INTERMEDIATE MATHEMATICS (STEM C) 45c) 2m 8 16 1 2m 3 4 0 3 8m 0 2 2 2 2 2 2 3 8m 0 8m 3 3 m 8 Example 5 Solve for x. a) x 3 x 1 2 2 48 b) 2x 3 x 2 33 2 4 0 Solution: a) x 3 x 1 2 2 48 x 3 x 1 x x x 2 2 2 2 48 2 8 2 2 48 Let 2 a 8a 2a 48 6a 48 a 8 x x x 3 Since 2 a So 2 8 2 2 x 3 b) 2x 3 x 2 33 2 4 0 2 3 x x x 2 2 2 33 2 4 0 Let y 2 8y 33y 4 0 8y 1 y 4 0 1 y ,y 4 8 Since x y 2 x x 2 3 x 3 1 1 2 and 2 28 2 2 2 x 3, x 2
INTERMEDIATE MATHEMATICS (STEM C) 46Solve the following equations without using calculator. a) 9 2n 2 n 2 16 2 b) x 1 2x 9 3 3 c) x 2x 3 27 3 243 d) 4x 19 2 81 e) 2 1 3 3 3x 5x 2 0 f) 2x 2 x 3 26 3 3 0
INTERMEDIATE MATHEMATICS (STEM C) 47Put log at both sides. With the use of a calculator Put log at log at both sides. Factor x Example 6 Evaluate the following equations using calculator: a) x 3 8 b) x x x 2 2 6 9 Solution: a) b) x x x 2 2 6 9 x x 2 x x 2 2 6 9 12 9 x x 2 log12 log9 xlog12 x 2 log9 xlog12 xlog9 2log9 xlog12 xlog9 2log9 x log12 log9 2log9 x 1.0792 0.9542 2 0.9542 0.125x 1.9084 1.9084 x 15.2672 0.125 x x 3 8 log3 log8 xlog3 log8 0.4771x 0.90309 0.90309 x 0.4771 1.8928 y y y a b a b
INTERMEDIATE MATHEMATICS (STEM C) 48Evaluate the following with the use of a calculator. a) x 3 2 5 0 b) x 1 2 3 8 c) x 2 5 9 1 d) x x x 4 4 3 9
INTERMEDIATE MATHEMATICS (STEM C) 492.3.2 Solving equation involving logarithms. A logarithmic equation is an equation which involves logarithms. a) If a log x y , then y x a b) If a a log x log y , then x y There are 2 techniques for solving logarithms function: Technique 1 : Comparing the bases of the log Example 1 Solve for x. a) log 2x 1 log x 7 3 3 Solution log 2x 1 log x 7 3 3 2x 1 x 7 2x x 7 1 x 8 1. Comparing the bases of the log. 2. Rewrite the equation in exponential form. By comparing the base, use the fact that: If a a log x log y , then x y
INTERMEDIATE MATHEMATICS (STEM C) 50b) 5 5 5 log 3x 2 log 3x 3log 2 Solution 5 5 5 3 5 5 5 5 2 2 log 3x 2 log 3x 3log 2 log 3x 2 3x log 2 log 3x 2 3x log 8 3x 2 3x 8 9x 6x 8 9x 6x 8 0 3x 2 3x 4 0 2 x , 3 4 x 3 2 x 3 Technique 2: Rewrite the equation in exponential form Example 1 Solve for x. a) 4 1 log x 2 Solution 4 1 2 1 log x 2 x 4 x 2 Ignore the negative value. Use Product Rule of Logarithm: a a a log xy log x log y Rewrite in exponential form. Use the fact: If a log x y , then y x a
INTERMEDIATE MATHEMATICS (STEM C) 51b) 2log x 3 log x 6 2 2 Solution 2 2 2 2 2 2 2 2 3 2 2 2 2log x 3 log x 6 log x log x 6 3 x log 3 x 6 x 2 x 6 x 8 x 6 x 8 x 6 x 8x 48 2 x 8x 48 0 x 12 x 4 0 x 12,x 4 x 12 Let's Master Mathematics! Solve for x. a) 2log x 4 0 3 b) x log 1 x 2 Use Quotient Rule of Logarithm: a a a x log log x log yy Ignore the negative value. Use the fact: If a log x y , then y x a
INTERMEDIATE MATHEMATICS (STEM C) 52c) log 2x 1 2 log x 7 3 3 d) 2 5 5 1 log 10x 1 2log x 12 2.3.3 Solving simultaneous equations involving exponents and logarithms. Example 1 Solve the following simultaneous equations. 2x 2y 7 2x 4y 3 3 (9) 3 5 (5) 5 Solution: 2x 2y 7 2x 4y 3 3 (9) 3 (1) 5 (5) 5 (2) Rewrite equation (1) and (2) 2x 2 2y 7 2x 4y 7 2x 4y 7 2x 4y 3 2x 4y 3 3 (3 ) 3 (1) 3 3 3 3 3 2x 4y 7 (3) 5 (5) 5 (2) 5 5 2x 4y 3 (4)
INTERMEDIATE MATHEMATICS (STEM C) 53Solve equation (3) and (4) 2x 4y 7 (3) 2x 4y 3 (4) Eliminate x : (3) (4) 2x 4y 7 (2x 4y 3) 8y 4 1 y 2 Substitute 1 y 2 into (3) 1 2x 4 72 2x 2 7 2x 55 x 2 Example 2 Solve the following simultaneous equations. 8 8 log x log y 1 x y 12 Solution: 1 8 8 x log log 8 y x 8 y x 8y (1) x y 12 (2) Substitute (1) into (2) 8y y 12 9y 12 4 y 3 Substitute 4 y 3 into (1) 4 x 8( ) 3 32 x 3
INTERMEDIATE MATHEMATICS (STEM C) 54Let's Master Mathematics! Solve the following simultaneous equations: a) 2 2 log y log 5 3 2x 7y 15 b) y 2 2x 9 27 log y log x 1 22 2
INTERMEDIATE MATHEMATICS (STEM C) 55EXERCISE INDEX FUNCTION 1. Simplify the following expressions. a) 3 6 x x b) 5n 3n 4 4 c) 2n 3 n n 6 6 6 d) 3 7 5 2 4x y 3x y e) 5 10 3 3 2 z 2 z f) 2 5 2 3 3ab 4a b 8a b g) 3 5 3 3 2 t 2r r t h) 4n 1 n 1 3 27 9 2. Evaluate the following without using calculator. a) 2 27 3 125 b) 3 4 2 3 34 9 27 81 c) 2 23 4 16 64 d) 2 1 3 2 12 8 9 25 3. Simplify and give your answer in positive exponent only. a) 3 3 2 2y z 16yz
INTERMEDIATE MATHEMATICS (STEM C) 56b) 2 5 2 mh h m c) 6 2 3 2 6 9 9 x y 3x y d) 2n n 1 2 3n 3 81 9 4. Solve the following equations. a) x 6 36 b) 2x 1 2x 3 27 c) n 1 2n 9 3 3 d) x x 1 8 2 4 5. Solve the following equations. a) 2x x 3 4 3 3 0 b) x 3 x 2 33 2 50 0 6. Evaluate the following with the use of a calculator. a) x 4 8 b) x x x 3 3 7 5 c) 1 x 2 6 24 LOGARITHM FUNCTION 1. Write each of the following as a single logarithm. a) 3 log x 4 log y 2 log z 2 2 2 b) 4 log 5 3 2 log z a a c) 1 2 log4 log9 log6 2 2. Simplify the following without using a calculator. a) 6 log 36 b) 7 3 log 49 log 27 3. Given that b log 3 0.792 and b log 5 1.161 . Find the value of: a) b log 15
INTERMEDIATE MATHEMATICS (STEM C) 57b) b 3 log 5 c) b 25 log 27 4. Without using calculator, compute the value of: a) 4 3 4 log 16 log 81 3log 2 . b) 5 5 5 1 1 log 3log 25 log 1 625 3 5. Solve the following equation: a) log x 2 5 2 b) 16 3 log 2x 4 c) log x 2 log 3x 8 3 3 d) 2log 3 log 4 2 x x e) 3 3 log x 8 2 log 4 f) 3 27 5 log x log 3 3 g) log3 logx log x 5 log2 6. Solve the simultaneous equations: a) log x log 1 y 1 4 4 y x 9 27 3 b) 2 2 log 2x log y 7 x log y 2
INTERMEDIATE MATHEMATICS (STEM C) 58Answers: Index Function 1. a) 9 x b) 2n 4 c) 3 6 d) 8 9 12x y e) 2 7 2 z f) 3 3ab 2 g) 4 9 32t r h) 2n 3 2. a) 9 25 b) 81 c) 6 4 d) 8 3. a) 8 5 y z 2 b) 11 mh c) 1 d) 1 4. a) x 2 b) 1 x 4 c) n 1 d) 1 x 3 5. a) x 0,x 1 b) x 1 6. a) 1.5 b) 1.7414 c) 3.5474
INTERMEDIATE MATHEMATICS (STEM C) 59Logarithm Function 1. a) 3 4 2 2 x y log z b) a 3 1250 log a c) log8 2. a) 2 b) 5 3. a) 1.953 b) -0.369 c) -0.054 4. a) 9 2 b) 2 5. a) x 30 b) x 4 c) x 3 d) x 6 e) x 28 f) x 9 g) x 1 6. a) 2 13x , y 7 14 b) x 4, y 16
INTERMEDIATE MATHEMATICS 60 CHAPTER 3 INEQUALITY Objectives At the end of this chapter the student should be able to: • Identify the types of interval notation, inequality notation, set notation and real number line. • Know the law of inequality. • Solve the inequality. 3.1 Introduction to Inequality Let a and b represent two real numbers with a < b. In each of these definitions, a is called the left endpoint and b the right endpoint of the interval. Interval Form Inequality Form Real Number Line The open interval (a,b) a < x < b a b The closed interval [a,b] a ≤ x ≤ b a b The half–open interval (a,b] a < x ≤ b a b The half–closed interval [a,b) a ≤ x < b a b The interval [a,∞) x ≥ a a The interval (a,∞) x > a a The interval (− ∞, a] x ≤ a a The interval (− ∞, a) x < a a The interval (− ∞,∞) All real numbers −∞ ∞
INTERMEDIATE MATHEMATICS (STEM C) 61 # A closed interval, denoted by [a,b], consists of all real numbers x for which a ≤ x ≤ b # An open interval, denoted by (a,b), consists of all real numbers x for which a < x < b # The half open, or half closed, interval are (a,b], consisting of all real numbers x for which a < x ≤ b and [a,b), consisting of all real numbers x for which a ≤ x < b Example 1 Write each inequality using interval form. a) 2 x 5 Answer: [2,5] ≤ ≤ b) x 5 Answer: [- ,5] ≤ ∞ c) x 2 Answer: [2, ] ≥ ∞ d) 2 x 5 Answer: [2,5) ≤ < e) x 2 Answer: (2, ] > ∞ f) x 5 Answer: [- ,5) < ∞ Example 2 Write each interval form as an inequality involving x. a) [5,10] Answer: 5 x 10 ≤ ≤ b) (− ∞,−1] Answer: x 1 ≤ − c) [− 2,6) Answer: -2 x 6 ≤ < d) (4,7) Answer: 4 x 7 < < e) (3,∞) Answer: x 3 > f) (− 3,12] Answer: -3 x 12 < ≤ g) [6,9) Answer: 6 x 9 ≤ < h) [−1,∞) Answer: x 1 ≥ − Example 3 Write each of the following in the real number line. Questions Real Number Line a) − 3 ≤ x < 0 Answer: -3 0 b) 4 > x ≥ 1 Answer: 1 4 c) x ≤ 5 Answer: 5
INTERMEDIATE MATHEMATICS (STEM C) 62 Let's Master Mathematics! Complete the following table. Interval Form Inequality Form Real Number Line ( 1,5] - x < -2 - 4 −≤ < 4x3 [0,∞) -7 7 -1 < x ≤ 9 (- ,-4) ∞ 4 9 x 8 ≤ (0,6) 7 x 9 ≥ −
INTERMEDIATE MATHEMATICS (STEM C) 63 3.2 Laws of Inequality 1. Nonnegative Property 0a2 ≥ Example: 0)2( 2 ≥− 2. Transitive property If ,ba and << ,cb then < ca Example: 72 and 7 << 10 , then 2 < 10 3. Addition Property If < ,ba then +<+ cbca If > ,ba then +>+ cbca Example: <− ,21 then +<+ 3231- > ,35 then +>+ 1315 4. Subtraction property If < ,ba then −<− cbca If > ,ba then −>− cbca Example: < ,40 then <− 2-420 > ,36 then −>− 5356 5. Multiplication Properties If < ba and if > ,0c then < bcac If < ba and if < ,0c then > bcac If > ba and if > ,0c then > bcac If > b.a and if < ,0c then ac < bc
INTERMEDIATE MATHEMATICS (STEM C) 64 Example: 31 and 4 >< 0, then 1( )( < 3()4 )( )4 < 31 and 4- > 0, then 1( )( 3()4 )(−>− )4 45 and 2 >> 0, then 5( )( > 4()2 )( )2 > 45 and 2- > 0, then 5( )( 4()2 )(−<− )2 6. Division property If c b c a < ba and c is positive, then < If c b c a < ba and c is negative, then > If b 1 a 1 < , ba and a and b both positive or negative, then > Example: 2 6 2 4 < 64 and c is 2, then < 2 6 2 4 64 and c is 2, then − > − < − 9 1 3 1 3a and <== ,93 ,9b then > 7. Reciprocal Property If 0 a 1 > ,0a then > If 0 a 1 < ,0a then < Example: 0 10 1 10 > ,0 then > 0 10 1 10 ,0 then < − >−
INTERMEDIATE MATHEMATICS (STEM C) 65 Let's Master Mathematics! Question 1 a) Express as an inequality the result of multiplying each side of the inequality < 52 by 3 Answer: _______________________ b) Express as an inequality the result of the multiplying each side of the inequality > 1-by 48 Answer: _______________________ Question 2 a) Express as an inequality the result of the addition each side of the inequality < 43 with 2 Answer: _______________________ b) Express as an inequality the result of the addition each side of the inequality > 1013 with 5 Answer: _______________________ Question 3 a) Express as an inequality the result of the subtraction each side of the inequality > 26 with 4 Answer: _______________________ b) Express as an inequality the result of the subtraction each side of the inequality < 91 with 7 Answer: _______________________ Question 4 a) Express as an inequality the result of the division each side of the inequality < 8 16 with 2 Answer: _______________________ b) Express as an inequality the result of the division each side of the inequality < 9 6 with 3- . Answer: _______________________
INTERMEDIATE MATHEMATICS (STEM C) 66 3.3 Solving Inequality (Linear Inequality) Solving inequalities is very like solving equations which is a value we can put in place of a variable (such as x) that makes the equation true. Example 1: =− 85x When we put 13 in place of x we get 13 →=− 85 True So, x = 13 But we must also pay attention to the direction of inequality. These things do not affect the direction of the inequality: a) Add (or subtract) a number from both sides. b) Multiply (or divide) both sides by a positive number c) Simplify a side. Example 2: +< 36x4 We can simplify +< 36x4 without affecting the inequality. 4 9 x 4 9 4 x4 9x4 < < < However, these things can change the direction of the inequality: a) Multiply (or divide) both sides by a negative number b) Swapping left and right sides Example 3: 2y3 <+ 10 When we swap the left- and right-hand sides, we must also change the direction of the inequality. 10 +> 2y3
INTERMEDIATE MATHEMATICS (STEM C) 67 Example 4: >− 7y25 1y 2 2 2 y2 2y2 575)y25( −< − < − − >− −>−− Example 5: Solve +>−+ 4x5)1x(3 . Interpret the solution geometrically. 3x 2 6 2 x2 6x2 x3 24x 4x2x3 4x53x3 > > > +>− +>− +>−+ Example 6: Solve ≤+<− 72x56 . Interpret the solution geometrically. Method 1 Method 2 6 5x 2 7 6 5x 2 and 5x 2 7 6 2 5x 7 2 6 2 5x 5x 7 2 8 5x 5 −< + ≤ −< + +≤ −− < ≤ − −−< ≤− −< ≤ − 8 x x 1 5 8 x 1 5 < ≤ −<≤ Interval Form: − 1, 5 8 Real Number Line: 8 5 − 1 Inequality: 1x 5 8 ≤<− → divide aby negative number, we must also change the direction Interval Form: ( ) ,3 ∞ Real Number Line: 3 Inequality Form: > 3x
INTERMEDIATE MATHEMATICS (STEM C) 68 Let's Master Mathematics! Solve the following inequalities and interpret the solution geometrically. 1) 8x 3 13x 7 +≤ + Solution: 2) 2 4x 6 5 < +≤ Solution: 3) +<−<− 8xx433x2 Solution: 4) 5 10 +<−<− x62x Solution: 5) 3 x x 5 x −≤<− Solution: 5) 2(x 1) 3(2x 1) 8 3(4 x) +− + ≤− + Solution:
INTERMEDIATE MATHEMATICS (STEM C) 69 3.4 Quadratic Inequality & Rational Inequality Other types of inequalities are quadratic and rational. To solve these inequalities, the following steps can be used as guidelines. i. Simplify the inequality so that 0 is on the right. ii. Factorized the left side. iii. Set it to be equal to 0 and solve for x. iv. Separate the solution(s) into intervals. v. Choose a test value for each interval (different from x) and test for the plus and minus signs. vi. Build a table and enter the intervals and signs. vii. Check the original inequality for confirmation. Example 1: Solve for x2 − 4x > 25 Solution: x 2 − 4x > 5 x 2 − 4x − 5 > 0 (x − 5) (x + 1) = 0 x = −1, 5 Interval Test Value Sign of (x − 5)(x +1) (− ∞,−1) -2 (−)(−) = + (−1,5) 0 (−)(+) = − (5,∞) 6 (+)(+) = + Since x2 − 4x − 5 > 0, hence the answer are Interval form: (− ∞,−1)∪(5,∞) Inequality form: x < −1∪ x > 5 Real Number Line: ∪ -1 5 Quadratic Form 2 Ax Bx C + +
INTERMEDIATE MATHEMATICS (STEM C) 70 Example 2: Solve for 0 1x x25 ≤ − − Solution: *Skip step ii and iii. Let the numerator and denominator equal to 0. 1x 2 5 x 0x25 x 01- = = =− = Interval Test value Sign of 1x x25 − − ( ) ∞− 1, 0 −= − + )( )( 2 5 ,1 2 += + + )( )( ,∞ 2 5 3 −= + − )( )( Since 0 1x x25 ≤ − − , Hence the answer are Interval Form: ( ) ,∞∪∞− 2 5 1, Inequality Form: 2 5 x1x ≥∪< Real Number Line: ∪ 1 5 2 Note that ≠ 1x Rational Form P(x) Q(x)
INTERMEDIATE MATHEMATICS (STEM C) 71 Example 3: Solve 2 4x 2x ≥ − + Solution: x 10 x 4 10 0x x 04- 0 4x 10 x 0 4x 8x22x 0 4x )4x(22x 222 4x 2x 2 4x 2x = = =− = ≥ − − ≥ − −−+ ≥ − +−+ −≥− − + ≥ − + Interval Test value Sign of 4x 10 x − − ( ) ∞− 4, 3 −= − + )( )( ( ) ,410 5 += + + )( )( ( ) 10,∞ 11 −= + − )( )( Since 2 4x 2x ≥ − + , Hence the answer are Interval Form: (4,10] Inequality Form: x4 ≤< 10 Real Number Line: 4 1 Note that ≠ 4x
INTERMEDIATE MATHEMATICS (STEM C) 72 Let's Master Mathematics! Solve the following inequalities. 2 1) 3x 4x 1 8 + +≥ Solution: 2 2) x 7x 8 − ≤ Solution: 2 3) 3 x 12 + ≤ Solution: 4) 3 x2 x > − Solution: 4x 5 5) 3 x 2 + ≤ + Solution: 4x 8 6) 1 4 x − ≥ − Solution:
INTERMEDIATE MATHEMATICS (STEM C) 73 3.5 Absolute Value (Linear) The absolute value is defined by Definition: Example 1 : Solve 5 2x 4 − = Solution : By Definition 2 9 x 2 1 x x21 9 2x x245 45 2x 4x25 Or 4x25 = = = = =− =+ =− −=− Theorem: i) If < , ax then << axaii) If > , ax then x a x <∪> a - Example 2: Solve the following inequalities a) <+ 94x4 Solution: x x x x x 94 49 94 4 4494 13 4 5 13 4 5 4 44 13 5 4 4 -< + < -- < + - < - -< < - < < - << a if 0a aa if 0a <= ≥= b) ≥− 1x76 Solution: x 1 7 5 x 7 7 7- - 7x 7 5 7- - 7x - 7x -5 - 7x -7 -6-6 7x 6-1 -6-6 7x -1 6- 1x76 Or -6 7x -1 ≤ ≥ − − ≥ − − ≤ ≥ ≤ ≥ ≤ ≥− ≤
INTERMEDIATE MATHEMATICS (STEM C) 74 Let's Master Mathematics! Solve the following absolute inequalities and interpret the solution geometrically. 1) 3x 2 4 + ≤ Solution: 2) x 2 1 −− ≥ Solution: 3) 1 3x 9 − < Solution:
INTERMEDIATE MATHEMATICS (STEM C) 75 EXERCISE Exercise 3.1 Mark the following inequalities using interval notation. a) x 10 > − b) −≤ ≤ 3x2 c) x 5 ≤ d) > 6x e) x 8 ≤ f) 7x 4 ≥ >− Exercise 3.2 Complete the table below with TRUE or FALSE. No Questions True / False a. If ba and >> c,b then > ca b. If a < b, then ±<± cbca c. If > ba and c is negative, then > cbca d. If < ba and c is positive, the < cbca e. c b c a If < ba and c is negative, then < f. c b c a If > ba and c is positive, then < g. b 1 a 1 If a and b are both positive or both negative, and a < b, then >
INTERMEDIATE MATHEMATICS (STEM C) 76 Exercise 3.3 Solve the following inequalities. a) 2 x 9x 10 − > b) 1x2 3x 1 − + ≤ c) 3 1x2 3x ≥ − + d) 3 x 5 x3 +≥ e) −<−<− x271x2x4 f) 24 2 x3 4 ≤− − <− g) x 4 3x x −< − h) ≥− 4x51 i) <− 5x32 j) ≤− 95x6
INTERMEDIATE MATHEMATICS (STEM C) 77 Answers: Exercise 3.1 a) ( ) 10,∞− b) [ ] − 2,3 c) ( ,5] −∞ d) 6x6x / ( ) ( ) −∞−∪∞−<∪> 6,,6 e) [ 8,8] − f) ( 4,7] − Exercise 3.2 a) True b) True c) False d) True e) False f) False g) True Exercise 3.3 a) ( , 1) (10, ) −∞ − ∪ ∞ b) 4, 2 1 c) 1 6, 2 5 d) ≥ .1x 8844, .0 8844 ≤≤ 0x e) 2x 3 5 << f) <≤− 3x9 g) ( ) ( ) ∪− 3,20,6 h) [ ∞∪ ) −∞− ,1 5 3 , i) 3 7 x1 <<− j) 3 7 x 3 2 ≤≤−
INTERMEDIATE MATHEMATICS (STEM C) 78 CHAPTER 4 CIRCULAR MEASURE Objectives At the end of this chapter the student should be able to • Understand the definition of a radian, and use the relationship between radians and degrees. • Use the formulae rs θ= and θ= 2 r 2 1 A in solving problems of the arc length and sector area of a circle. 4.1 Introduction of converting measurement of angle An angle is measured in degrees ( ) or radians (rad). 1 radian is the angle subtended at the centre of a circle by the arc of length which is equal to the radius of the circle. A radius is a straight line from the center to the circumference of a circle or sphere Relationship between degrees and radians degrees 180 1rad radians 180 1 180 radians 360 2 radians π = π = π= π=
INTERMEDIATE MATHEMATICS (STEM C) 79 Example 1 : Convert 60 degrees to radians Solution : rad or 1.05 rad 3 180 6060 π = π ×= Example 2 : Convert π 3 2 radians to degrees Solution : 120 180 3 2 rad 3 2 = π ×π=π Example 3 : a) The diagram below shows a circle with cintre O. The angle of the major sector MON is 260 . Find the value of θ in radians. Solution : 360 260 100 5 100 180 9 °°° ° − = π × =π θ N M O
INTERMEDIATE MATHEMATICS (STEM C) 80 C O A 50○ B 1) Convert 125 to radians Solution: 2) Convert 305 to radians Solution: 3) Convert 4 3 π radians to degrees Solution: 4) Convert 7 12 π radians to degrees Solution: 5) In the diagram below, CBA is a semicircle with centre O. Determine the angle BOC in radians in terms of π . Solution: 4.2 Arc Length and Chord of a Sector
INTERMEDIATE MATHEMATICS (STEM C) 81 chord θ P Q R O 8 cm An arc is a part of a curve. It is a fraction of the circumference of the circle. A sector is part of a circle enclosed between two radii. A chord is a line joining two points on a curve. The diagram above shows a circle with centre O and radius r, AB is a minor arc and ACB is a major arc. The length of the arc AB is directly proportional to the angle θ . Hence, Length of arc AB: rs θ= Area of minor sector AOB: θ= 2 r 2 1 A The angle subtended by the arc ACB at the centre of the circle is −π θ)2( radians. Example 1 : Figure below shows an arc PQR subtended at O with angle 72 . Find a) angle, θ in radians b) the arc length of PQR Solutions : r θ A B O C where θ is in radians
INTERMEDIATE MATHEMATICS (STEM C) 82 a) 180 72 π ×=θ = π 5 2 1.2566 rad or b) rs θ= arc PQR ×= .18 2566 = 10.05 cm Example 2 : The diagram below shows a circle with centre O and radius 9 cm. Calculate the length of the mjor arc AB Solution : Let the angle subtended by the major arc AB at O be θ 38.48cm 9 4.276 r Length of the major arc AB 4.276 rad 180 245 245 115360 = ×= θ= = π ×= = −=θ Example 3 : Calculate the radius r of the following circles O 115○ B A
INTERMEDIATE MATHEMATICS (STEM C) 83 a) b) Solution : a) =θ 8.0 rad, = cm 6 s 7.5 cm 0.8 6 s r = = θ = b) π−π=θ 3 2 2 π= 3 4 cm 9 3 4 37.7 s r = π = θ = r O 0.8 rad s = 6 r O s =37.7 cm
INTERMEDIATE MATHEMATICS (STEM C) 84 1) The diagrams shows a circle with centre O. Given that the length of the major arc PQ is 39.8 cm, find the radius of the circle. Solution: 2) Find the value of θ of the following circle Solution: 3) The diagram shows a circle with centre O and radius 5.2 cm. Given that length of the minor arc AB is 6 cm, find the angle of the major sector OAB. Solution: 4) PQ is an arc of a circle with centre O and radius 8.2 cm. The angle of the sector OPQ is 2.36 radians. Find the length of the major arc. Solution: 4.3 Area and Perimeter of Sector and Segment O 42.6 cm r = 7 cm O 5.2 cm B A 1.65 rad P Q
INTERMEDIATE MATHEMATICS (STEM C) 85 Figure 4.3 Figure 4.3 shows a sector AOB centered at O with angle θ radian in a circle. The area of the sector, A is propotional to the angle θ subtended at the centre of the circle: π θ = π = r 2 A Angle subtended by the circle at O Angle subtended by the arc AB at O Area of circle Area of sector AOB 2 Hence, The area of a segment is the area of the sector minus the area of triangle, Example 1 segment r r A B O 1 2 A r where must be in radians. 2 =θ θ ( ) θ−θ= θθ= θ r sin 2 1 r sin wher e must be in radian. 2 1 - r 2 1 A 2 2 2 A O B The segment of a circle is the region bounded by a chord and the arc subtended by the chord.
INTERMEDIATE MATHEMATICS (STEM C) 86 Find the area of the following sector: a) Solution: 11.67cm 6.0 7 s r rs == θ = θ= ( ) ( ) 2 2 2 40.86cm 11.67 6.0 2 1 r 2 1 A = = θ= b) Solution: ( ) 2 2 2 23.95cm 180 7 56 2 1 r 2 1 A = π = × θ= Example 2 7 cm 0.6 rad A B O A B O 7 cm Convert θ into radian
INTERMEDIATE MATHEMATICS (STEM C) 87 The area of the sector ABC is 2 πcm7 and BC = cm5 . Find ∠ABC . Solution: ( ) .1 759rad 75 2 1 7r 2 1 7A 2 2 =θ π=θ π=θ π= Example 3 The diagram below shows a circle with center O and radius 10 cm. A and B are points on the circumference such that arc AB makes an angle of 130 at O. Calculate the area of the shaded segment. Solution: ( ) ( ) 2 2 2 75.148cm ) sin130 180 10 (130 x 2 1 r sin 2 1 Area of shaded segment = − π = θ−θ= O B A Convert θ into radian
INTERMEDIATE MATHEMATICS (STEM C) 88 1) Find the radius of a sector if the angle subtended at the center of the circle is 3 radians and the area of the sector is 2 39 cm . Solution: 2) Find the area of a sector with subtended angle of 65ο and radius of 6 cm . Solution: 3) The diagram shows a sector OAB with centre O. OAC is a right triangle. Find a) angle θ , in radians. b) area of the shaded region, ABC. Solution: 4) The diagram shows two arcs, AD and BC of two sectors with centre O and having radius OA and OB. Find the a) angle θ , in radians. b) perimeter of the shaded region, ABCD. Solution:
INTERMEDIATE MATHEMATICS (STEM C) 89 EXERCISE Exercise 2.1 1. Convert the following angles to degree a) 1. 25 rad b) 0.8 rad c) rad 2 π d) rad 3 5 π 2. Convert the following angles to radians a) 27 b) 136 2. c) 205 d) 84 7. 3. Convert the following angles to radians in terms of π a) 65 b) 100 c) 260
INTERMEDIATE MATHEMATICS (STEM C) 90 θ R O P Q Q θ R O P S Exercises 2.2 1. The diagram shows a circle with centre O and radius 8.2 cm. Given that the length of the minor arc AB is 18 cm, find the angle of the majoe sector OAB 2. The diagram below shows a semicircle with centre O. Given that OP = cm 6.6 and arc length PQ = cm 5.4 , find the value of θ in radian. 3. PS and QR are the arcs extended by the central angle θ of the same circle with centre O. Given SRPQ , =θ= 0.6 radian, the arc = 4.8QR cm and the arc PS = cm 3 . Find the length of OP and OQ. A B A O r 18 cm
INTERMEDIATE MATHEMATICS (STEM C) 91 B C O A D rad 3 2 4. The diagram below shows two sectors of circles centre at O. Given that rad and the perimeter of ABCD 16 cm 3 2 AB DC 3 cm, AOD =∠== = . Find the length of OA. 5. The diagram shows two sectors of circles centre at O. If 10OA cm, DOC =∠= 60 and arc length cm 3 7 CD π= . Find the length of DA. Exercise 2.3 1. Figure below shows a sector of circle with center at O with radius 12cm and arc length of AB 20cm. Find a) the angle of AOB in radians. b) the angle of AOB in degrees. c) the area of sector AOB. A B C D O 12 cm 20 cm A B O
INTERMEDIATE MATHEMATICS (STEM C) 92 2. ORS is a sector of a circle with radius 6.5cm has an area of 15cm2. Calculate a) the angle, θ , in degree. b) the arc length of the sector. 3. Figure shows a sector of a circle OPS and a square PQRS. Given that the arc length PS is 9.5cm and QR = cm6 . If =∠ 70POS , find a) the radius of the circle, in cm. b) the area of shaded region. c) perimeter of shaded region. 4. Figure shows a sector MON subtends at angle =θ 35 with a radius of 10cm. Calculate the area of shaded region. 5. A sector OBC subtends at center O with radius OB and OC. ∠OAB and ∠COD are right angles as shown below: 70o O
INTERMEDIATE MATHEMATICS (STEM C) 93 Given OA = cm4 , AB = cm3 ,CD = 13cm and the area of sector OBC is 2 cm 3 25 π . Find a) the angle of COB. b) the perimeter of the shaded region to 3 decimal places. 6. AOB is a minor sector with a radius of 10cm. Find ∠AOB of the sector if the area is 53cm2. 7. Figure below shows a square PQRS and a sector OPS with centre at O. Given = 10OP cm , PS = cm4.8 and =∠ 50POS . Find : a) the length of arc PS. b) the area of the sector OPS. c) the area of the shaded region. 8. An arc of length 15cm and it subtends an angle of 52 at the center of a circle. Calculate the radius and area of a segment of a circle. 9. Given that the area of minor sector AOB is 84cm2. Find the radius of the sector if =∠ 100AOB .
INTERMEDIATE MATHEMATICS (STEM C) 94 10. Given the radius and arc length of a sector is 9cm and 5.7cm respectively. Find the subtended angle, θ , of the sector in radians. 11. Figure below shows a sector OPQ with centre O. OPR is a right triangle and length of OP is 5cm and RQ is 2cm. Find the : a) angle in radians. b) perimeter of the shaded region. 12. Find the area of a sector with subtended angle of 138 and radius is 8cm. 13. Figure below shows a circle centered at O. If =θ 3.2 radian and the area of the major sector OABC is 93.2cm2. Find the : a) radius of the circle. b) The length of the straight line AC. c) The perimeter of OABC. A C O B