500, ii) 3600 , −300,  iii) −1800 , 5400, iv) −4050 , 6750,  v) 8600 , 5800, vi) 9000 , −9000, 

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1) Solution :- 2100 , −1500,  2100 − −1500 = 3600 = 1 × 3600,  Which is an integral multiple of 3600,  ∴ given pair of angles is coterminal.,  2) Solution :- 3600 , −300,  3600 − −300 = 3900,  Which is not an integral multiple of 3600,  ∴ given pair of angles is not coterminal., 

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3) Solution :- −1800 , 5400,  −1800 − 5400 = −7200 = (−2) × 3600,  Which is an integral multiple of 3600,  ∴ given pair of angles is coterminal,  4) Solution :- −4050 , 6750,  −4050 − 6750 = −10800 = (−3) × 3600,  Which is an integral multiple of 3600,  ∴ given pair of angles is coterminal, 

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, , , , , , B) Draw the angles of the following measures, and determine their quadrants., i) −1400 ii) 2500, iii) 4200, iv) 7500 v) 9450, vi) 11200, vii) −800 viii) −3300, ix) −5000, x)−8200

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, , i) m∡𝑋𝑂𝑃 = −1400, Y, O, , −1400, p, , X

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Q. 4 Express the following angles in degree, minute and second., i) 183.7, , 0, , ii) 245.33, , 0, , iii), , 1 𝑐, 5, ,  (i) (183.7)0 = 1830 + 0.70, , = 1830 , (0.7 × 60)′, , = 1830 , 42′,  ii) 245.33 0 = 2450 + 0.330, , = 2450 , (0.33 × 60)′, , = 2450 , (19.8)′, , = 2450 , 19′ + 0.8′, , = 2450 , 19′ , (0.8 × 60)′′, , = 2450 , 19′ , 48′′

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1 𝑐, 5, , 1, 5, , = × 1𝑐, , , , iii), , , , 1, 5, , ≈ × (57.32)°, , , , ≈ (11.46)0 = 11° + (0.46)°, ≈ 11° , (0.46 × 60)′, ≈ 11° ,27.6′ = 11° ,27′ +(0.6)′, ≈ 11° ,27′ , (0.6 × 60)′′, ≈ 11° ,27′ , 36′′, , , , , , , , 1 𝑐, 5, , ≈ 11° ,27′ , 36′′

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Q.13 Find the degree and radian measure of exterior and interior angle of, a regular, i) Pentagon, ii) Hexagon, iii) Septagon, iv) Octagon, , , SOLUTION:- Exterior angle of a regular n-sided polygon= 360°, 𝑛, , i) Pentagon (side=n=5), 360°,  Exterior angle of a regular Pentagon =, = 72°, 5, , , 𝜋, × 180, , 𝑐, , , , Exterior angle of a regular Pentagon = 72° = 72, , , , we now that Exterior angle + Interior angle =180°, ∴ Interior angle =180° −Exterior angle=180° − 72° =108°, , , , , ∴ Interior angle of a regular Pentagon =108° = 108 ×, , =, , 2𝜋 𝑐, 5, , 𝜋 𝑐, 180, , =, , 3𝜋 𝑐, 5

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, , SOLUTION:- Exterior angle of a regular n-sided polygon= 360°, 𝑛, , , , ii) Hexagon (side=n=6), Exterior angle of a regular Hexagon = 360°, = 60°, 6, , , , 𝜋, × 180, , 𝑐, , , , Exterior angle of a regular Pentagon = 60° = 60, , , , we now that Exterior angle + Interior angle =180°, ∴ Interior angle =180° −Exterior angle=180° − 60° =120°, , , , , , ∴ Interior angle of a regular Pentagon =120° = 120, , =, , 𝜋 𝑐, 3, , 𝜋 𝑐, ×, 180, , =, , 2𝜋 𝑐, 3

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Q.14 Find the angle between hour hand and minute hand in a clock at, i) ten past eleven, ii) twenty past seven, iii) thirty five past one, iv) quarter to six, v) 2: 20, vi) 10: 10, , , SOLUTION:- i) ten past eleven, , , , The Angle between two consecutive marks of a clock is, , 360°, 12, , = 30°, ,  Angle between marks 11 and 2 = 3× 30° = 90°, , , Angle trace by hour-hand in one minute =, , 1 0, 2, 1 0, 2, , , , ∴ Angle trace by hour-hand in 10 minutes = 10 ×, , , , ∴ Angle between the hour-hand and minute-hand of a clock, at ten past eleven is = 90° − 5° = 85°, , , , = 5°

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, , SOLUTION:- ii) twenty past seven, , , , The Angle between two consecutive marks of a clock is, , 360°, 12, , = 30°, ,  Angle between marks 4 and 7 = 3× 30° = 90°, , , Angle trace by hour-hand in one minute =, , 1 0, 2, 1 0, 2, , , , ∴ Angle trace by hour-hand in 20 minutes = 20 ×, , , , ∴ Angle between the hour-hand and minute-hand of a clock, at twenty past seven = 90° + 10° = 100°, , , , = 10°

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, , SOLUTION:-, , , , iv) quarter six, 360°, Angle between two consecutive marks =, = 30°, 12, ∴ Angle between marks 6 and 9 = 3 × 30° = 90°, , , , , , , , , , ∴ Angle trace by hour-hand in one minute =, , 1 0, 2, , 1 0, 2, , ∴ Angle trace by hour-hand in fifteen minutes = 15 ×, = 7.5°, ∴ Angle between the hour-hand and a minute-hand of a clock, at quarter six = 90° + 7.5° = 97.5° = 97°, 30’

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THANK YOU, , Next lecture we continue……… →, , Trigonometry, -I