The first is simple angle measuring, with accurate answers provided. Three short worksheets of questions for pupils to consolidate.These are all animated to a high standard and should help pupils avoid developing any misconceptions about how to use a protractor. It also includes examples of subtle ‘problem’ questions like the answer being between two dashes on the protractor’s scale or the lines of the angle being too short to accurately read off the protractor’s scale. The range of examples includes measuring all angle types using either the outer or inner scale. An extended set of examples, intended to be used as mini whiteboard questions, where an angle is shown and then a large protractor is animated, leaving pupils to read off the scale and write down the angle. The angles are all very close or equal to 90 degrees, so pupils have to come up with a way (using the gridlines) to decide. Designed to come after pupils have been introduced to acute, obtuse and reflex angles and they can already estimate angles.Ī nice set of problems where pupils have to judge whether given angles on a grid are acute, 90 degrees or obtuse. Includes lots of large, clear, animated examples that make this fiddly topic a lot easier to teach. The intersection of the diameter and the chord at 90 degrees can be very close to the centre and so the two lengths coming from the point of intersection to the radius are assumed to be equal, but they aren’t.A complete lesson on how to use a protractor properly. Incorrect assumption of isosceles triangles.This also includes the inverse trigonometric functions. The incorrect trigonometric function is used and so the side or angle being calculated is incorrect. The missing side is calculated by incorrectly adding the square of the hypotenuse and a shorter side, or subtracting the square of the shorter sides. The only case of this is when both angles are 90^o. Opposite angles are the same for a cyclic quadrilateralĪs angles in the same segment are equal, the opposing angles in a quadrilateral are assumed to be equal.Angle at the centre is supplementary to opposing angleĪs the shape is a quadrilateral, the angle at the centre is assumed to be supplementary and add to 180^o.The angle ABC = 56^o as it is in the alternate segment to the angle CAE. Here, angle ABC is incorrectly calculated as 180 - 56 = 124^o. The angle is taken from 180^o which is a confusion with opposite angles in a cyclic quadrilateral. Opposite angles in a cyclic quadrilateral.Top tip: Use arrows to visualise which way the alternate segment angle appears: The chord BC is assumed to be parallel to the tangent and so the angle ABC is equal to the angle at the tangent. Parallel lines (alternate segment theorem).The angle at the circumference is assumed to be 90^o when the associated chord does not intersect the centre of the circle and so the diagram does not show a semicircle. They should total 90^o as the angle in a semicircle is 90^o. The angles that are either end of the diameter total 180^o as if the triangle were a cyclic quadrilateral. Look out for isosceles triangles and the angles in the same segment. Make sure that you know when two angles are equal. The angle at the centre is always larger than the angle at the circumference (this isn’t so obvious when the angle at the circumference is in the opposite segment). Make sure you know the other angle facts including:īy remembering the angle at the centre theorem incorrectly, the student will double the angle at the centre, or half the angle at the circumference. Below are some of the common misconceptions for all of the circle theorems:
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