Hits: 150
Hits: 150 Question i. Show that ii. Hence, showing all necessary working, solve the equation For . Solution i. We are given the expression; We have the algebraic identity; We have the trigonometric identity; Since ; We can write; Therefore; ii. We are required […]
Hits: 321 Question The diagram shows an arc BC of a circle with centre A and radius 5 cm. The length of the arc BC is 4 cm. The point D is such that the line BD is perpendicular to BA and DC is parallel to BA. i. Find angle BAC in radians. […]
Hits: 325 Question The diagram shows a triangle ABC in which BC = 20 cm and angle ABC = 90o. The perpendicular from B to AC meets AC at D and AD = 9 cm. Angle BCA = . i. By expressing the length of BD in terms of in each of the triangles […]
Hits: 243 Question The functions f and g are defined by for , for i. Solve the equation . ii. Sketch the graph of . Solution i. We are given; for for We are required to solve equation ; To solve this equation for , we can substitute . Hence, Since given […]
Hits: 596 Question The diagram shows a triangle OAB in which angle ABO is a right angle, angle AOB = radians and AB = 5 cm. The arc BC is part of a circle with centre A and meets OA at C. The arc CD is part of a circle with centre O and meets […]
Hits: 192 Question i. Show that the equation may be expressed as ii. Hence solve the equation for . Solution i. We are given the equation; We have the trigonometric identity; From this we can substitute in above equation; ii. We are required to solve the equation for . From […]
Hits: 812 Question The diagram shows a sector POQ of a circle of radius 10 cm and centre O. Angle POQ is 2.2 radians. QR is an arc of a circle with centre P and POR is a straight line. i. Show that the length of PQ is 17.8 cm, correct to 3 significant […]
Hits: 219 Question a. Express the equation as a quadratic equation in and hence solve the equation for . b. The diagram shows part of the graph of , where and are constants and . Find the value of and the value of . Solution a. We are required to solve this quadratic equation […]
Hits: 240 Question A straight line cuts the positive x-axis at A and the positive y-axis at B(0, 2). Angle radians, where O is the origin. i. Find the exact value of the x-coordinate of A. ii. Find the equation of the perpendicular bisector of AB, giving your answer in the form y […]
Hits: 426 Question a. i. Express in the form , where and are constants to be found. ii. Hence, or otherwise, and showing all necessary working, solve the equation For . b. The diagram shows the graphs of and for . The graphs intersect at the points A […]
Hits: 1003 Question The diagram shows a triangle OAB in which angle OAB = 90o and OA = 5 cm. The arc AC is part of a circle with centre O. The arc has length 6 cm and it meets OB at C. Find the area of the shaded region. Solution It is evident from […]
Hits: 723 Question i. Solve the equation for . ii. Sketch, on the same diagram, the graphs of and for . iii. Use your answers to parts (i) and (ii) to find the set of values of x for for which . Solution i. We have the equation; We know that […]
Hits: 2134 Question The diagram shows points A and B on a circle with centre O and radius r. The tangents to the circle at A and B meet at T. The shaded region is bounded by the minor arc AB and the lines AT and BT. Angle AOB is radians. i. In the […]
Hits: 732 Question The function f, is such that for . It is given that and . i. Find the values of the constants a and b. ii. Find the set of values of k for which the equation f(x) = k has no solution. Solution i. We are given the function […]
Hits: 1477 Question The diagram shows a circle with centre O and radius r cm. The points A and B lie on the circle and AT is a tangent to the circle. Angle radians and OBT is a straight line. i. Express the area of the shaded region in terms of r and . […]
Hits: 604 Question i.Prove the identity ii.Hence solve the equation for . Solution i. We have the trigonometric identity; ii. We have the equation; From (i) we know that left hand side of given equation can be written as; Therefore; […]