Hits: 4113
Hits: 4113 Question Solve the equation , for Solution i. We have the equation; Dividing both sides of the equation by ; We have the relation , therefore, To solve this equation, we can substitute . Hence, Since given interval is , for interval can be found as follows; Multiplying both sides of the inequality with […]
Hits: 424 Question i. Solve the equation for . ii. How many solutions has the equation for ? Solution i. We are given; To solve this equation for , we can substitute . Hence, Since given interval is , for interval can be found as follows; Multiplying the entire inequality with 2; Since ; Hence the given interval for is . To solve equation […]
Hits: 1102 Question In the diagram, D lies on the side AB of triangle ABC and CD is an arc of a circle with centre A and radius 2 cm. The line BC is of length 2√3 cm and is perpendicular to AC. Find the area of the shaded region BDC, giving your answer in terms of and […]
Hits: 5470 Question The diagram shows a sector of a circle with centre and radius 20 cm. A circle with centre and radius cm lies within the sector and touches it at P, Q and R. Angle POR = 1.2 radians. i. Show that , correct to 3 decimal places. ii. Find the total area of the three parts […]
Hits: 524 Question i. Prove the identity ii. Use this result to explain why for . Solution i. We have the equation; We have the relation , therefore, We have the trigonometric identity; We can rewrite the identity as; Therefore; Again using the relation , we can rewrite; ii. It is evident that in equation for […]
Hits: 248 Question Solve the equation , for . Solution We have the equation; We have the trigonometric identity; We can rewrite the identity as; Therefore; To solve this equation for , we can substitute . Hence, Now we have two options; Since; NOT POSSIBLE Using calculator we can find the values of . We utilize the symmetry property of to […]
Hits: 396 Question i. Show that the equation can be written as a quadratic equation in . ii. Solve the equation , for . Solution i. We have the equation; To write the given equation in terms of , we utilize the trigonometric relation; Hence; Multiplying both sides with ; We have the trigonometric identity; We can rewrite the identity as; […]
Hits: 1140 Question i. Solve the equation , for . ii. The smallest positive solution of the equation , where is a positive integer, is . State the value of and hence find the largest solution of this equation in the interval . Solution i. We have the equation; We have the trigonometric identity; We can rewrite the identity as; […]
Hits: 3630 Question The diagram shows a sector OAB of a circle with centre O and radius . Angle AOB is radians. The point C on OA is such that BC is perpendicular to OA. The point D is on BC and the circular arc AD has centre C. i. Find AC in terms of and . […]
Hits: 1904 Question In the diagram, AB is an arc of a circle with centre O and radius . The line XB is a tangent to the circle at B and A is the mid-point of OX. i. Show that angle radians. Express each of the following in terms of, and : ii. the perimeter of the shaded region, iii. the area of […]
Hits: 525 Question i. Prove the identity ii. Solve the equation for . Solution i. We have the equation; We have the relation , therefore, We have the trigonometric identity; Therefore; ii. Solve the equation for . We can rewrite the given equation as; As demonstrated in (i), we can rewrite the given equation as; To […]
Hits: 1701 Question In the diagram, ABC is an equilateral triangle of side 2 cm. The mid-point of BC is Q. An arc of a circle with centre A touches BC at Q, and meets AB at P and AC at R. Find the total area of the shaded regions, giving your answer in terms of and . […]