NYGH 2014-S3EOY-IM1 with ans

[Turn Over 1 1(a) If tan A = 24 7 , such that 90° < A < 180°, find, without using a calculator, the value of (i) sin A, [2] (ii) cos A, [1] (iii) tan ( A – 90°). [1] (b) Find the area of the following tria ngle. Express your answer in the form ba , where a and b are positive integers. [2] 2 On separate axes, sketch the graph of (i) y2 + x = 0, [2] (ii) xy = 1, for x < 0, [2] (iii) y = kx3, where k is a negative constant. [2] 3 A, B, C, D and E are points on the circumference of a circle, centre O. KAX is a tangent to the circle. Given that  20ADE and  120ABC , calculate (i) ,ACE [1] (ii) ,CAX [2] (iii) .OEC [4] State all the reasons clearly. 135° 6c m 4c m  K A X E D O C B 120° 20°

2 4 In the diagram, the circle, with centre C, passes through the points P(–3, 7) and Q(6 , 4). The vertical line x = –3 is a tangent to the circle at P. (i) Explain why the y-coordinate of C is 7. [2] (ii) Find the equation of the circle. [4] 5 In the diagram, the coordinates of P, R and Q are (1, –1), (3, 3) and (7, 1) respectively. (a) T is a point on the line segment PR such that PT : PR is 3 : 5. Find (i) the coordinates of T, [2] (ii) the area of triangle PQR. [2] (b) Find the coordinates of S if PQRS is a parallelogram. [2]    R Q O y x P y C x Q(6, 4) P(–3, 7)   0  x = – 3

[Turn Over 3 6 Three points S, Y and E, are on level ground. Y is the foot of a building XY. S is 40 m due south of Y and E is due east of Y. (a) The angle of depression of S from X is 55°. Calculate the height of the building. [2] (b) The bearing of E from S is 064°. Calculate the distance ES. [2] (c) A man walks from E to S and stops at a point from which the angle of elevation of X is the greatest. Find the size of this angle. [4] S Y E X 40 m

4 7 W, X, Y and Z are four points on level ground with W due west of X. It is given that WY = 55 km, ZY = 25 km, WZ = 70 km,  50YWX and  48YXW . Find (i) the length of WX, [3] (ii) ,WYZ [2] (iii) the bearing of W from Y, [2] (iv) the bearing of Z from W, [3] (v) how far W is west of Y. [2] Z Y W X North 50° 48° 55 km 70 km 25 km

[Turn Over 5 8 The diagram below shows part of a straight line graph drawn to represent the equation ysx rx  , where r and s are constants. (a) Calculate the value of (i) s, [3] (ii) r. [2] (b) A point (9, k) lies on the straight line. Find the value of k. [2] (c) Find the values of x when y = 5, giving your answers to 2 decimal places. [2] (d) Find the gradient and the vertical intercept of the straight line that can be drawn on the diagram to solve the simultaneous equations ysx rx  , .2 20 yxx  [2]   xy x2 0 (6 , 0) (12 , 2)

6 9 The solution of this question by accurate drawing will not be accepted. In the diagram, the coordinates of A and B are (5 , 6) and (0 , –4) respectively. AD is the perpendicular bisector of BC. (i) Explain why AB = AC. [1] (ii) Given that the gradient of BC is 4 3 , find the equation of BC. [1] (iii) Show that the equation of AD is 3y + 4x = 38. [2] (iv) Hence, find the coordinates of D. [3] (v) Find the length of AD. [2]     y x A B D C

[Turn Over 7 10 Answer the whole of this question on the INSERT provided. The graph of xxy 48  is partially shown on the INSERT. (a) By drawing a tangent, estimate the gradient of the graph at the point where x = 1. [3] (b) Find the coordinates of the point on the curve at which the gradient is 0. [1] (c) By drawing suitable straight lines on the graph, find (i) the range of values of x for which 1 ≤ y ≤ 3, [2] (ii) the solutions of 2 x2 – 11x + 4 = 0. [3] Bonus Question 11 In triangle ABC, the lengths of the sides are 3 consecutive positive integers and the largest angle is twice the smallest angle. Find the length of the longest side. Hint: sin 2x = 2 sin x cos x. [3] End of Paper

8 INSERT Question 10 Tie this page behind the last sheet of your answer paper. (a) Ans: _______________________ (b) Ans: _______________________ (c)(i) Ans: _______________________ (c)(ii) Ans: _______________________

1 Qn Solution 1ai 7 2 + 242 = 625 = 252 sin A = 25 7 1aii cos A = 25 24 1aiii tan (A – 90°) = 7 24 1b Area = 0.5(4)(6)sin 135° = 12 sin 45° = 12 × 2 1 = 6 2 cm2 2a y2 = - x 2b y = x 1 , x < 0 2c y = kx3, k is a negative constant x  y  0  x  y  0  x  0 y

2 Qn Solution 3i  20ACE ( s in the same segmt) 3ii   60 )(120180 segmtoppinsADC  60CAX (alt. segmt theorem) 3iii   160 2)6020(EOC ( at centre = 2at circumference)   10 ).,(2 160180 isosofsbaseofsumOEC 4 (i) The radius, PC, is perpendicular to the tangent x = – 3, (or radius perpendicular to tangent) Since x = – 3 is a vertical line, PC is horizontal or parallel to the x-axis, so C has the same y-coord as P. (ii) (6 – a)2 + (4 – 7)2 = (a + 3)2 36 – 12a + a 2 + 9 = a 2 + 6a + 9 36 = 18a a = 2 r = 2 + 3 = 5 or r2 = 25 Eqn: (x – 2)2 + (y – 7)2 = 25 or x2 + y2 – 4x – 14y + 28 = 0 5ai T(x,y) =       5 )1(2)3(3,5 )3(3)1(2 =      5 7,5 11 or (2.2 , 1.4) 5aii Area = 1 1 3 3 1 7 1 1 2 1  = 10 square units 5b S(–3, 1)