146 PRIME Opt. Maths Book - IX v) A 3 B O X’ X Y Y’ 4. Find the equation of straight line under the following conditions. i) Having y-intercept 3 and slope = 2. ii) Having y-intercept –4 and making an angle 30° with x-axis in positive direction. iii) Having y-intercept 2 and making an angle 120° with x-axis in negative direction. iv) Making y-intercept –3 and equally inclined on the axes. v) Making equal angles on the axes and passes through a point (0, 4). 5. Find the equation of straight line under the following conditions. i) P = 3 and a = 30° ii) Perpendicular distance from the origin = 4 and a = 150° iii) A 30° B C O X’ X Y Y’ 2 units iv) OC = 2 units A 60° B C O v) A B C O X’ X Y Y’ OC = 2 units \BAO = 30° 6. Find the equation of straight line under the following conditions. i) Find the equation of straight line having perpendicular distance from the origin is 4 units and angle made by the perpendicular to the x-axis a the origin is 150°. Also prove that it passes through a point (–3 3, –1)
PRIME Opt. Maths Book - IX 147 Answer 1. Show to your teacher. 2. i) 4x + 3y –12 = 0 ii) 4x – 3y + 24 = 0 iii) x + y – 5 = 0 iv) x – y – 7 = 0 v) x + y –3 = 0 & x – y – 3 = 0 3. i) 2x + y – 4 = 0 ii) x – 3y –3 3 = 0 iii) x – y + 4 = 0 iv) 3x + y – 4 = 0 v) x + y + 3 = 0 4. i) 2x – y + 3 = 0 ii) x – 3y – 4 3 = 0 iii) 3x – y + 2 = 0 iv) x – y – 3 = 0 and x + y + 3 = 0 v) x + y – 4 = 0 and x – y + 4 = 0 5. i) 3x + y – 6 = 0 ii) 3x –y + 8 = 0 iii) 3x – y + 4 = 0 iv) x + 3y – 4 = 0 v) x – 3y + 4 = 0 6. i) 3x – y + 8 = 0 ii) x + 2y – 8 = 0 iii) 3x + y + 5 = 0 iv) x + y – 7 = 0 or 4x + 3y – 24 = 0 v) x + y – 3 = 0 or 2x + y – 4 = 0 7. i) 3x – 2y + 12 = 0 ii) 6x + y – 9 = 0 iii) 2x + y – 4 = 0 iv) 2x – y + 8 = 0 v) 2x + 3y – 12 = 0 ii) Find the equation of straight line passes through a point (2, 3) which makes x-intercept double the y-intercept. iii) Find the equation of straight line passes through a point (–3, 4) which makes y-intercept thrice the x-intercept. iv) Find the equation of straight line passes through a point (3, 4) which makes sum of the intercepts on the axes is 14. v) Find the equation of straight line passes through a point (1, 2) which makes sum of the intercepts on the axes is 6. 7. PRIME more creative questions. i) Find the equation of straight line passes through a point (–2, 3) which bisects the line intercepted between the axes. ii) Find the equation of straight line passes through a point (1, 3) which cuts the line intercepted between the axes in the ratio 1:2. iii) Find the equation of straight lien passes through a point (1, 2) which bisects the line intercepted between the axes. iv) Find the equation of straight line passes through a point (–3, 2) which cuts the line intercepted between the axes in the ratio 1:3. v) Find the equation of straight line passes through a point (3, 2) where the area of triangle made by the line between the axes is 12 square units.
148 PRIME Opt. Maths Book - IX 4.6 Equation of straight line in point form: 4.6.1 Equation of straight line in slope-point form: (When a point and the slope of a straight line are given.) Let, a straight line AB makes an angle q with x-axis in positive direction which passes through a point B(x1 , y1 ). X’ X Y P(x, y) B(x1 , y1 ) Y’ A N q q M Q P(x, y) be any point in the line AB. Then, slope of straight line AB(m) = Tanq. Draw the perpendiculars PM^OX, BN^OX and BQ^PM \ \PBQ = \PAX = q. Now, In right angled DPQB, Tanq = run rise or, m = x x y y – – 2 1 2 1 or m = x x y y – – 1 1 or, y – y1 = m(x – x1 ) It is the required equation of straight line AB. Alternative method Let ‘m’ be the slope, A(x1 , y1 ) be a given point on the straight line. Let p(x, y) be any point on the straight line. \ Slope of straight line = slope of AP m = x x y y – – 2 1 2 1 or, m = x x y y – – 1 1 or, y – y1 = m(x – x1 ) Which is the required equation. A(x1 , y1 ) P(x, y)
PRIME Opt. Maths Book - IX 149 4.6.2 Equation of straight line in double point form: Let, AB be a straight line joining the points A(x1 , y1 ) and B(x2 , y2 ) which makes an angle q with x-axis in positive direction. i.e. \BPX = q Slope of line AB(m) = Tanq. Tanq = run rise or, m = x x y y – – 2 1 2 1 Then, Equation of straight line AB is, y = mx + C ........................ (i) It passes through the points A(x1 , y1 ) So, y1 = mx1 + C .......................... (ii) Subtracting equation (i) and (ii) y – y1 = (mx + c) – (mx1 + c) or, y – y1 = mx + C – mx1 – C or, y – y1 = m(x – x1 ) \ y – y1 = x x y y – – 2 1 2 1 (x – x1 ) It is the required equation of straight line. Alternative method: Let P(x, y) be any point on the straight line joining the points A(x1 , y1 ) and B(x2 , y2 ). \ Slope of AP = Slope of AB or, x x y y – – 1 1 = x x y y – – 2 1 2 1 [ a m = x x y y – – 2 1 2 1 ] \ y – y1 = x x y y – – 2 1 2 1 (x – x1 ) Which is the required equation. 1. Find the equation of straight line passes through a point (1, 2) which makes an angle 30° with x-axis in positive direction. Solution : Angle made by straight line, (q) = 30° \ Slope (m) = Tan30° = 3 1 It passes through a point (1, 2). Worked out Examples X’ X B(1, 2) 30° A Y Y’ X’ X Y P(x, y) A(x1 , y1 ) B(x2 , y2 ) Y’ P O M q q N A(x1 , y1 ) B(x2 , y2 ) P(x, y)
150 PRIME Opt. Maths Book - IX Now, Equation of straight line in slope point form y – y1 = m(x – x1 ) or, y – 2 = 3 1 (x – 1) or, x – 1 = 3y – 2 3 ∴ x – 3y + 2 3 – 1 = 0 It is the required equation. 2. Find the equation of straight line passes through the points (3, –2) and (–2, 2). Solution : The given points are A(3, –2) and B(–2, 2) Now, The equation of straight line in two points form is, y – y1 = x x y y – – 2 1 2 1 (x – x1 ) or, y + 2 = 2 3 2 2 – – + (x – 3) or, 5y + 10 = –4x + 12 ∴ 4x + 5y – 2 = 0 is the required equation. 3. Find the equation of median drawn from the �irst vertex of a triangle having vertices A(3, 4), B(–3, 1) and (7, –3). Solution : Let, AD be the median of ∆ABC having vertices A(3, 4), B(–3, 1) and C(7, –3). Now, Using mid-point formula, (x, y) = , x x y y 2 2 1 2 + 1 + 2 a k or, D(x, y) = , 2 3 7 2 – + 1 – 3 a k = (2, –1) Again, Equation of median AD is, y – y1 = x x y y – – 2 1 2 1 (x – x1 ) or, y – 4 = 2 3 1 4 – – – (x – 3) or, y – 4 = 5x – 15 ∴ 5x – y – 11 = 0 is the required equation. D(2, –1) A(3, 4) B(–3, 1) C(7, –3)
PRIME Opt. Maths Book - IX 151 4. If A(a, b)lies in the straight line x + 2y – 7 = 0 and B(b, a)lies in the straight line 3x – y + 2 = 0 �ind the length and equation of AB. Solution, The point A(a, b) lies in the equation of straight line x + 2y – 7 = 0 i.e. a + 2b – 7 = 0 ∴ a = 7 – 2b ........................... (i) The point B(b, a) lies in the equation of straight line 3x – y + 2 = 0 i.e. 3b – a + 2 = 0 or, 3b – (7 – 2b) + 2 = 0 [a From equation (i)] or, 3b – 7 + 2b + 2 = 0 or, 5b = 5 ∴ b = 1 Substituting the value of ‘b’ in equation (i) a = 7 – 2 × 1 = 5 ∴ The points becomes A(5, 1) and B(1, 5) Then, using distance formula, d = (x2 – x1 )² + (y2 – y1 )² ∴ d(AB) = (1 – 5)² + (5 – 1)² = 16 + 16 = 4 2 units. Also equation of straight line AB is, y – y1 = x x y y – – 2 1 2 1 (x – x1 ) or, y – 1 = 1 5 5 1 – – (x – 5) or, y – 1 = – x + 5 ∴ x + y – 6 = 0 is the required equation. 5. In what ratio does the line joining the points (–4, 2) and (6, 8) is divided by the line 3x + 4y – 20 = 0? Solution : Let, the straight line 3x + 4y – 20 = 0 cuts the line joining the points A(–4, 2) and B(6, 8) in the ratio k:1. Here, using section formula, R(x, y) = , m m m x m x m m m y m y 1 2 1 2 2 1 1 2 1 2 2 1 + + + + a k = ( ) , ) k k k k 1 6 1 4 1 × – × 8 1× 2 + + + + a k = , k k k k 1 6 4 1 – 8 2 + + + a k
152 PRIME Opt. Maths Book - IX Exercise 4.6 1. Find the equation of straight line AB under the following conditions. i) Write down the formula to find equation of straight line in single point form. ii) Write down the equation of straight line in double intercepts form. iii) Having slope 2 and passes through a point (1, –2). iv) Which makes an angle 45° with x-axis in negative direction and passes through a point (–3, 2) v) X’ X A O Y Y’ 30° B(4, 5) 2. Find the equation of straight lien passes through the following pairs of points. i) (3, –2) and (5, 1) ii) origin and (3, 5) iii) (a, b) and (a, –b) iv) X’ X O Y Y’ B(4, 6) A(–2, 1) v) X’ X O Y Y’ A(–4, –5) 3. Find the equation of straight the under the following conditions. i) equation of line joining the mid-point of line joining the points A(1, –2) and B(–4, –7) and the origin. ii) Equation of median of a triangle drawn from first vertex having vertices A(3, –2), B(–1, 3) and C(7, 5). It lies in the given straight line PQ, 3x + 4y – 20 = 0 or, 3 k k 1 6 4 – + a k + 4 k k 1 8 2 + + a k= 20 or, 18k – 12 + 32k + 8 = 20k + 20 or, 30 k = 24 or, k = 30 24 5 4 \ k = 5 4 The required ratio is 4:5.
PRIME Opt. Maths Book - IX 153 Answer 1. i) Show to your teacher. ii) Show to your teacher. iii) 2x – y – 4 = 0 iv) x + y + 1 = 0 v) x – 3y + 5 3 = 4 2. i) 3x – 2y – 13 = 0 ii) 5x – 3y = 0 iii) x – a = 0 iv) 5x – 6y + 16 = 0 v) 5x – 4y = 0 3. i) 3x – y = 0 ii) x – 3 = 0 iii) y – 1 = 0 iv) x + 3y = 0 4. i) 2x – y = 0 ii) 1:3 iii) 1:3 iv) 5:14 v) 23:14 5. i) x – 10y = 0 ii) x + y – 8 = 0 and 2 2 units iii) x + y – 5 = 0; 2 units iii) Equation of median of a triangle drawn from second vertex having vertices (3, 5), (1, 1) and (5, –3). iv) Equation of line joining the origin and centroid of a triangle having vertices (1, –2), (3, 5) and (5, –6). v) Prove that the points (3, 4), (7, 7) and (11, 10) are collinear points. 4. i) Find the equation of straight line joining the intersecting point of x + 2y – 5 = 0 and 3x – y – 1 = 0 and the point (–3,–6). ii) In what ratio does the line having equation 11x + 5y = 12 divide the line joining the points (–4, 2) and (10, 8)? iii) In what ratio does the line joining the points (–2, 2) and (8, 6) is divided by the straight line having equation 2x + y – 4 = 0 ? iv) In what ratio does the line joining the points (1, 2) and (9, 8) is divided by the line joining the points (3, 4) and (5, –4) ? v) In what ratio does the line joining the points (2, –6) and (8, 4) divides the line joining the points (4, 5) and (6, –4) ? 5. PRIME more creative questions: i) Find the equation of line joining the origin and a point which cuts the line joining the points (3, –2) and (–4, 3) in the ratio 1:2. ii) If P(a, b) lies int he straight line 3x – y – 2 = 0 and Q(b, a) lies in the line x + 3y = 10 find the length and equation of PQ. iii) If a point A(m, n) lies in the straight line x + 2y – 7 = 0 and B(n, m) lies in 5x – 2y – 4 = 0, find the length and equation of AB. iv) If the pints (a, 0), (m, n) and (0, b) are collinear points, prove that : a m b n + =1. v) Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear points.
154 PRIME Opt. Maths Book - IX 4.7 Reduction of linear equation Ax + By + C = 0 in standard form of the equations. i) Slope-intercept form: The linear equation is, Ax + By + C = 0 or, By = – Ax – C or, y = B – – Ax C or, y = B A ` j – x + – B C ` j Comparing it with y = mx + c, we get, slope (m) = B A– y – intercept (C) = – B C ii) Double intercepts form: The linear equation is, Ax + By + C = 0 or, Ax + By = – C Dividing both sides by ‘–C’ or, C Ax – + C By C C – –– = or, A C x – + B C y – = 1 Comparing it with a x b y + =1 we get, x - intercept (a) = A C – y - intercept (b) = – B C iii) Perpendicular form: The linear equation is, Ax + By + C = 0 or, Ax + By = – C or, Ax + By = – C Dividing both sides by ± (Coeff. of x)² + (Coeff. of y)² = ± ( ) – – A B ( ) 2 2 + = ± A B 2 2 + X’ X B A O Y Y’ Ax + By + C = 0 b = – B C a = A C – X’ X O Y Y’ Ax + By + C = 0 q = Tan–1( B A– ) c = – B C
PRIME Opt. Maths Book - IX 155 or, A B Ax 2 2 ! + + A B By 2 2 ! + = – A B C 2 2 ! + (*) or, – A B Ax 2 2 ! + – A B By 2 2 ! + = + A B C 2 2 ! + [a sign of perpendicular distance A B C 2 2 ! + is always positive] or, A B A – x 2 2 ! + c m + A B B – y 2 2 ! + c m = A B C 2 2 ! + Let us choose the sign (out of ±) in such a way that the right hand side of (*) becomes positive is the length, so it is always positive. (\ p = A B C ± – 2 2 + ) Comparing it with xCosa + ySina = p, we get, Cosa = A B –Ax 2 2 ! + Sina = A B –By 2 2 ! + P = A B C 2 2 ! + Here, P = A B C 2 2 + always should be positive. From Cosa and Sina value of angle made by perpendicular to x-axis should be calculated. 4.7.1 Perpendicular distance between a point (x1, y1) & a line Ax + By + C = 0. Let, Equation of a straight line AB is xCosa + ySina = p where \ROX = a OR = perpendicular distance = P Again, Equation of straight line PQ which is parallel to AB is xCosa + ySina = P’ where, \SOX = a OS = P’(perpendicular distance) It passes through a point C(x1 , y1 ) \ x1 Cosa + y1 Sina = P’ is the equation of PQ Then, Distance between two parallel lines is, CD = OS – OR = P’ – P = x1 cosa + y1 sina – P ......................... (i) X B Q S A D P R O Y C(x1 , y1 )
156 PRIME Opt. Maths Book - IX Again, For the straight line Ax + By + C = 0, or, Ax + By = – C Dividing both sides by ± A B 2 2 + or, A B Ax 2 2 ! + + A B By 2 2 ! + = A B –C 2 2 ! + Comparing it with xCosa + ySina = P Cosa = ± A B A 2 2 + Sina = ± A B B 2 2 + OP = ± A B –C 2 2 + Putting the value of ‘P’ in equation (i), we get, CD = x1 cosa + y1 sina – P = x1 A B A 2 2 + + y1 A B B 2 2 + + A B –C 2 2 + = ± A B Ax By C 2 2 1 1 + + + [\ Choose the sign in such a way that it becomes positive] But distance is always positive Hence, perpendicular distance = A B Ax By C 2 2 1 1 + + + 1. Express the equation 2x – 5y – 10 = 0 in double intercepts form. Solution : The given equation is, 2x – 5y + 10 = 0 or, 2x – 5y = – 10 Dividing both sides by – 10. or, x 10 2 – – y 10 5 – = 10 10 – – or, x y 5 2 1 – + = Comparing it with a x b y + =1, we get, x - intercept (a) = – 5 y - intercept (b) = 2 Worked out Examples
PRIME Opt. Maths Book - IX 157 2. Find the area of a triangle bounded by a straight line of equation 4x + 3y – 12 = 0 with the axes. Solution : The given equation is, 4x + 3y – 12 = 0 or, 4x + 3y = 12 or, x y 12 4 + 3 = 1 or, x y 3 4 + =1 Comparing it with a x b y + =1, we get, x - intercept (a) = 3 = OA y - intercept (b) = 4 = OB Again area of ∆OAB = 2 1 × b × h = 2 1 × OA × OB = 2 1 × 3 × 4 = 6 square units. 3. Find the perpendicular distance from a point (1, –2) to the straight line having equation 3x + 4y – 15 = 0. Solution : Perpendicular distance from a point (1, –2) to the straight line 3x + 4y – 15 = 0 is, P = A B Ax By C 2 2 1 1 + + + = ( ) ( ) 3 4 3 1 4 –2 –15 2 2 + + = 5 3 – 8 – 15 = 5 –20 = –4 = 4 units ∴ Required distance is 4 units. X’ X B A O Y Y’ 4x + 3y – 12 = 0
158 PRIME Opt. Maths Book - IX 4. Find the perpendicular distance between the two parallel lines having equation x + y + 7 2 = 0 and 3x + 3y + 2 2 = 0. Solution : The given two parallel straight lines are : x + y + 7 2 = 0 ............................ (i) and 3x + 3y + 2 2 = 0 ..................... (ii) From equation (i), taking y = 0, or, x + 0 + 7 2 = 0 ∴ x = –7 2 Again, perpendicular distance from a point (–7 2 , 0) to the line (ii) is, P = a b ax by c 2 2 1 1 + + + = 3 –7 2 3× 0 20 2 2 2 + + + = 18 –21 2 + 20 2 = 3 2 –21 2 + 20 2 = 3 2 – 2 = 3 1– = 3 1 units [length is always positive] ∴ The distance between the parallel lines is 3 1 units. 5. Express the equation 3 x – y + 6 = 0 in normal form. Solution : The given equation is, 3 x – y + 6 = 0 or, 3 x – y = – 6 or, – 3 x + y = 6 Dividing both sides by A B 2 2 + = (– 3) (1) 2 2 + = 2 or, – x 2 3 + y 2 = 2 6 or, x 2 3 c– m + y 2 1 = 3 Comparing it with xcosα + ysinα = P,
PRIME Opt. Maths Book - IX 159 We get, Cosa = – 2 3 Sina = 2 1 P = 3 units Then, a = 180° – 30° = 150° \ The required form of the equation is xcos150° + ysin150° = 3 6. Express m x n y + = 1 in perpendicular form. Also prove that P 1 m 1 n 1 2 2 = + 2 . Solution : The given equation of straight line is, m x n y + =1 or, mn nx my 1 + = or, nx + my = mn Dividing both sides by A B 2 2 + = n m 2 2 + = m n 2 2 + Then, or, m n nx 2 2 + + m n my 2 2 + = m n mn 2 2 + or, m n n x 2 2 + a k + m n m y 2 2 + a k = m n mn 2 2 + Comparing it with xCosa + ySina = p Cosa = m n n 2 2 + Sina = m n m 2 2 + P = m n mn 2 2 + Again, P2 = m n m n 2 2 2 2 + or, P 1 2 = m n m n 2 2 2 2 + = m n m m n n 2 2 2 2 2 2 + = n m 1 1 2 2 + \ p m n 1 1 1 2 2 = + 2 is proved.
160 PRIME Opt. Maths Book - IX Exercise 4.7 1. i) Write down the formula to find perpendicular distance from a point to a straight line. ii) Find slope and y - intercept from the equation y = 3x + 3. iii) Find the angle made 10y a straight lien y = x 3 1 – 4 to the x- axis. iv) Find the intercepts made by the straight line 3x + 2y = 1 on the axes. v) Find the angle made by the perpendicular from the origin to the straight line having equation x 2 3 + 1 2 y = 4. 2. Express the following equations in double intercepts form. i) 3x + 4y – 12 = 0 ii) 2x – 5y – 7 = 0 iii) 6y – 8x + 12 = 0 iv) x – 3y + 4 = 0 v) 3x + 2y – 2 3 = 0 3. Express the following equation in slope-intercepts form. Also find the inclination of the line with x-axis. i) 3x – y + 5 = 0 ii) x + 3y – 7 = 0 iii) 3x – 3y + 12 = 0 iv) x – y – 12 = 0 v) 2 3x + 6y – 9 = 0 4. Express the following equations in perpendicular form. i) x + y – 2 = 0 ii) x + 3y + 4 = 0 iii) 3x – y – 2 = 0 iv) – x + 3y + 4 2 = 0 v) 3x + y = 2 2 5. Find the perpendicular distance from a point to the given straight line. i) 3x + 4y + 4 = 0; (1, 2) ii) 4x – 3y + 2 = 0; (3, –2) iii) 6x – 8y + 7 = 0; (–4, 1) iv) 3x + y = 0; ( 3, 1) v) ax + by + c = 0; (0, 0) 6. Find the distance between the two parallel lines for the followings: i) 4x + 3y – 6 = 0 and 8x + 6y – 2 = 0 ii) 3x + y – 4 = 0 and 3 3x + 3y = – 12 iii) 3x – 4y + 9 = 0 and 4y – 3x + 6 = 0 iv) x + y = 2 and 2x + 2y + 6 = 0 v) 3x + 4y – 14 = 0 and 9x + 12y + 18 = 0 7. PRIME more creative questions a. i) Reduce a x b y + =1 in perpendicular form. Also prove that : p a b 1 1 1 2 2 = + 2
PRIME Opt. Maths Book - IX 161 Answer 1. Show to your teacher. 2. i) 60°, 3, 5 ii) – 3 1 , 3 7 , 150° iii) 3 1 , – 3, 30° iv) 1, – 12, 45° 3. Show to your teacher. 4. i) xCos45° + ySin45° = 2 ii) xCos240° + ySin240° = 2 iii) xCos330° + ySin330° = 1 iv) xCos300° + ySin300° = 2 2 v) xCos30° + ySin30° = 2 5. i) 3 units ii) 4 units iii) 2.5 units iv) 2 units v) a b c 2 2 + units 6. i) 1 units ii) 4 units iii) 3 units iv) 4 units v) 4 units 7.a. ii) 6 sq. units, (2, 2 3 ) iii) ±24 b. i) ±6 ii) (–4, –3) and (3, –4) iii) 2 7 sq. units. iv) 3x – y = 0 and 4x – 3y = 0 v) a = 2, b = 9 ii) If a straight line 3x + 4y – 12 = 0 cuts the axes at A and B, find the area of DOAB and mid-point of line AB. iii) If a straight line 4x + 6y – k = 0 cuts the axes at A and B and area of DOAB = 12 square units, find the value of ‘k’. iv) Reduce the equation ax + by = a2 + b2 into normal form and prove that p2 = a2 + b2 . v) Reduce y = mx + c in perpendicular form and prove that p 1 2 = ( ) c m 1 2 1 2 + b. i) If a straight line of equation 3x + 4y – 2m = 0 cuts the axes at A and B and area of DOAB = 6 sq. units, find the value of ‘m’. ii) If P and Q two points are on a straight line having equation x – y + 1 = 0 which are 5 units distant from the origin, find the co-ordinate of P and Q. iii) If A and B two points are on a straight line having equation x + y – 1 = 0 which are 5 units distant from the origin, find the area of DOAB. iv) Find the equation of straight line passes through the origin and points of trisection of the line joining the points (4, 2) and (1, 8) v) If a straight line ax + by = 12 forms an area between the axes is 4 square units and its x-intercept is 6, find the value of ‘a’ and ‘b’. 7. Project work Prepare a report of formula used in co-ordinate in a chart paper and present the activity in your classroom.
162 PRIME Opt. Maths Book - IX 4.8 Area of triangle The space occupied by a triangle on any surface is called area of triangle. The area of triangle can be calculated by using various method in geometry, trigonometry as well as in co-ordinate geometry also. Here we are going to discuss for finding area of triangle using three vertices. 4.8.1 Area of triangle having co-ordinate of the vertices. Here, A(x1 , y1 ), B(x2 , y2 ) and C(x3 , y3 ) are the vertices of a triangle in co-ordinate system. Draw the perpendiculars, AM^OX, BN^OX and CP^OX Then, NM = OM – ON = x1 – x2 MP = OP – OM = x3 – x1 NP = OP – ON = x3 – x2 BN = y2 AM = y1 CP = y3 Now, We have, Area of DABC = Area of trapezium (AMNB + AMPC – BNPC) = 1 2 × NM(BN + AM) + 1 2 × MP(AM + CP) – 1 2 × NP(BN + CP) = 1 2 (x1 , x2 )(y2 + y1 ) + 1 2 (x3 – x1 )(y1 + y3 ) – 1 2 (x3 – x2 )(y2 + y3 ) = 1 2 (x1 y2 + x y1 1 – x y2 2 – x2 y1 + x3 y1 + x y3 3 – x y1 1 – x1 y3 – x3 y2 – x y3 3 + x y2 2 + x2 y3 = 1 2 [(x1 y2 – x2 y1 ) + (x2 y3 – x3 y2 ) + (x3 y1 – x1 y3 )] \ DABC = 1 2 x1 x2 x3 x1 y2 y1 y3 y1 • Arrow ( ) shows the positive sign and multiplication and ( ) shows the negative sign and multiplication. • Similarly as above area of quadrilateral can be derived. Which can be written as below. Area of quadrilateral = 1 2 x1 x2 x3 x4 x1 y1 y2 y3 y4 y1 A B C A(x1 , y1 ) C (x3 , y3 ) B(x2 , y2 ) X’ X O N M P Y Y’
PRIME Opt. Maths Book - IX 163 • Area of triangle and quadrilateral are always positive. • Square units is the unit of area. • If area of the figure of the given points becomes zero, the given points must be collinear. A B C 1. Find the area of a triangle having vertices A(1, –2), B(3, 5) and C(6, –1). Solution : The given vertices of a triangle are (1, –2), B(3, 5) and C(6, –1) Then, Using area of triangle, DABC = 2 1 x1 x2 x3 x1 y1 y2 y3 y1 = 2 1 1 3 6 1 –2 5 –1 –2 = 2 1 [{1 × 5 – (–2) × 3} + {3(–1) – 5 × 6} + {6(–2) – (–1) × 1} ] = 2 1 [(5 + 6) + (–3 – 30) + (–12 + 1)] = 2 1 [11 – 33 – 11] = – 2 33 square units. [Area is always positive] \ Area of DABC = 2 33 square units. 2. Find the area of a quadrilateral having vertices (–2, 4), (–3, –5), (4, –3) and (6, 2). Solution : The given vertices of a quadrilateral as A(–2, 4), B(–3, –5), C(4, –3) and D(6, 2). Now, Using area of a quadrilateral quad. ABCD = 2 1 x1 x2 x3 x4 x1 y1 y2 y3 y4 y1 = 2 1 –2 –3 –4 6 –2 4 –5 –3 2 4 = 2 1 [(10 + 12) + (9 + 20) + (24 + 7)] = 2 1 [22 + 29 + 31] = 2 1 × 82 = 41 square units. \ Area of quadrilateral ABCD is 41 square. Worked out Examples
164 PRIME Opt. Maths Book - IX 3. Prove that the points (1, 2), (4, 6) and (7, 10) are collinear points. Solution : The given points are A(1, 2), B(4, 6) and C(7, 10) Now, using area of a triangle, DABC = 2 1 x1 x2 x3 x1 y1 y2 y3 y1 = 2 1 1 4 7 1 2 6 10 2 = 2 1 [(6 – 8) + (40 – 42) + (14 – 10)] = 2 1 [– 2 – 2 + 4] = 2 1 × 0 = 0 Hence, Area of triangle so formed is zero. It means the points are collinear. 4. If area of a triangle having vertices (–2, 4), (x, –5) and (4, –3) is 6 square units, find the value of ‘x’. Solution, The given vertices of DABC are A(–2, 4), B(x, – 5) and C(4, – 3) Area of DABC = 6 square units, We have, Area of DABC = 2 1 x1 x2 x3 x1 y1 y2 y3 y1 or, 6 = 2 1 –2 x 4 –2 4 –5 –3 4 or, 6 = 2 1 [(10 – 4x) + (– 3x + 20) + (16 – 6)] or, 6 = 2 1 [–7x + 40] or, 12 – 40 = –7x or, 7x = 28 \ x = 4
PRIME Opt. Maths Book - IX 165 5. If P(x, y) be any point in a triangle having vertices A(3, 2), B(4, 7) and C(6, –3), prove that : ABC PBC T T = x y 10 5 2 + – 7 Solution : P(x, y) is any point in a triangle having vertices A(3, 2), B(4, 7) & (6, –3) Then, ABC PBC T T = 1 2 x1 x2 x3 x1 y1 y2 y3 y1 1 2 x1 x2 x3 x4 y1 y2 y3 y4 = 1 2 x 4 6 x y 7 –3 y 1 2 3 4 6 3 2 7 –3 2 = (7x – 4y) + (–12 – 42) + (6y + 3x) (21 – 8) + (–12 – 42) + (12 + 9) = x y 13 54 21 10 2 54 – – + + = ( ) x y 20 2 5 + – 27 [\ Area is always +ve] \ ABC PBC T T = x y 10 5 2 + – 7 proved
166 PRIME Opt. Maths Book - IX Exercise 4.8 1. i) What do you mean by area of triangle? ii) Write down the calculating formula of area of a triangle having three vertices. iii) Write down the formula to find the area of a quadrilateral having co-ordinate of the vertices. iv) In what condition the three or more points will be collinear? v) If area of a triangle formed by any three points becomes zero, What information do you get? 2. Find the area of the triangle for the followings. i) Having vertices A(1, 7), B(–3, 2) and C(5, –1). ii) Having vetices A(–1, 2), B(1, –5) and C(–4, –3). iii) Having vertices P(3, 2), Q(5, –4) and R(7, 0). iv) (a, b), (b, c), (c, a) are the vertices. v) Having vertices (4, –3), (–4, –5) & (–3, 6) 3. Find the area of the quadrilateral having vertices. i) (2, –3), (3, –4), (6, 0) and (5, 3) ii) (–2, –1), (1, 5), (5, 5) and (2, –1) iii) (3, 4), (3, –2), (6, –2) and (6, 4) iv) (–4, 3), (2, 3), (2, –3) and (–4, –3) v) (5, 2), (2, 4), (3, –3) and (5, –4) 4. Prove that the following points are collinear. i) (3, 4), (7, 7) and (11, 10) ii) (1, –2), (4, 2) and (7, 6) iii) (–3, 5), (0, 0) and (3, –5) iv) (a, b + c), (b, c + a) and (c, a + b) v) (k – 3, k – 5), (k, k – 2) and (k + 3, k + 1) 5. i) If area of a triangle having vertices (2, 1), (m, –5) and (6, 3) is 12 square units, find the value of ‘m’. ii) If area of a quadrilateral having vertices A(2, –3), B(3, –4), C(P, 0) and D(5, 3) is 11 square units, find the value of p. iii) If the points A(1, 3), B(k, 5) and C(5, 7) are collinear, find the value of k. iv) If the points (a, 0), (x, y) and (0, b) are collinear, prove that a x b y + =1. v) If the points (P, 0), (0, q) and (r, r) lie in a same straight line prove that p q r 1 1 1 + = . 6. i) Find the area of a triangle formed by a straight line 4x + 3y – 24 = 0 with the axes by finding the co-ordinate of the points on the axes made by given line.
PRIME Opt. Maths Book - IX 167 Answer 1. Show to your teacher. 2. i) 26 sq. units ii) 15.5 sq. units iii) 10 sq. units iv) 2 1 [ab + bc + ca – a2 – b2 – c2 ] v) 43 sq. units 3. i) 11 sq. units ii) 24 sq. units iii) 18 sq. units iv) 36 sq. units v) 15.5 sq units 5. i) 2 ii) 6 iii) 3 6. i) 24 sq. units ii) 2 27 sq. units v) (2, 6) 7. i) –1, 2 1 ii) (7, 2) v) 4:3 ii) If a straight line having equation x – 3y + 9 = 0 cuts the axes at P and Q, find the area of DOPQ by finding the co-ordinate of P and Q. iii) If the point P(x, y) is any point in a triangle having vertices A(6, 3),. B(–3, 5) and C(4, –2), prove that ABC PBC T T = x y 7 + – 2 . iv) If A(1, 5), B(–2, –3) and C(5, –2) are the vertices of a triangle and P, Q & R are the mid-point of sides of DABC, prove that DABC = 4DPQR. v) If P(6, 3), Q(–3, 5) and R(4, –2) are the vertices of a triangle and S(k, 3k) be any point where PQR SQR T T = 7 6 find the co-ordiante of ‘S’. 7. PRIME more creative questions: i) If the points (k, 2 – 2k), (–k + 1, 2k) and (–4 – k, 6 – 2k) are the collinear points, find the value of ‘k’. ii) If A(3, 4) and B(5, –2) are the two vertices of an isosceles triangle PAB where PA = PB. Find the co-ordinate of the vertex p where area of DPAB = 10 square units. iii) If the points (– m c , 0), (0, c) and (x, y) are the collinear points, prove that y = mx+ c. iv) If the points ( cos p a , 0), (x, y) and (0, sin p a ) are the collinear points, prove that xCosa + ySina = p. v) The vertices of DABC are A(1, 7), B(–3, 3) and C(5, –1) where P and Q are the mid-point of sides AB and AC respectively compare the area of DABC and quadrilateral BPQC. 8. Project work: Put your optional mathematics book in a sheet of graph paper and find the coordinate of the vertices of the book. Also find its area using the co-ordinates and compare it with A = l × b.
168 PRIME Opt. Maths Book - IX Co-ordinate Geometry Unit Test - 1 Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. If a straight line makes an angle of 60° in negative direction, find the slope of the straight line. 2. a. Find the Co-ordinate of a point on X-axis which is 5 units distant from the point (–2, 3). b. If the points (a, 4), (7, 7) and (11, 10) are collinear, find the value of ‘a’. c. Find the perpendicular distance between a point (1, 5) and a straight line having equation 4x + 3y + 6 = 0. 3. a. Find the equation of straight line passes through a point (2, – 1) which bisects the line intercepted between the axes. b. Find the equation of locus of a point which moves so that it is equidistant from the points (1, 2) and (3, 1). 4. Prove that x Cosa + Y Sina = P as the equation of straight line. Unit Test - 2 Time : 30 minutes [1 × 1 + 3 × 2 + 2 × 4 + 1 × 5 = 20] Attempt all the questions: 1. Write down the formula to find the perpendicular distance from a point (a, b) to the straight line mx + ny + c = 0 2. a. Find the equation of straight line AB from the given diagram where OC = 4 units, ∠OAC = 30°. b. Find the equation of locus of a point which moves so that its distance from the origin is always 4 units. 3. a. Prove that y = mx + c is the equation of straight line in slope intercept form. b. If A(3, 2), B(–4, 5) and C(–5, –6) are the three vertices of a parallelogram find the co-ordinate of the fourth vertex. 4. If P and Q are any two points lies in a straight line 3x + 4y + 20 = 0 which are 5 units distant from the origin, find the area of DOPQ. B C A 30° 4 O
PRIME Opt. Maths Book - IX 169 Unit 5 Trigonometry Specification Grid Table K(1) U(2) A(4) HA(5) TQ TM Periods No. of Questions 2 3 3 – 8 20 35 Weight 2 6 12 – K = Knowledge, U = Understanding, A = Application, HA = Higher ability, TQ = Total Question, TM = Total Marks Objectives : At the end of the lesson • Students are able to identify and can find the system of measurement of angles and their interrelationship. • Students are able to find the right angled triangle and trigonometric ratios. • Students are able to find the values of trigonometric ratios for standard angles, complementary angles and other form of angles. • Students are able to prove the trigonometric identities. • Studentscan solve the problems involving compound angles and applications. Materials Required: • Chart of interrelation ship between the system of measurement of angles. • Chart paper. • Flash card. • Graph paper. • Chart of finding trigonometric ratios and its application. • Chart of value of the trigonometric ratios for standard angles. • Chart of the concept of quadrant system.
170 PRIME Opt. Maths Book - IX 5.1 Trigonometry The word trigonometry is defined using three words; tri → three, gones → angles, metron → measurement. The word trigonometry is said to originated from the three words of samskrit as below. Tri – lq gono – sf]0f metry – dfqf The measurement of three angles of a triangle is called trigonometry. The trigonometrical ratios are defined by using right angled triangle only. It is useful to find the angles of a right angled triangle and to find the length of the sides of it. It is used by engineers to find the height and distance during construction of structures which is useful to estimate the constructing materials and cost. It is also useful for various purposes like in physics, mathematics, statistics as well as other scientific purposes. 5.1.1 Measurement of angles Here, we discuss the three important systems of measurement viz, sexagesimal (Britrsh System), Centesimal (French System) and Circular system. i. Sexagesinal measurement: (In degree) The right angle is divided into 90 equal parts called degree, whose one part can be divided into 60 equal parts called minute and the one part of it can be divided into again 60 equal parts called Seconds. Thus, One right angle = 90° (degrees) \ 1° = 60’ (minutes) \ 1’ = 60’’ (seconds) ii. Centesimal measurement: (In grade) One right angle is divided into 100 equal parts in this system called grades. Again, each grade is divided into 100 equal parts called minutes and one minute is divided into 100 equal parts called Seconds. Thus, One right angle = 100g (degrees) \ 1g = 100’ (minutes) \ 1’ = 100’’ (seconds)
PRIME Opt. Maths Book - IX 171 iii. Circular measurement: (In radian) We know c = 2πr i.e. 2π times of radius (r) is equal to the length of circumference. So, the angle subtended by circumference at centre is 2πc . Radian angle : The central angle formed by an arc of a circle equal to the radius of the circle is called radian angle (1c ). Here, Radius OA = arc length AB $ \AOB = one radian angle (1c ). Two right angles is denoted by the greek letter π in radian measurement. i.e. 2 right angle = πc . It is called circular measurement. Relationship table between the measurement of angles. 2rt. \ = 180° = 200g = πc 1 rt. \ = 90° = 100g = 2 c π 1° = 9 10 g a k = 180 c π a k 1g = 10 9 c a k = 180 π c ` j 1c = 180 π c ` j = 200 g π ` j 1. Convert 25g 32’16” into seconds Solution: 25g 32’16” = (25 × 100 × 100 + 32 × 100 + 16)” = (250000 + 3200 + 16) = (253216)” 2. Convert 36°15’36” into centesimal measurement. Solution : 36°15’36” = 36 60 15 60 60 36 × + + c a k = (36 + 0.25 + 0.01)° = (36.26)° = 36.26 9 10 × g a k = (40.2888)g Worked out Examples B O A radian angle (1c )
172 PRIME Opt. Maths Book - IX = 40g + (0.2888 × 100)’ = 4g 28’ + (0.88 × 100)” = 40g 28’88” 3. Convert 250 3 c r a k into sexagesimal measurement. Solution: 250 3 c r a k = 250 3 180 × r r a k = (2.16)° = 2° + (0.16 × 60)’ = 2°9.6’ = 2°9’ + (0.6 × 60)” = 2°9’36” 4. If any two angles of a triangle are 70° and 76°, find the third angle in grades. Solution : Let, ABC be a triangle, Where, \A = 70° \B = 76° \C = ? We have, \A + \B + \C = 180° [a Being int. triangles of a angle in degrees.] or, 70° + 76° + \C = 180° or, \C = 180° – 146° \ \C = 54° \ Third angle = (54 × 9 10 )g = 60g 5. One angle of a triangle is 54° and the Second angle is 8 3 c r a k , find the third angle in grades. Solution : Let, ABC be a triangle where, \A = 54° = (54 × 9 10 )g = 60g \B = 8 3 c r a k = 8 3 200 × g r r a k = 75g \C = ? We have, \A + \B + \C = 200g [a Being int. triangles of a angle in grades.] or, 60g + 75g + C = 200g or, C = 200g – 135g \ C = 65g \ The third angle is 65g . 70° B C A 76° ? 54° B C A 8 ? 3 c r a k
PRIME Opt. Maths Book - IX 173 6. Find the angles of a triangle in grades which are in the ratio 2:3:5. Solution : Let, ABC be a triangle, Where, \A = 2xg \B = 3xg \C = 5xg We have, \A + \B + \C = 200g or, 2x + 3x + 5x = 200g or, x = 20g Then, the angles are, \A = 2 × 20g = 40g \B = 3 × 20g = 60g \C = 5 × 20g = 100g 7. The angles of a triangle in degrees, grades and radius respectively are in the ratio 288 : 280 : p. Find the angles in degrees. Solution : Let, PQR be a triangle where the angles taken in degrees, grades and radians respectively are in the ration 288 : 280 : p. i.e. \P = 288x° \Q = 280xg = 280x 10 9 × c a k = 252x° \R = xc r = x 180 r × r c ` j = 180x° Then, we have \P + \Q + \R = 180° or, 288x + 252x + 180x = 180° or, 720x = 180° or, x = 4 1 Then, the angles of the 3PQR are, \P = 288 × 4 1 = 72° \Q = 252 × 4 1 = 63° \R = 180 × 4 1 = 45° 2xg 3xg 5xg B C A 288x° Q R P 280xg pxc
174 PRIME Opt. Maths Book - IX 8. The difference between two acute angles of a right angled triangle is 5 2 c r a k . Find the angles of the triangle in degrees. Solution : Let, 3ABC is the right angled at \A. Where, \A = 90° \B – \C = 5 2 c r a k or, \B – \C = 5 2 180 × r r a k° or, \B – \C = 72° or, \B = \C + 72° Then, We have \A + \B + \C = 180° or, 90° + \C + 72° + \C = 180° or, 2\C = 180° – 162° or, 2\C = 18° ` \C = 9° Then, The angles of 3ABC are, \A = 90° \B = 9° + 72° = 81° \C = 9° 9. Find the angle formed between two hands of a clock at 3:40 o’clock in circular measure. Solution : At 3 : 40 O’ clock Time difference between two hands= 4hrs + 20 minutes = 4 hrs 60 20 a + k = hrs 3 13 a k We have, Hour hand turns in 12 hours by 360° Hour hand turns in 1 hour by 12 360 ° ` Hour hand turns in hrs 3 13 a k by 12 360 × 3 13 = 130° = 130 180 × r c ` j = 18 13 c r a k ` Angle formed between the two hands of a clock at 3:40 O’clock is 18 13 c r a k . A C B
PRIME Opt. Maths Book - IX 175 Exercise 5.1 1. i) What is trigonometry? Write down its importance. ii) What is sexagesimal measurement. What down the relations used in it. iii) What is centesimal measurement? Write down the relation of angles used in it. iv) Write down the value of pc in sexagesimal and centesimal measurement. v) What is the value of sum of the angles of a triangle in sexagesimal and centesimal measurement. 2. Express the following angles in sexagesimal Seconds. i) 25° 26’ 27’’ ii) 36° 25’’ iii) 45’ 36’’ iv) 75° 50’ 55’’ v) 42° 45’ 3. Express the following angles in centesimal Seconds. i) 42g 35’’ 42’’ ii) 75g 65’ 72’’ iii) 36g 38’’ iv) 55g 36’ v) 25’ 32’’ 4. Express the following angles in degrees. i) 22° 36’ 45’’ ii) 72° 42’ 54’’ iii) 50° 27’’ iv) 55° 45’ v) 80g vi) 65g 75’ 42’’ vii) 42g 36’ 45’’ viii) 5 2 c r a k ix) 3 r c ` j x) 45 7 c r a k 5. Convert the following angles in grades. i) 85g 52’ 45’’ ii) 27g 52’ 46’’ iii) 44g 45’’ iv) 47g 27’ 35’’ v) 72° vi) 66° 42’ vii) 56° 24’ 36’’ viii) 10 3 c r a k ix) 40 r c ` j x) 25 2 c r a k 6. Convert the following angles into radian. i) 72° ii) 240° iii) 144° iv) 125g v) 400g vi) 350g 7. Convert the followings into sexagesimal measurement. i) 18g 24’45” ii) 50g 42’75” iii) 375 5 c r a k iv) 24g 85’ v) 250 7 c r a k 8. Convert the following into centesimal measurement. i) 35°24’54” ii) 74°42’30’’ iii) 500 3 c r a k iv) 375 5 c r a k v) 44°27”
176 PRIME Opt. Maths Book - IX 9. Answer the following problems: i) Find the ratio of the angles 48° and 80g . ii) Find the sum of the angles 5 2 c r a k and 28° in degrees. iii) Find the difference of the angles 60g and 10 3 c r a k in degree measurement. iv) Find any two angles whose sum is 40° and difference is 20°. v) Find any two angles in degrees whose sum is 50° and difference is 14°. 10. Find the angles of the triangle from the followings. i) Find the angles of a triangle which are in the ratio 3:4:5 in degree measurement. ii) If two angles of a triangle are in the ratio 3:5 and the third angle is 60°, find the angles in degrees. iii) One angle of a triangle is 5 3 of a right angle and the second angle is 3 2 of a right angle, find the third angle in degrees. iv) If two angles of a triangle are 40° and 68° respectively, find the third angle in grades. v) Find the angles of the right angled isosceles triangle in grades. 11. Solve the problems given below. i) One angle of a triangle is 60g , Second angle is 36°. Find the third angle in degrees. ii) Two angles of a triangle are in the ratio 2:7 and the third angle is 30g , find the angles in degrees. iii) One angle of a triangle is 20 7 c r a k , Second angle is 81°, find the third angle in grades. iv) One angle of a triangle is 3 2 of a right angle and the Second angle is 5 3 c r a k find the third angle in degrees. v) The sum of any two angles of a triangle is 100g and their difference is 10 r c ` j . Find the angles of the triangle in degrees. 12. PRIME more creative questions: a. i) The angles of a triangle in degrees, grades and radius are respectively , x x 5 3 3 2 g c a k a k and x 75 r c ` j . Find the angles in degrees. ii) The first angle of a triangle in degrees, second angle in grades and third angle in radian are in the ratio 144:140: 2 r respectively. Find the angle of the triangle in degrees. iii) One angle of a triangle is 3 2 of 2 right angles where the other angle exceeds the third angle by 20g . Find the angles in degrees. iv) One angle of a triangle is greater than the second angle by 5 r c ` j and the third angle is less than the first angle by 27°. Find the angles in degrees. v) The angles of a triangle in degrees, grades and radians respectively are in the ratio 72:120:p, find the angles in grades. b. i) The first angle of a triangle in sexagesimal measurement, second angle in
PRIME Opt. Maths Book - IX 177 centesimal measurement and the third angle in circular measurement are respectively x 5 2a k, x and x 3 2 60 r . Find the angles in degrees. ii) The difference between two acute angles of a right angled triangle is 10 3 c r a k , find the angles of the triangle in centesimal measurement. iii) The sum of any two angles of a triangle in circular measure is 9 5 c r a k and their difference is 9 200 g a k . Find the angles of the triangle in degrees. iv) The angles of a triangle are (a – d)°, a° and (a + d)° where the greatest angle is twice the least. Find the angle of the triangle in degrees. v) The angles of a quadrilateral are in the ratio 3:4:5:6. Find the angles indegres. c. i) If D be the number of degrees and G be the number of grades of same angle, prove that G = D + 9 1 D. ii) If x be the number of sexagesimal seconds and y be the number of centesimal seconds of same angle, prove that x:y = 81:250. iii) If two angles of a triangle are 37°15’36’’ and 68°24’45’’, find the third angle in sexagesimal measure. iv) If P and Q denotes the number of sexagesimal and centesimal minutes of same angle, prove that P 50 = Q 27 . v) The difference of the acute angles of a right angled triangle in centesimal measure is 63g88’33’’. Find the angle in sexagesimal measurement. Answer 1. Show to your teacher. 2. i. 91587” ii. 129625” iii. 2736” iv. 273055” v. 153900” 3. i. 423542” ii. 756572” iii. 360038” iv. 553600” v. 2532” 4. i. 22.6125° ii. 72.715° iii. 50.00075° iv. 55.0125° v. 72° vi. 59.15878° vii. 38.12805° viii. 72° ix. 60° x. 28° 5. i. 85.5245g ii. 27.5246g iii. 44.0045g iv. 47.2735g v. 80g vi. 74.1111g vii. 62.6778g viii. 60g ix. 5g x. 16g 6. i. 5 2 c π a k ii. 3 4 c π a k iii. 5 4 c π a k iv. 8 3 c π a k v. (2p)c vi. 4 7 c π ` j 7. i. 16° 25’ 12” ii. 45° 23’ 5” iii 2° 24” iv. 22° 21’ 54” v. 5° 2’ 2.4” 8. i. 39g 35’ ii. 83g 9’ 25.5” iii. 1g 20’ iv. 2g 66’ 67” v. 48g 89’ 72” 9. i. 2 : 3 ii. 100° ii. 0 iv. 10° , 30° v. 32° , 18° 10. i. 45°, 60°, 75° ii. 45°, 75° iii. 66° iv. 80g v. 100g , 50g , 50g 11. i. 90° ii. 34°, 119° iii. 40g iv. 12° v. 54°, 36°, 90° 12.a. i. 30°, 30°, 120° ii. 72°, 63°, 45° iii. 120°, 39°, 21° iv. 81°, 45°, 54° v. 36°, 54°, 90° b. i. 18°, 27°, 135° ii. 100g , 80g , 20g iii. 40°, 60°, 80° iv. 40°, 60°, 80° v. 60°, 80°, 100°, 120° c. iii. 74° 13’ 39” v. 73°, 44’ 51”, 90° and 16°15’9”
178 PRIME Opt. Maths Book - IX 5.2 Central angle of a circle : The angle formed by an arc at the centre of a circle is called central angle of the circle. We measure the angle either in radian measure or convert to other system. In previous unit, we define 1c (radiant) angle on a circle by an arc equal to the length of the radius. Here, we discuss the angle substended by an arc of any length. Let us consider O is the centre of a circle, A O B q OA is radius of the circle, AB is an arc Then \AOB = q is called the central angle of the circle by the arc AB $. Relation between radius of a circle, arc length and central angle of the circle. Here, O is the centre of a circle, OA is radius (r), AB $ is an arc length (l), \AOB is a central angle (q). Draw an arc AC $ which is equal to radius OA of the circle. i.e. AC $ = radius OA = r \ \AOC = 1c Then, Taking the ratio of central angle w.r.t. their corresponding arcs, AOC AOB AC AB \ \ = $ $ or, r l 1c i = or, r × q = l × 1c or, q = r l × 1c \ q = r l c a k It is the formula to find the central angle of a circle in circular measurement. A O C r l 1c q B r
PRIME Opt. Maths Book - IX 179 Polygon: The geometrical closed figures made with straight lines are called polygons where the straight lines so used are called sides of the polygon and the angles are called interior angles of the polygon. Triangle Quadrilateral Pentagon Hexagon ......... etc The polygons are started from the three sides polygon called triangle and so on Polygons are named according to the number of sides as, 3 sides = Triangle 4 sides = Quadrilateral 5 sides = Pentagon 6 sides = Hexagon 7 sides = Heptagon (Septagon) 8 sides = Octagon 9 sides = Nonagon 10 sides = Decagon 11 sides = Undecagon 12 sides = Dodecagon 13 sides = Tridecagon 14 sides = Tetradecagon 15 sides = Quindecagon The polygon having all the sides equal (also all the interior angles are equal) are called regular polygons. • Equilateral triangle, square are called regular polygons. • Measurement of the sum of the interior angles of a polygon. 3 sides polygon = 180° = (3 – 2) × 180° 4 sides polygon = 360° = (4 – 2) × 180° 5 sides polygon = (5 – 2) × 180° 6 sides polygon = (6 – 2) × 180° Similarly, Sum of the interior angles of ‘n’ sides polygon = (n – 2) × 180° • One interior angle of a regular polygon, q = × ° n n 2 180 – • Sum of the exterior angles of a polygon is 360° • One exterior angle of a regular polygon is n 360° .
180 PRIME Opt. Maths Book - IX 1. Find the central angle of a circle of radius 70 mm with arc of 8.8cm in degrees. Solution: Arc length (l) = 8.8cm Radius of circle (r) 70 mm = 10 70 cm = 7cm. Central angle (q) = ? We have, Central angle (q) = r l c a k = . 7 8 8 180 × ° r a k = . 7 8 8 22 180 7 × × a k ° = 72° \ Central angle of a circle is 72°. 2. A pendulum of length 8.4cm swings making an angle of 60° at the centre, what distance is the bob traveled? Solution: Radius (r) = Length of the pendulum = 8.4cm Central angle (q) = 60° = 60 180 × r c ` j = 7 180 60 22 × × c a k Arc length (l) = Distance travelled by the bob = ? We have, q = r l or, l = r × q = 8.4 × 7 180 60 22 × × 3 = 8.8 cm. \ Distance travelled by the bob is 8.8cm. 3. A goat is tethered on a stake for grazing where the goat travelled 27.5m distance by making the rope tight and the rope makes an angle of 70° at the stake, find the length of the rope. Solution: Central angle (q) = 70° = 70 180 × r c ` j = 7 180 70 22 × × c a k Worked out Examples A 70 mm 8.8cm O B q=? 8.4cm ? 60° Stake ? 70°
PRIME Opt. Maths Book - IX 181 And Length (l) = Distance diseribed by goat = 27.5m Radius (r) = Length of the rope = ? We have, q = r l c a k or, r = l i = . 7 180 70 22 27 5 × × = . 70 22 27 5 7 180 × × × = 22.5 m. \ Length of the rope is 22.5 m. 4. The moon subtends an angle of 15.75’ at a point on the earth surface where distance between the earth and the moon is 384000 km, find the diameter of the moon. Solution: 15.75’ 384000km Earth moon Radius (r) = Distance between the earth and the moon. = 384000 km central angle (q) = 15.75’ = . 60 15 75 c a k = . 60 15 75 180 × c r a k = . 60 180 7 15 75 22 × × × c a k Arc length (l) = Diameter of the moon = ? We have, q = r l or, l = r × q = 384000 × . 60 180 7 15 75 22 × × × = 1760 km \ Diameter of the moon is 1760 km. diameter of the moon
182 PRIME Opt. Maths Book - IX 28min 15cm 5. How long does the tip of minute hand of a clock of length 15cm moves in 28 minutes. Solution: Radius (r) = length of the minute hand of a clock. = 15cm. Time taken = 28 minutes. Here, In 60 minutes, minute hand makes 360° angle In 1 minute, minute hand makes 60 360° angle. In 28 minutes, minute hand makes = 60 360° 6 × 28 angle = 168 180 × r c ` j \ Central angle (q) = 180 7 168 22 × × c a k Arc length (l) = Distance travelled by minute hand = ? We have, q = r l or, l = r × q = 15 × 180 7 168 22 × × 12 24 2 = 44cm \ Required distance = 44 cm . 6. Find the interior angle of a regular decagon in centesimal and circular measurement. Solution: No. of sides of regular decagon (n) = 10 Interior angle (q) = ? We have, Interior angle (q) = n n 2– × 180° = 10 10 – 2 × 180° = 144° = 144 9 10 × g a k = 160g = 160 180 × c r a k = 9 8 e r a k ` Interior angle of decagon is 144° or 160g or 9 8 c r a k .
PRIME Opt. Maths Book - IX 183 7. How many sides are there in a regular polygon whose interior angle is 5 3 c r a k ? Solution: Interior angle of a regular polygon, θ = 5 3 c r a k = 5 3 180 × ° r a k = 108° No. of sides (n) = ? We have, Interior angle (θ) = n n – 2 × 180 or, 108 = n n – 2 × 180° or, 180 108 5 3 = n n – 2 or, 5n – 10 = 3n or, 2n = 10 ∴ n= 5 ∴ No. of sides of regular polygon is 5. 8. The number of sides of regular polygons are in the ratio 3:2 where difference of their interior angles is 12°. Find the number of sides of the polygons. Solution: Let, No. of sides of 1st regular polygon (n1 ) = 3x No. of sides of 2nd regular polygon (n2 ) = 2x Difference of their interior angles = 12° Then, Interior angle of 1st – Interior angle of 2nd = 12° or, n n – 2 1 1 × 180 – n n – 2 2 2 × 180° = 12° or, x x 3 3 – 2 × 180° – x x 2 2 – 2 × 180° = 12° or, x x x 180 6 30 6 –4 – 6 + 6 ; E = 12° or, 60 = 12x or, x = 5 ∴ No. of sides of 1st regular polygon = 3 × 5 = 15 No. of sides of 2nd regular polygon = 2 × 5 = 10.
184 PRIME Opt. Maths Book - IX Exercise 5.2 1. i. What is radian angle? Show with diagram. ii. Write down the formula to find the central angle of a circle with usual meaning. iii. Write down the formula to find interior and exterior angles of a regular polygon. iv. What do you mean by regular polygon? v. Prove that the relation q = r l c a k 2. i. Find the central angle of a circle having radius 21cm and length of the arc is 44 cm and convert in sexagesimal measurement. ii. Find the central angle of a circle in centesimal measurement formed in an arc of length 6.6cm of a circle of radius 4.2cm. iii. Find the central angle of a circle in degrees formed in an arc of length 0.44m where length of the radius is 84cm. iv. An arc of length 22cm subtends 120° angle at the centre of a circle. Find the length of the radius. v. Find the central angle in centesimal measurement formed in an arc of length 176mm of a circle of radius 16.8cm. 3. i. An arc of length 3m 8cm is described with an angle of 20° at the centre of a circle. Find the radius of the circle. ii. Find the length of the side OP from the adjoining diagram. iii. An angle of 3 400 g a k is formed at the centre of a circle of radius 21cm, find the arc length for the angle so formed. iv. Arc length is 7 11 th part of the radius of a circle. What angle in grades formed by that arc at the centre? v. What distance travelled by the tip of 42cm long minute hand of a clock in 20 minutes? 4. i. Minute hand of a clock of length 10.5cm moves in 40 minutes. What distance travelled by it in that time. ii. A pendulum described an angle of 40° by swinging a distance of 66cm. Find the length of the pendulum iii. A minute hand of a clock moves a distance of 44cm in 30 minutes. Find the length of the minute hand. iv. A cow is tethered to a stake for grazing a field. The cow grazed by making the rope tight and described 56° angle while moving 8.8m. Find the length of the rope. v. A train is travelling in a circular track of radius 7km with a speed of 66 km per hour. Find the angle substandard at the centre by moving the train in 7 minutes. 33cm Q O P 6 5 c π a k
PRIME Opt. Maths Book - IX 185 5. Find the interior angle in degrees and grades of the following regular polygons. i. Pentagon ii. Hexagon iii. Decagon iv. Dodecagon v. Quindecagon 6. Find the number of sides of the regular polygon whose interior angles are given below. i. 108° ii. 160g iii. 3 2 c r a k iv. 156° v. 3 500 g a k 7. i. The exterior angle of a regular polygon is equal to 4 1 of the interior angle of a regular hexagon. Find the number of sides of the regular polygon. ii. The difference between interior angle and exterior angle of a regular polygon is 90° find the number of sides of the polygon. iii. The difference between interior angle of a regular polygon in centesimal measurement and sexagesimal measurement is 15. Find the number of sides of the polygon. iv. One regular polygon has twice as many sides as another and ratio of their interior angles is 5:4, find the number of sides of the regular polygons. v. The difference between the interior angles of two regular polygons having their number of sides in the ratio 4:3 is 15. Find the number of sides of the polygons. 8. PRIME more creative questions: i. The angle subtended by the sun at a point on the earth surface which are at a distance of 150 million km is 31.85’, find the diameter of the sun. ii. The radius of the earth is 6400km and distance between the earth and the moon is 60 times the radius of the earth. If the moon subtends an angle of 31.5’ at a point on the earth surface. Find the radius of the moon. iii. The number of sides of two regular polygons are in the ratio 2:3 and the sum of the interior angles of them is 260°, find the number of sides of the polygons. iv. The number of sides of two regular polygons are in the ratio 5:4 and difference of their interior angles is 10g , find the number of sides of the polygons. v. If ‘a’ is the radius of the circle and ‘b’ is the length of an arc of a circle. Prove that the central angle formed in that arc is a b c a k . vi. An athlete is running around a circular track of radius 90m and makes an angle of 56° at the centre of the track in 11 seconds. Find the speed of him at that time in km/h.
186 PRIME Opt. Maths Book - IX Answer 1. Show to your teacher. 2. i. 120° ii. 100g iii. 30° iv. 10.5 cm v. 66g 66’ 67” 3. i. 8.82m ii. 12.6cm iii. 44cm iv. 100g v. 88cm 4. i. 44cm ii. 94.5cm iii. 14cm iv. 9m v. 63° 5. i. 108°, 120g ii. 120°, 3 400 g a k iii. 144°, 160g iv. 150°, 3 500 g a k v. 156°, 3 520 g a k 6. i. 5 ii. 10 iii. 6 iv. 15 v. 12 7. i. 12 ii. 8 iii. 8 iv. 12, 6 v. 8, 6 8. i. 1.39 × 106 km ii. 1760km iii. 6, 9 iv. 10, 8 vi. 28.8km/hr
PRIME Opt. Maths Book - IX 187 5.3 Trigonometrical ratios The triangle having an angle of 90° is called right angled triangle, rest two angles are acute angles. The longest side of right angled triangle is called hypotenuses (opposite to right angle) and other two sides are taken as perpendicular and base. The acte angle of a right angled triangle considered for calculation is called the angle of reference. The side opposite to angle of reference is called perpendicular, the side opposite to right angle is hypotenuse and rest side is called the base. Here, In a right angled triangle PQR, \Q = 90° (right angle) \ R = q = reference angle PR = hypotenuses (h) = opposite to right angle. PQ = perpendicular (p) = opposite to reference angle. QR = base (b) = side which joins right angle and angle of reference. Also, h2 = p2 + b2 for the solution i.e. PR2 = PQ2 + QR2 (Pythagoreans theorem) Ratio of any two sides of a right angled triangle are called trigonometical ratios. Ratio of perpendicular and hypoteneous is taken as Sine. i.e Sinq = h p = PR PQ Ratio of base and hypotenuses is taken as Cosine. i.e. Cosq = h b = PR QR Ratio of perpendicular and base is taken as Tangent. i.e. Tanq = b p = QR PQ Receprocal ratio of Cosine is Secant i.e. Secq = b h = QR PR Receprocal ratio of Sine is Co–Secant i.e. Cosecq = p h = PQ PR Receprocal ratio of Tangent is Co-tangent i.e. Cotq = p b = Q PQ R Trick : Some Person Has Curly Black Hair To Produce Beauty. P Q R q Symbols used in trigonometry are: Theta – q, Gamma – g, Delta – d, Alpha – a, Pi – p, Sai – y, Beta – b, Kappa – k Phai – f
188 PRIME Opt. Maths Book - IX 5.3.1 Algebraic operation of trigonometrical ratios: Here are the discussion of addition, subtraction, multiplication, division and factorisation as the operations used in algebra. Look at the examples given below. • As a + a = 2a, Sinq + Sinq = 2Sinq • As 3a – a = 2a 3Sinq – Sinq = 2Sinq • As 2a × 3a = 6a2 2Sinq × 3Sinq = 6Sin2 q • As 6a2 ÷ 2a = 3a 6 Sin2 q ÷ 2Sinq = 3Sinq • As a2 – b2 = (a + b)(a – b) Sin2 q – Cos2 q = (Sinq + Cosq) (Sinq – Cosq) • As 4a2 + 2a = 2a(2a + 1) 4Sin2 q + 2Sinq = 2Sinq(2Sinq + 1) • As a2 + 2a + 1 = (a + 1)2 Sin2 q + 2Sinq + 1 = (Sinq + 1)2 • As 3a2 + 4ab – 4b2 = 3a2 + (6 – 2)ab – 4b2 = 3a2 + 6ab – 2ab – 4b2 = 3a(a + 2b) – 2b(a + 2b) = (a + 2b) (3a – 2b) 6Sin2 q – 5SinqCosq – 4Cos2 q = 6Sin2 q – (8 – 3)Sinq.Cosq – 4Cos2 q = 6Sin2 q – 8SinqCosq + 3SinqCosq – 4Cos2 q = 2Sindq(3Sinq – 4Cosq) + Cosq(Sinq – 4Cosq) = (3Sinq – 4Cosq) (2Sinq + Cosq) Some of the important NOTES. • Sinq ≠ Sina • SinA . SinB ≠ SinAB ≠ Sin2 AB • Sin(A + B) ≠ SinA + SinB • (SinA)2 ≠ SinA2 but (SinA)2 = Sin2 A
PRIME Opt. Maths Book - IX 189 1. Find the product of (2Sinθ + Cosθ)(3Sinθ – 2Cosθ) Solution : (2Sinθ + Cosθ)(3Sinθ – 2Cosθ) = 2Sinθ(3Sinθ – 2Cosθ) + Cosθ(3Sinθ – 2Cosθ) = 6Sin2 θ – 4SinθCosθ + 3SinθCosθ – 2Cos2 θ = 6Sin2 θ – Sinθ.Cosθ – 2Cos2 θ. 2. Simplify : Sin Sin 1 – θ θ – Cos Cos 1 – θ θ Solution : Sin Sin 1 – θ θ – Cos Cos 1 – θ θ θ θ θ θ = Sinθ – Cosθ (1 – Sinθ)(1 – Cosθ) 3. Factorise : Tan4 α – Sin4 α Solution : Tan4 α – Sin4 α = (Tan2 α)2 – (Sin2 α)2 = (Tan2 α + Sin2 α)(Tan2 α – Sin2 α) = (Tan2 α + Sin2 α) (Tanα + Sinα) (Tanα – Sinα) 4. Factorise : 8Cos2 θ – 14Cosθ + 3 Solution : 8Cos2 θ – 14 Cosθ + 3 = 8Cos2 θ – (12 + 2)Cosθ + 3 = 8 Cos2 θ – 12Cosθ – 2Cosθ + 3 = 4 Cosθ(2Cosθ – 3) – 1(2Cosθ – 3) = (2Cosθ – 3) (4Cosθ – 1) 5. Simplify : (SinA – CosA)2 – (CosA + SinA)2 Solution : (SinA – CosA)2 – (CosA + SinA)2 = (Sin2 A – 2SinA.CosA + Cos2 A) – (Cos2 A + 2CosA.SinA + Sin2 A) = Sin A 2 – 2SinA.CosA + Cos A 2 – Cos A 2 – 2SinA.CosA – Sin A 2 = – 4SinA.CosA Worked out Examples
190 PRIME Opt. Maths Book - IX Exercise 5.3 1. i) What is angle of reference? Show in right angled triangle. ii) Write down any five symbols used in trigonometry with their name according to Latin language. iii) Write down the basic trigonometric ratios taking the reference angle ‘x’ in a right angled DPQR. iv) Write down the reciprocal ratios taking the reference angle ‘q’ in a right angled DKLM. v) Write down perpendicular, base & hypotenuses of a right angled DPQR taking R = 90° and P as reference angle. 2. Multiply the followings. i) (2Sinq + 3) (3Sinq – 2) ii) (Sinq + Cosq) (Sin2 q – Sinq.Cosq + Cos2 q) iii) (Sec2 q + Tan2 q) (Sec2 q – Tan2 q) iv) (Sin2 q + Cos2 q)(Sinq – Cosq) (Sinq + Cosq) v) (2Sinq + 3) (2Sinq – 3) (4Sin2 q + 9) 3. Simplify the followings. i) (Sinq + Cosq)2 – (Sinq – Cosq)2 ii) (Tanq + Secq)2 – (Secq – Tanq)2 iii) Sinq 1 + Sinq + Sinq 1 – Sinq iv) Cosq Cosq – Sinq + Sinq Sinq – Cosq v) Sin3 q Sinq – Cosq – Cos3 q Cosq – Sinq 4. Factorise the followings. i) Sinq + 2Sinq.Cosq + 2Cosq + 4Cos2 q ii) 2 – 2Sinq – 3Cosq + 3Sinq.Cosq iii) Sin2 q – 4Cos2 q iv) 8Cos3 q – Sin3 q v) Sin4 q – Cos4 q 5. Find the factors of the followings. i) Sec8 q – Cosec8 q ii) 2Sin2 q – 7Sinq + 6 iii) 4Cos2 q – 5Cosq – 6 iii) 3Sin2 q + 5Sinq.Cosq – 12Cos2 q v) Cos2 A – 2CosA – 15
PRIME Opt. Maths Book - IX 191 Answer 1. Show to your teacher. 2. i) 6Sin2 q + 5Sinq – 6 ii) Sin3 q + Cos3 q iii) Sec4 q – Tan4 q iv) Sin8 q – Cos8 q v) 16Sin4 q – 81 3. i) 4(Sin2 q + Cos2 q) ii) 4Secq.Tanq iii) Sin Sin 1 2 – q q iv) 1 v) Sin2 q + Sinq.Cosq + Cos2 q 4. i) (1 + 2Cosq) (Sinq + 2Cosq) ii) (1 – Sinq) (2 – 3Cosq) iii) (Sinq + 2Cosq) (Sinq – 2Cosq) iv) (2Cosq – Sinq) (4Cos2 q + 2Cosq.Sinq + Sin2 q) v) (Sinq + Cosq) (Sinq – Cosq) (Sin2 q + Cos2 q) 5. i) (Secq + Cosecq) (Secq – Cosecq) (Sec2 q + Cosec2 q) (Sec4 q + Cosec4 q) ii) (Sinq – 2) (2Sinq – 3) iii) (Cosq – 2) (4Cosq + 3) iv) (Sinq + 3Cosq) (3Sinq – 4Cosq) v) (CosA – 5) (CosA + 3) 6. i) (Sinq + Cosq)3 ii) 1 Sin 2 + θ iii) (Sin2 q + Sinq.Cosq + Cos2 q) (Sin2 q – Sinq.Cosq + Cos2 q) iv) (Cos2 q + Cosq + 1) (Cos2 q – Cosq + 1) v) 2Cosq(3Sin2 q + Cos2 q) 6. PRIME more creative questions: i) Find the product of (Sinq + Cosq) and (Sin2 q + 2Sinq.Cosq + Cos2 q) ii) Simplify : 1 Sin 1 + i + 1 Sin 1 – i + Sin Sin 1 2 – 2 i i iii) Factories : Sin4 q + Sin2 q.Cos2 q + Cos4 q iv) Find the factors of : Cos4 q + Cos2 q + 1 v) Simplify : (Sinq + Cosq)3 – (Sinq – Cosq)3
192 PRIME Opt. Maths Book - IX 1. Receprocal relations: i) SinA.CosecA = h p × p h = 1 \ SinA = CosecA 1 & CosecA = SinA 1 ii) CosA.SecA = h b × b h = 1 \ CosA = SecA 1 & SecA = CosA 1 iii) TanA.CotA = 1 \ TanA = CotA 1 & CotA = TanA 1 2. Quotient relations : i) TanA = b p = h b h p = CosA SinA ii) CotA = p b = h p h b = SinA CosA iii) SinA = h p = b h b p = SecA TanA iv) CosA = h b = p h p b = Co secA CotA v) CosecA = p h = b p b h = TanA SecA v) SecA = b h = p b p h = sec CotA Co A 5.4 Trigonometrical Identities Relation between the trigonometrical ratios. In a right angled DABC. \B = 90° \A = reference angle B A C SinA.CosecA = 1 CosecA.SecA = 1 TanA.CotA = 1 TanA = SinA CosA CotA = CosA SinA
PRIME Opt. Maths Book - IX 193 3. Pythagorean relations : i) Sin2 A + Cos2 A = h p h b 2 2 ` j + a k = h p b2 2 2 + = h h 2 2 = 1 ii) Sec2 A – Tan2 A = – b h b 2 p 2 a k ` j = b h p – 2 2 2 = b b 2 2 = 1 iii) Cosec2 A – Cot2 A = – p h p b 2 2 a a k k = p h b – 2 2 2 = p p 2 2 = 1 Note : Students are requested to memorise (remember the all above relations up to their tip of the tongue.) Sin2 A + Cos2 A = 1 Sin2 A = 1 – Cos2 A Cos2 A = 1 – Sin2 A SinA = 1 – Cos A2 CosA = 1 – Sin A2 Sec2 A – Tan2 A = 1 Sec2 A = 1 + Tan2 A Tan2 A = Sec2 A – 1 SecA = 1 Tan A2 + TanA = Sec A – 1 2 Cosec2 A – Cot2 A = 1 Cosec2 A = 1 + Cot2 A Cot2 A = Cosec2 A – 1 CosecA = 1 Cot A2 + CotA = Co sec A 1– 2
2 2 Co TanA n A n A c A 2 2 + 194 PRIME Opt. Maths Book - IX Worked out Examples 1. Prove that the followings: i) Sin2 A – Sin2 ACos2 A = Sin4 A L.H.S. = Sin2 A – Sin2 A.Cos2 A = Sin2 A(1 – Cos2 A) = Sin2 A.Sin2 A = Sin4 A = R.H.S. proved ii) Co sec A – 1 2 . 1 – Cos A 2 = CosA L.H.S. = Co sec A – 1 2 . 1 – Cos A 2 = CotA.SinA = . SinA CosA SinA = CosA = R.H.S. proved iii. Se Co sec A = TanA + CotA L.H.S.= Sec A Co sec A 2 2 + = 1 Tan A 1 Cot A 2 2 + + + = Tan A 2 Cot A 2 2 + + = Ta 2×1 Cot A 2 2 + + = Ta 2TanA.CotA Cot A 2 2 + + = + CotA = TanA + CotA = R.H.S. proved iv. Tan2 A – Sin2 A = Sin2 A.Tan2 A L.H.S.= Tan2 A – Sin2 A = Cos A Sin A 2 2 – Sin2 A = – . Cos A Sin A Sin A s A 2 2 2 2 = ( – ) Cos A Sin A 1 Cos A 2 = Cos A Sin A 2 2 . Sin2 A = Tan2 A.Sin2 A = R.H.S. Proved
θ θ θ θ PRIME Opt. Maths Book - IX 195 v. Sin 1– Cos 4 4 i i = 1 + 2Cot2 θ L.H.S.= Sin 1 – Cos 4 4 i i = Sin 1 4 i – Sin Cos 4 4 i i = Cosec4 θ – Cot4 θ = (Cosec2 θ)2 – (Cot2 θ)2 = (Cosec2 θ – Cot2 θ) (Cosec2 θ + Cot2 θ) = 1 (1 + Cot2 θ + Cot2 θ) = 1 + 2Cot2 θ = R.H.S. proved vi. Cos Cos 1 1– i i + = Cosecθ – Cotθ L.H.S. = Cos Cos 1 1 – i i + = × Cos Cos Cos Cos 1 1 1 – 1 – – + = ( ) Cos Cos 1 1 – – 2 2 i i = ( ) Sin 1 – Cos 2 2 i i = Sin 1 – Cos i i = – Sin Sin 1 Cos i i i = Cosecθ – Cotθ = R.H.S. proved 2. If a = xCosα + ySinα and b = xSinα – yCosα, prove that a2 + b2 = x2 + y2 . Solution : L.H.S. = a2 + b2 = (xCosα + ySinα)2 + (xSinα – yCosα)2 = x2 Cos2 α + 2xy sina. cosa + y2 Sin2 α + x2 Sin2 α – 2xy sina. cosa + x2 Sin2 α = x2 (Sin2 α + Cos2 α) + y2 (Sin2 α + Cos2 α) = x2 × 1 + y2 × 1 = x2 + y2 = R.H.S. proved