THIRD TERM E-LEARNING NOTE
SUBJECT: MATHEMATICS CLASS: SS2
SCHEME OF WORK
| WEEK | TOPIC |
| 1 | Circle Theorem: Tangent properties of circle; Angles in alternate segment; Two tangent from a circle at external point. |
| 2 | Trigonometry: Derivative of sine rule and cosine rule and their applications. |
| 3 | Bearing and Distances; Elevation and Depression. |
| 4 | Statistics: Class boundaries, class mark, and cumulative frequencies of grouped data, and histogram. |
| 5 | Statistics: Cumulative frequency curve (Ogive); Using ogive to calculate the median, quartile, percentile and decile. |
| 6 | Review of the First Half Term Work and Periodic Test. |
| 7 | Statistics: Mean, median, and mode of grouped data. |
| 8 | Probability: Introduction; use of dice, coins and playing cards. |
| 9 | Probability: Addition and multiplication rules of probability; Mutually exclusive, independent, and complementary events; Experiment with or without replacement. |
| 10 | Revision |
REFERENCE BOOKS
1.New General Mathematics SSS2 by M.F. Macrae etal.
2. Essential Mathematics SSS2 by A.J.S. Oluwasanmi.
3. Exam Focus Mathematics.
WEEK ONE
TOPIC:TANGENTS FROM AN EXTERNAL POINT
Theorem:
The tangents to a circle from an external point are equal.
Given: a point T outside a circle, centre O, TA and TB are tangents to the circle at A and B.
To prove: |TA| = |TB|
Construction: Join OA, OB and OT
In ∆s OAT and OBT
OAT = OBT = 900 (radius tangent)
|OA| = |OB| (radii)
|OT| = |OT| (common side)
∆OAT = ∆OBT (RHS)
|TA| = |TB|
Note that <AOT = <BOT and <ATO = <BTO hence the line joining the external point to the centre of the circle bisects the angle between the tangents and the angle between the radii drawn to the points of contact of the tangents.
Example:
1.In the figure below O is the centre of the circle and the TA and TB are tangents if <ATO = 390, calculate < TBX
In ∆TAX
AXT = 900 (Symmetry)
TAX = 180 – (900 + 390) sum of angles of ()
1800 – 1290 = 510
TBX = 510 (symmetry)
OR
∆ ATB is an Isosceles triangle
|AT| =|BT| (tangents from external point)
<ATO = <BTO = 390 (symmetry)
< ATB = 2(39) = 780
<TAX = < TBX (base angle of Isos ∆ )
2TBX = 1800 – 780 (sum of angle in a
2 TBX = 1020
TBX= 1020
2
TBX = 510
2.PQR are three points on a circle Centre O. The tangent at P and Q meet at T. If < PTQ = 620 calculate PRQ.
Solution
Join OP and OQ
In quadrilateral TQOP
<OQT = <OPT = 900 (radius 1 tangent)
POQ = 3600 – (900 + 900 + 620) sum of angle in a quadrilateral)
POQ = 3600 – 2420
POQ = 1180
PRQ = 1180 = 590 (2x angle at circumference = angle at centre)
2
PR1QR is a cyclic quadrilateral
R + R1 = 1800 (opp. angles of a cyclic quadrilateral )
R1 = 1800 – R
R1 = 1800 – 590
R1 = 1210
PRQ = 590 or 1210
Evaluation
1. ABC are three points on a circle, centre O such that <BAC = 370, the tangents at B and C meet at T. Calculate < BTC.
GENERAL EVALUATION/REVISION QUESTIONS
1. AB is a chord and O is the centre of a circle. If AOB = 780 calculate the obtuse angle between AB and the tangent B.
1 The dimension of a cuboid metal is 24cm by 21cm by 10cm, if the cuboid is melted and used in making a cylinder whose base radius is 15cm find the height of the cylinder.
2 The volume of a cylinder is 3600cm3 and its radius is 10cm calculate its
(a) curve surface area
(b) total surface area
READING ASSIGNMENT
Essential Mathematics, pages149-151, numbers 11-20.
WEEKEND ASSIGNMENT
Use the diagram below to answer the questions.
1.If < ATO = 360 ,calculate < ABO.
(a) 360 (b) 720 (c) 180 (d) 440
2.If <ABT = 570, calculate < AOT (a) 1140 (b) 570 (c) 330 (d) 1230
3.If< BTO = 440, calculate <TAX (a) 880 (b) 440 (c) 460 (d) 920
4.If |AB| = 18cm and |TB| = 15cm, calculate |TX|
(a) 180 (b) 330 (c) 780 (d) 120
5.If < AOT = 470, calculate ABO (a) 470 (b) 940 (c) 1330 (d) 430
THEORY
1.O is the centre of a circle and two tangents from a point T touch the centre at A and B. BT is produced to C. If <AOT = 670.calculate < ATC.
2.AD is a diameter of a circle,AB is a chord and AT is a tangent. a) State the size of <ADBb)If BAT is an acute angle of x0,find the size of DAB in terms of x.