Part (1) Second : Trigonometry (1) Complete the following table : The angle Ratio 42 12 \ Sin 0.3214 Cas 0.5321 Tan 2.0625 (2) Complete the following : 1) 46 36 \ 24 \\ =. In degrees. 2) 44.125 = in degrees, minutes, seconds. 3) If tan = 1.42 where is the measure of an acute angle. Then =... 4) If Sin = 0.63 where is the measure of an acute angle, then =.. 5) If Sin x = where x is an acute angle then m ( x) =.. 6) If cos = where x is an acute angle then m ( x ) =. 7) Sin 60 + Cos 30 - Tan 60 =. 8) Cos 60 + Sin 30 - Tan 45 =... 9) 2 sin 30 x Cos 60 - Tan 45 =... 10) Sin 2 30 + Cos 2 30 =. 11) If tan (x + 10) = where x is an acute angle then m ( x) =. 12) If tan 3x = where x is an acute angle then m ( x) =.. 1
(3) In the opposite figure:- ABC is a triangle,, AC = 12 cm, BC = 16 cm and m ( C) = 30 Complete the following sin 30 = AD =.. x Sin 30 =. cm The area of ABC = x AD x BC The area of ABC = x x = cm 2 Can you calculate the height of the triangle which is drawn from the point B on? Explain your answer showing the steps of solution (4) Choose the correct answer form those given:- 1) 4 cos 30 Tan 60 = a) 3 b) 2 c) 6 d) 12 2) If cos 2x = where x is an acute angle then m ( x) =.. a) 15 b) 30 c) 45 d) 60 3) If tan = 1 where x is acute angle then m ( x) =.. a) 10 b) 30 c) 45 d) 60 4) 2 Tan 45 - = a) zero b) c) d) 1 5) If cos = where x is an acute angle then sin x =. a) b) c) d) 2
6) In ABC : If m ( A) = 85, Sin B = Cos B, then m ( C) =. a) 30 b) 45 c) 50 d) 60 (5) Find the value of the following:- 1) (cos 30 - cos 60 ) (sin 30 + sin 60 ) 2) sin 2 45 tan 2 60 - sin 2 60 tan 2 30 3) Sin 45 cos 45 + sin 30 cos 60 - cos 2 30 4) (6) Prove that: 1) cos 60 = 2 cos 2 30-1 2) tan 60 (1 tan 2 30 ) = 2 tan 30 3) tan 2 60 - tan 2 45 = 4 sin 30 4) tan 60 = 5) = 8 (7) Find the value of x in each of the following:- 1) x cos 30 = tan 60 2) x sin 2 45 = tan 2 60 3) 4x = cos 2 30 tan 2 30 tan 2 45 4) x sin 30 cos 2 45 = cos 2 30 5) x sin 45 cos 45 tan 60 = tan 2 45 - cos 2 60 6) tan x = 3
15 cm 3 rd Preparatory (8) Find m ( ) where is an acute angle : 1) Sin 2 45 = cos tan 30 2) 2 sin = tan 2 60-2 tan 45 3) Sin = sin45 cos30 + cos45 sin45 4) Sin sin 2 60 = 3sin 2 45 cos 45 cos60 5) tan = 3 (sin 30 + cos 30 ) 4 (sin 3 60 + cos 3 60 ) 6) 3 tan 2 = 4 sin 2 30 + 8 cos 2 60 (9) In the opposite Figure:- ABCD is a rectangle where AB = 15cm. AC = 25cm. Find: First: m ( ACB) D C Second: The surface area of the rectangle ABCD A B (10) In the opposite figure:- ABCD is an isosceles trapezium where AB = AD = DC = 5cm. A 5 cm // D BC = = 11cm, find First : m( B), m ( A) Second: the area of the trapezium ABCD. B 11 cm C 4
Third geometry 1) Complete each of the following;- 1) The distance between the two points (9, 0), (4, 0) is 2) The distance between the two points (0, -11), (0, -5) is 3) The distance between the points (4, -3) and the origin point is 4) The distance between the points (5, 0), (0, -12) is 5) The diameter length of the circle whose centre is (8, 5) and passes through the point (4, 2) equals 6) If the distance between the two points (a, 0) and (0, 1) is one length unit then a =. 7) The distance between the points (3, - 4) and the X axis = length unit. 8) In the square ABCD: If A (2, -5), B (-1, -1) then the perimeter of the square is.. length unit and its area is.. square unit. 2) Answer the following questions:- 1) Find the length of in each of the following cases: a) M (2, -1), N (5, 3) b) M (-3, -5), N (5, 1) c) M (7, -8), N (2, 4) d) M (7, -3), N (0, 4). 2) Prove that the points A (3, -1), B (-4, -2), C (2, -2) which belong to an orthogonal Cartesian co-ordinates plane lie on the circle whose centre M (-1, 2) then find the circumference of the circle. 3) Find the value of a in each of the following a) If the distance between the two points (a, 7) and (-2, 3) equals 5. b) If the distance between the two points (a, 7) and (3a -1, -5) equals 13 5
4) If A (x, 3), B (3, 2), C (5, 1) and if AB = BC find the value of x. 5) If the distance between the point (x, 5) and the point (6, 1) equals 2 find the value of x. 6) Identify the type of the triangle whose vertices are A (-2, 4), B (3, -1), C (4, 5) due to its sides lengths. 7) Prove that triangle whose vertices A (5, -5), B (-1, 7), C (15, 15) is right angled at B, then calculate its area. 8) Prove that the points (5, 3), (6, -2), (1, -1), (0, 4) are vertices of a rhombus. Then find its area. 6
Part Two (1) Complete each of the following : 1) The mid point of the line segment joining the two points (2, 5) and (4, 3) is the point 2) If (2, 1) is the mid-point of where A (3, -4) and B (m, 6) then m= 3) If the origin point is the mid-point of the line segment where A (5, -2) then the co-ordinates of B (, ) 4) If // and the slope of = 0.75 then the slope of is 5) If and the slope is 0.5 then the slope of equals.. 6) The slope of the straight line parallel to the straight line passing through the two points (2, 3) and (-2, 3) equals. 7) If the straight line is parallel to X-axis where A (8, 3) and B (2, K) then K =.. 8) If the straight line is parallel to the Y-axis where C(m, 4) and D (-5,7) then m=.. 9) ABC is a right angled triangle at B where A (1, 4) and B(-1, -2) then the slope of = 10) If the straight line which passes through the two points (a, 0) and (0, 3) and the straight line which makes an angle of measure 30 with the positive direction of the X-axis are perpendicular then a=. 7
11) If y= m x + c represents the equation of a straight line given its slope and the length of the intercepted part of the Y-axis then (a) The equation of the straight line when m= 1 and c=3 is (b)the equation of the straight line when m= -2 and c=1 is (c) The equation of the straight line when m= 3 and c=0 is 12) In the opposite figure: C (3, 4) is the mid-point of a) OA =.. length unit. Y b) OB =. Length unit. c) The slope of is d) The slope of is `x e) The slope of is f) The slope of is B 0 `Y /. C(3.4) / A X g) C is the centre of the circle which passes through the points,, h) The area of OAB is square unit. i) The perimeter of OAB =. unit length. j) The equation of the is. k) The equation is.. 8
Choose the correct answer from those given: 1- The distance between the point (4, -3) and the X-axis equals a) -3 b) 3 c) 4 d) 5 2- A circle of centre at the origin point and its radius is 2 unit length which of the following points belongs to the circle? a) (1, 2) b) (-2, 1) c) (, 1) d) (, 1) 3- If (4, 3) is the mid-point of where A (3, 4) then the co-ordinates of B is. a) (5, -2) b) (2, 5) c) (5, 2) d) (3.5, -3.5) 4- The straight line whose equation is 2x 3y 6 =0 intercepts from the Y-axis a part of length. a) -6 b) -2 c) d) 2 5- If the two straight lines 3 x 4 y - 3 = 0 and k y + 3 x 8 = 0 are perpendicular then k= a) -4 b) -3 c) 3 d) 6- If the two straight lines x + y = 5 and k x + 2y = 0 are parallel then k= a) -2 b) -1 c) 1 d) 2 7- The area of the triangle bounded by the straight lines 3x 4y = 12, x=0 and y=0 is square unit equal. a) 6 b) 7 c) 12 d)15 9
8- is straight line passing through the two points (2, 5) and (5, 2) which of the following points a) (1, 6) b) (2, 3) c) (0, 0) d) (3, -4) 9- The points (0, 0), (3,0) and (0,4) a) form an obtuse angles triangle b) form an octue angled triangle. c) form a right angled triangle. d) are collinear. 10- If A (0, 0), B(5, 7) and C(5, h) are the vertices of a right angled triangle at C then h =.. a) zero b) 5 c) 7 d) -5 Answer the following questions:- 1- Find the co-ordinates of the mid-points of in each of the following: a) A (2,4), B(6,0) b) A (7, -5), B (-3, 5) c) A (-3, 6), B (3, 0) d) A (7, -6), B(-1, 0) 2- If C is the mid-point of find x and Y in each of the following cases: First : A (1,5), B(3,7), C(x, y) Second: A (-3, y), B(9, 11), C(x, -3) Third: A (x, -6), B(9, -11), C(-3, y) Fourth: A(x, 3), B(6, y), C(4, 6) 10
3- Find the slope of the straight line which makes with the positive direction of the X-axis a positive angle of measure: a) 30 b) 45 c) 60 4- Using the calculator find the measure of the positive angle which is made by the straight line whose slope is m with the positive direction of the X-axis in each of the following cases: a) m = 0.3673 b) m = 1.0246 c) m = 3.1648 5- If the points (0, 1), (a, 3), (2, 5) are collinear find the value of a. 6- In which of the following cases, the points A, B and C are collinear? Explain your answer. First: A (-1, 5), B(0, -3), C(2, 1) Second: A (-2, 1), B(2, 3), C(4, 4) 7- Prove that the points A (-2, 5), B (3, 3), C (-4, 2) are not collinear and if D (-9, 4) prove that the figure ABCD is a parallelogram. 8- Let A (5, -6), B(3,7) and C(1, -3), Find the equation of the straight line which passes through A and the mid-point of 9- Find the equation of the straight line passing through the point (3, -5) and parallel to the straight line x + 2y 7 = 0 10- Find the equation of the straight line which intercepts the two axes two positive parts of lengths 4 and 9 for X and Y-axes respectively. 11- If A (1,-6), B (9, 2) find the co-ordinates of the points which divide into four equal parts in length. 11
12- Prove that the points A (6, 0), B(2, -4) and C(-4, 2) are vertices of a right angled triangle at B then find the co-ordinates of the point D which makes the figure ABCD a rectangle. 13- If the points A (3, 2), B (4, -3), C(-1, -2), D (-2, 3) are vertices of a rhombus find: a) The co-ordinates of the point of intersection of its two diagonals. b) The area of the rhombus ABCD. 14- If A (-1, -1), B(2, 3), C(6, 0), D(3, -4) are four points on an orthogonal Cartesian co-ordinates plane. Prove that and bisect each other. What is the name of this figure? 15- ABCD is a parallelogram where A (3, -4), B(2,-1), C(-4, 3), find the co-ordinates of point D then find the co-ordinates of point E such that the figure ABCE becomes a trapezium in which //, AE=2BC 16-If the straight line L 1 passes through the two points (3, 1) and (2, K), and the straight line L 2 makes with the positive direction of the X- axis and angle of measure 45, Find the value of K if: First : L 1 // L 2 Second: L 1 L 2. 17- Using the slope prove that the points A (-1, 3), B (5,1) C(6,4) D(0, 6) are vertices of a rectangle. 12
Model Answers (1) Complete the following table : Part (1) Second :Trigonometry The angle Ratio 42 12 \ 18 44 \ 51 \\ 57 51 \ 9 \\ 64 8 \ 1 \\ Sin 0.6717 0.3214 0.8467 0.8998 Cos 0.7408 0.9469 0.5321 0.4363 Tan 0.9067 0.3394 1.5912 2.0625 (2) Complete the following: 1) 46.6067 2) 44 7 \ 30 \\ 3) 54 50 \ 45 \\ 4) 39 3 \ 5) 30 6) = 30 x = 30 2 = 60 7) + - = 0 8) + 1 = 0 9) 2 1 = - 10) ( ) 2 + ( ) 2 = + = = 1 13
11) x + 10 = 60 x = 50 12) 3x = 60 x = 20 (3) In the opposite figure:- sin 30 = = AD = AC x Sin 30 = 12 x = 6cm The area of ABC = x AD x BC The area of ABC = x 6 x 16 = 48 cm 2 In ADC E BEC is right angled at E m ( C) = 30, BC = 16 cm BE = BC = 8 cm (4) Choose :- 1) 6 2) (30) 3) = 45 x = 30 4) Zero 5) = 30 x = 60 sin 60 = 6) sin B = cos B m ( B) = 45 m( C) = 50 (5) Find the value of the following:- 1) (cos 30 - cos 60 ) (sin 30 + sin 60 ) = ( ) ( ) = ( ) ( ) = 2) sin 2 45 tan 2 60 - sin 2 60 tan 2 30 = x ( ) 2 x ( ) 2 - x ( ) 2 ( ) 2 14
= x x 3 - x x = = - = 3) Sin 45 cos 45 + sin 30 cos 60 - cos 2 30 2 = + - = 0 4) = 1 (6) Prove that: 1) cos 60 = 2 cos 2 30-1 L.H.S. = Cos 60 = R.H.S. = 2cos 2 30-1 = 2 x ( ) 2 1 = 2x 1 = 1 = 2) tan 60 (1 tan 2 30 ) = 2 tan 30 L.H.S. = Tan60 (1-tan 2 30 ) = (1 ( ) ) = R.H.S. = 2 tan 30 = 2x = 3) tan 2 60 - tan 2 45 = 4 sin 30 L.H.S. = tan 2 60 - tan 2 45 = ( ) 2 (1) 2 = 2 R.H.S. = 4 sin 30 = 4 x = 2 4) tan 60 = L.H.S. = Tan 60 = R.H.S. = = = x = ( ) 15
5) [ ] [ ] [( ) ] = [ ] = = 8 (7) Find the value of x in each of the following:- 1) x cos 30 = tan 60 x = 2) x sin 2 45 = tan 2 60 x = 3 = 3 x 2 = 6 3) 4x = cos 2 30 tan 2 30 tan 2 45 4x = ( ) 2 ( ) 2 (1) 2 = x x 1 4x = x = 4) x sin 30 cos 2 45 = cos 2 30 x= 5) x sin 45 cos 45 tan 60 = tan 2 45 - cos 2 60 x= = x = 16
6) tan x = Tan x = x = 45 (8) Find m ( ) where is an acute angle: 1) Sin 2 45 = cos tan 30 Cos = = 30 2) 2 sin = tan 2 60-2 tan 45 2 sin = Sin = 3) Sin = sin45 cos30 + cos45 sin30 Sin = 4) Sin sin 2 60 = 3sin 2 45 cos45 cos60 Sin Sin = = 45 17
15 cm 3 rd Preparatory 5) tan = 3 (sin 30 + cos 30 ) 4 (sin 3 60 + cos 3 60 ) tan = 3 ( ( ) tan = = tan = 1 = 45 - ( ) = 6) 3 tan 2 = 4 sin 2 30 + 8 cos 2 60 3 tan 2 = 4 x Tan 2 = Tan = 1 = 45 (9) In the opposite Figure:- In ABC m ( B) = 90 sin ( ACB) = D A m ( (ACB) = 38 52 \ 12 \\ C B BC= = 20 Area of rectangle = L x W = 20 x 15 = 300cm 2 18
(10) In the opposite figure:- Construction: Draw A 5 cm D AD = AB = DC = 5cm EF = 5cm, BE + CF = 11 5 = 6 cm. BE = FC = 3cm. B E 11 cm F C In AEB m ( AEB) = 90, AB = 5cm, BE = 3 cm. Cos (B) = = m (B) = 53 7 \ 48 \\ 180 - (90 + 53 7 \ 48 \\ ) = 36 52 \ 12 \\ AE = 4cm "Pythagoras" Area of trapezium = x H = = 8 x 4 = 32 cm 2 Third geometry 1) Complete each of the following;- 1) D = D = length unit. 2) D = length unit. 3) D = length unit. 4) D = length unit. 19
5) r = length unit. Diameter = 2r = 10 length unit. 6) D = = = a 2 + 1 = 1 2 = a 2 = 1-1 a = 0 7) 4 = 4 length unit. 8) AB = AB = length unit. P. of square = side length x 4 = 4 x 5 = 20 length unit. Area = S 2 = 5 2 = 25 squared length unit. 2) Answer the following questions:- 1) a) MN = length unit. b) MN = length unit. c) MN = = 13 length unit. d) MN = 2) D= MA = length unit. MB = = 5 length unit. MC = = 5 length unit. 20
MA = MB = MC = r A, B, and C lie on the circle M circumference = 2 r = 2 x 3.14 x 5 = 31.4 length unit. 3) D = (5) = By squaring both sides. 25 = (a + 2) 2 + 16 (a + 2) 2 = 9 = + a + 2 = + 3 a + 2 = 3 or a + 2 = -3 a = 1 or a = -5 (b) 13 = 13 = 169 = (2a 1) 2 + 144 (2a 1) 2 = 169 144 = 25 = + 2a 1= + 5 2a 1 = 5 or 2a 1 = -5 2a = 6 a = 3 or 2a = -4 a = -2 21
4) AB = BC = By squaring both sides. (x 3) 2 + 1 = 5 (x 3) 2 = 4 = + x 3 = + 2 x 3 = 2 or x 3 = -2 x = 5 or x = 1 5) D= 2 = By squaring both sides. (2 ) 2 = ( ) 2 20 = ( x 6) 2 + 16 (x 6) 2 = 4 x 6 = 2 x 6 = 2 or x 6 = -2 x = 8 or x = 4 22
6) AB = BC = AC = AC = BC = ABC is an isosceles 7) AB = BC = CA = (AC) 2 = ( ) 2 = 500 (AB) 2 + (BC) 2 = 180 + 320 = 500 (AC) 2 = (AB) 2 + (BC) 2 ABC is right-angled at B 23
8) A (5, 3), B (6, -2), C (1, -1), D (0, 4) AB = BC = CD = DA = AB = BC = CD = DA. A, B, C, and D are vertices of Rhombus. AC = BD = Area of the rhombus = x d 1 x d 2 = x x = 24 (u.l. ) 2 24
Part Two (1) Complete each of the following : 1- midpoint = ( ) = ( ) 2- (2, 1) = ( ) m + 3 = 4 m = 4-3 = +1 3- B (-5, 2) because the origin point is the mid point. Or we can use the rule of mid point. 4-0.75 m 1 = m 2 5- m 1 = = -2 6- m 1 = m 2 = 7- If // x-axis its slope = 0 = 0 K 3 = 0 K = 3 8- // y axis The slope is undefined where the denominater equal to zero. Slope of -5 m = 0 m = - 5 25
9- Slope of = Slope of = - because, then m 2 = 10- m 1 = = - m 2 = tan 30 = m 1 x m 2 = -1 - x = -1 = 1 a = 11- a) y = x + 3 b) y = -2x + 1 c) Y = 3x 12- A (x, 0), B(0, y) C (3, 4) is midpoint ( ) = (3, 4) = 3 x = 6 = 4 y = 8 a) OA = 6 L.U. b) OB = 8 L.U. c) m = 26
d) m = e) m = 0 f) m = undefined g) A, B, O h) x 6 x 8 = 24 square unit i) AB = = = 10 Perimeter = 6 + 10 + 8 = 24 L.U. j) Slope of =, C = 8 y = - x + 8 k) Slope of =, C = 0 y = x Second : Choose 1) b 4) d 7) a 10) a 2) c 5) d 8) a 3) c 6) d 9) c Answer the following questions 1- Midpoint ( ) a) M = ( ) = (4, 2) b) M = ( ) = (2, 0) c) M = ( ) = (0, -3) d) ( ) = (3, -3) 27
2- a) (x, y) = ( )= (2, 6) X=2, y = 6 b) (x, 3) = ( ) = (3, x = 3, = -3 y + 11= -6 y = -17 c) (-3, y) = ( ) =( ) = -3 x+ 9 = -6 x= -15, y = -8.5 d) (4, 6) = ( ) 3- a) m= tan 30 = b) m= tan 45 = 1 c) m= tan 60 = = 4 x + 6 = 8 x = 2 = 6 3 + y = 12 y = 9 4- a) = tan -1 0.3673 = 20 10 / 6 // b) = tan -1 (1.0246) = 45 41 / 46 // c) = tan -1 3.1648 = 72 27 / 53 // 5- the points are collinear. the slope of = slope of where A (0,1), B(a,3), C(2,5) Slope of = = Slope of = = -2a = 2a-4 2a + 2a = 4 4a = 4 a = 1 28
6- First: Slope of = = Slope of = = 2 m 1 m 2 A, B and C are not collinear. Second: m 1 = Slope of, m 2 = Slope of m 1 m 2, B is a common point. A, B, and C are collinear. 7- Slope of = Slope of = slope of slope of A, B and C are not collinear Slope of = = Slope of = = slope of = slope of slope of = slope of //, // ABCD is a parallelogram. 8- midpoint of = ( ) D = ( ) = (2, 2) A(5, -6), D (2, 2) Slope of = = 29
A (5, -6) lies on the s.line A satisifies the equation of s.l. Or y = + C at A (5, -6) m = -6 = x 5 + c -6 = + c C= -6 + = - 3y = - 8 + 22 y = + y = 9- m 1 = = or +2y-7=0 2y=- + 7 y = + m 1 = - the two st. lines are parallel m 1 =m 2 = - y = mx + c y = - + C at (3, -5) -5 = - x 3 + C -5 = - + c c = -5 + = y = - - 30
10- the st. line cuts x-axis at point (4, 0) (4, 0) the st. line The s. line cuts y axis at point (0, 9) (0, 9) the st. line. m = (0,9) Y y = m + c y = - + c at (0, 9) c = 9 y = + 9 x / Y / (4,0) X 11- let C is the midpoint of C = ( ) Let D is the midpoint of 0 A 0 0 E C D 0 0 B D =( ) Let E is the mid-point of of E = ( ) = (3, - 4) 12- AB = = BC = = D C CA = (CA) 2 = 104 = M (AB) 2 + (BC) 2 = 32 + 72 = 104 A B (CA) 2 = (AB) 2 + (BC) 2 = 104 ABC is right angles at B 31
If M is the point of intersection of the two diagonals and M is a mid-point of and, Let D (x, y) M= ( ) = ( ) x + 2 = 2 x=0 y - 4 = 2 y=6 D = ( 0, 6) 13- let M is the point of intersection of its two diagonals. M is the mid point of m = ( ) AC = = l.u. BD = = l.u. Area of the rhombus = x d 1 x d 2 = x x = 24 square units. 14- Midpoint of = ( ) Midpoint of = ( ) The midpoint of = the mid point and bisect each other. AB = = l.u. BC = = l.u. CD = = l.u. DA = = l.u. 32
AC = = l.u. BD = = l.u. AB = BC = CD = DA AC = BD ABCD is a square. 15- M is the midpoint of and m(x 1, Y 1 ) = ( ) ( ) ( ) = ( ) = - x + 2 = -1 x = -3 = - y - 1 = -1 y = 0 D ( -3, 0) ABCD is aparallelogram. B / C AE = 2BC AD = BC = DE = AE D is mid point of / A D / E (-3, 0) = ( ) = -3 = 0 E (-9, 4) 33
16- First: m 1 = m 2 = tan 45 = 1 L 1 // L 2 = 1 k 1 = -1 k = 0 Second : L 1 L 2 m 1 m 2 = -1 x 1 = -1 K 1 = 1 k = 2 17- using the slope m = Slope of = = = - Slope of = = = 3 Slope of = = = - Slope of = = = 3 slope of = slope of, slope of = slope of //, // slope of slope of = ABCD is a rectangle. 34