Theory question 110
Algebraic Expressions(a) Make d the subject of the formula S = n/2 (2a + (n − 1) d)
(b) (i) In the diagram, O is the centre of the circle: A, B and P are points on the circumference. Prove that <AOB = 2 <APB. (ii) Find angles x, y and δ in the diagram below.
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(a) 2S = n(2a + (n − 1)d) = 2an + n(n − 1)d, so n(n − 1)d = 2S − 2an and d = (2S − 2an)/(n(n − 1)).
(b)(i) Join PO and produce it to a point X. OA = OP (radii), so <OAP = <OPA = a; the exterior angle <AOX = 2a. Likewise OB = OP gives <OBP = <OPB = b and <BOX = 2b.
So <AOB = <AOX + <BOX = 2(a + b) = 2<APB (or the difference, if P is placed so that both angles are on one side of PX).
(ii) The angle at the circumference 126° stands on the major arc, so the reflex angle at the centre δ = 2 × 126° = 252°.
x = 360° − 252° = 108° (angles at a point).
y = ½x = 54° (angle at centre is twice angle at circumference); or y = 180° − 126° = 54° (opposite angles of a cyclic quadrilateral).
The angle at the centre is twice the angle at the circumference standing on the same arc.