# A 23 alternate angles are equal b 64 alternate angles

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a = 23° (alternate angles are equal) b = 64° (alternate angles are equal) So f = 23° + 64° = 87° Exercise 9D 1 a 85° b 100° c 42° d 40° e 97° 2 a 65° b 35° c c = 62°, d = 56° d e = 64°, f = 116° e g = 58 1 2 °, h = 121 1 2 ° 3 a, b i a = 39° (interior), b = 141° (exterior) ii c = 59° (interior), d = 121° (exterior) iii e = 45° (interior), f = 135° (exterior) c The exterior angle is the sum of the two interior angles. 4 55°, 60° and 65° 5 25° and 130° or 77 1 2 ° and 77 1 2 ° 6 a 316° b 305° c 110° d 30° e e = 70°, f = 45°, g = 25° 7 a p = 49° (angles on a line = 180°) q = 97° (angles in a triangle = 180°) Maths Frameworking 3rd edition © HarperCollins Publishers Ltd 2014 Teacher Pack 1.3
24 r = 97° (opposite angles are equal) s = 40° (angles in a triangle = 180°) so c = 140° (angles on a line = 180°) b e = f = 47° (angles in an isosceles triangle) g = 133° (angles on a line = 180°) d = h = 23 1 2 ° (angles in an isosceles triangle) so d = 23 1 2 ° 8 a 9 z = 180°, z = 20°, so the three angles are 40°, 60° and 80° b 4 y + 120° = 180°, 4 y = 60°, y = 15°, so the three angles are 25°, 45° and 110° Problem solving: Calculating angles in triangles A Mark unknown angles on the diagram. Label them a and b for example. a = 86° (corresponding angles are equal) b = 52° (angles on a line = 180°) So e = 42° (angles in a triangle = 180°) B Mark unknown angles on the diagram. Label them a , b , c and d for example. a = 74° (angles on a line = 180°) b = 62° (angles on a line = 180°) c = 74° (corresponding angles are equal) d = 62° (corresponding angles are equal) So m = 44° (angles in a triangle = 180°) Note: there are other ways to work out the solution Exercise 9E 1 a 80° b 139° c 108° d 126° 2 a 109° b 108° c 68° d 94° 3 set b, the only set whose sum is 360° 4 a The missing interior angle = 80° (angles on a line = 180°), so a = 90° (angles in a quadrilateral = 360°) b The missing interior angle = 117° (angles in a quadrilateral = 360°), so a = 63° (angles on a line = 180°) 5 a 110° b Both pairs of interior angles between the parallel lines add up to 180°. 6 a + 2 a + 3 a + 4 a = 360°, 10 a = 360°, a = 36°. So the four angles are 36°, 72°, 108° and 144°. 7 a + a + 10° + a + 20° + a + 30° = 360°, 4 a + 60° = 360°, 4 a = 300°, a = 75° So the four angles are: 75°, 85°, 95° and 105°. 8 No because 3 × 90° = 270° and 360° – 270° = 90°, so the 4th angle must also be 90°. Problem solving: Triangles in quadrilaterals Label the angles in both triangles with different letters. So the sum of the interior angles of the quadrilateral = ( a + b + c ) + ( d + e + f ) = 180° + 180° (angles in both triangles = 180°) = 360° Maths Frameworking 3rd edition © HarperCollins Publishers Ltd 2014 Teacher Pack 1.3
25 Exercise 9F 1 AB = AC = BC, ABC = ACB = BAC 2 a BC and CD b AD and BC c AD and CD or BC and CD 3 AB = BC = CD = AD, AD is parallel to BC, AB is parallel to CD and the diagonals bisect each other at 90° 4 a square, rhombus b rectangle, parallelogram, kite, arrowhead c square, rectangle, parallelogram, rhombus d trapezium e square, rhombus, kite, arrowhead f square, rectangle g square, rectangle, parallelogram, rhombus h square, rhombus, kite i arrowhead 5 a A square is a rectangle with equal sides. b A rhombus is a parallelogram with equal sides. c An arrowhead is a kite with an interior reflex angle.