{"id":13068,"date":"2021-03-25T22:39:09","date_gmt":"2021-03-25T22:39:09","guid":{"rendered":"http:\/\/Enoteprime.com\/?p=13068"},"modified":"2021-04-13T00:29:18","modified_gmt":"2021-04-13T00:29:18","slug":"Enote-on-mathematics-ss2-third-term","status":"publish","type":"post","link":"https:\/\/enoteprime.com\/Enote-on-mathematics-ss2-third-term\/","title":{"rendered":"Enote on Mathematics SS2 Third Term"},"content":{"rendered":"\n

Third Term Mathematics Enotes – Enoteprime.com<\/a><\/h2>\n\n\n\n

<\/p>\n\n\n\n

THIRD TERM <\/strong><\/p>\n\n\n\n

SUBJECT:  MATHEMATICS  <\/strong><\/p>\n\n\n\n

CLASS: SS2<\/strong><\/p>\n\n\n\n

SCHEME OF WORK<\/strong><\/p>\n\n\n\n

WEEK<\/strong>S<\/td>   TOPIC<\/strong>S<\/td><\/tr>
1<\/td>Circle Theorem: Tangent properties of circle; Angles in alternate segment; Two tangent from a circle at external point.<\/td><\/tr>
2<\/td>Trigonometry:  Derivative of sine rule and cosine rule and their applications.<\/td><\/tr>
3<\/td>Bearing and Distances; Elevation and  Depression.<\/td><\/tr>
4<\/td>Statistics: Class boundaries, class mark, and cumulative frequencies of grouped data, and histogram.<\/td><\/tr>
5<\/td>Statistics: Cumulative  frequency curve (Ogive); Using ogive to calculate the median, quartile, percentile and decile.<\/td><\/tr>
6<\/td>Review of the First Half Term Work and Periodic Test.<\/td><\/tr>
7<\/td>Statistics: Mean, median, and mode of grouped data.<\/td><\/tr>
8<\/td>Probability: Introduction; use of dice, coins and playing cards.<\/td><\/tr>
9<\/td>Probability: Addition and multiplication rules of probability; Mutually exclusive, independent, and complementary events; Experiment with or without replacement.<\/td><\/tr>
10<\/td>Revision<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

REFERENCE BOOKS<\/strong><\/p>\n\n\n\n

1.New General Mathematics SSS2 by M.F. Macrae etal.<\/p>\n\n\n\n

2. Essential Mathematics SSS2 by A.J.S. Oluwasanmi.<\/p>\n\n\n\n

3. Exam Focus Mathematics.<\/p>\n\n\n\n

WEEK ONE<\/strong><\/p>\n\n\n\n

TOPIC:TANGENTS FROM AN EXTERNAL POINT<\/strong><\/p>\n\n\n\n

Theorem<\/strong>:<\/p>\n\n\n\n

The tangents to a circle from an external point are equal.<\/strong><\/p>\n\n\n\n

<\/td><\/tr>
<\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Given: a point T outside a circle, centre O, TA and TB are tangents to the circle at A and B.<\/p>\n\n\n\n

To prove<\/strong>:  |TA| = |TB|<\/p>\n\n\n\n

Construction<\/strong>:  Join OA, OB and OT<\/p>\n\n\n\n

In  \u2206s OAT and OBT<\/p>\n\n\n\n

OAT = OBT = 900<\/sup> (radius    tangent)<\/p>\n\n\n\n

|OA| = |OB|    (radii)<\/p>\n\n\n\n

|OT| =  |OT|    (common side)<\/p>\n\n\n\n

\u2206OAT = \u2206OBT (RHS)<\/p>\n\n\n\n

|TA| =  |TB|<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

Example:<\/strong><\/p>\n\n\n\n

1.In the figure below O is the centre of the circle and the TA and TB are tangents if <ATO = 390<\/sup>, calculate     < TBX<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

In  \u2206TAX<\/p>\n\n\n\n

AXT = 900<\/sup> (Symmetry)<\/p>\n\n\n\n

 TAX = 180 \u2013 (900<\/sup> + 390<\/sup>) sum of angles of ()<\/p>\n\n\n\n

1800<\/sup> \u2013 1290<\/sup> = 510<\/sup><\/p>\n\n\n\n

 TBX = 510<\/sup> (symmetry)<\/p>\n\n\n\n

OR<\/p>\n\n\n\n

\u2206 ATB is an Isosceles  triangle<\/p>\n\n\n\n

|AT| =|BT|    (tangents from external point)<\/p>\n\n\n\n

<ATO = <BTO  = 390<\/sup> (symmetry)<\/p>\n\n\n\n

< ATB = 2(39)  =  780<\/sup><\/p>\n\n\n\n

<TAX  = < TBX  (base angle of Isos \u2206 )<\/p>\n\n\n\n

2TBX = 1800 <\/sup>\u2013 780<\/sup> (sum of angle in a<\/p>\n\n\n\n

2 TBX   =    1020<\/sup><\/p>\n\n\n\n

     TBX= 1020<\/sup><\/p>\n\n\n\n

                  2<\/p>\n\n\n\n

TBX  =  510<\/sup><\/p>\n\n\n\n

2.PQR are three points on a circle Centre O. The tangent at P and Q meet at T. If   < PTQ = 620<\/sup> calculate PRQ.<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

Join OP and OQ<\/p>\n\n\n\n

In quadrilateral TQOP<\/p>\n\n\n\n

<OQT = <OPT = 900<\/sup> (radius 1 tangent)<\/p>\n\n\n\n

POQ = 3600<\/sup> \u2013 (900<\/sup> + 900<\/sup> + 620<\/sup>) sum of angle in a quadrilateral)<\/p>\n\n\n\n

POQ = 3600<\/sup> \u2013 2420<\/sup><\/p>\n\n\n\n

POQ = 1180<\/sup><\/p>\n\n\n\n

PRQ = 1180<\/sup> =  590<\/sup> (2x angle at circumference = angle at centre)<\/p>\n\n\n\n

              2<\/p>\n\n\n\n

PR1<\/sup>QR is a cyclic quadrilateral<\/p>\n\n\n\n

R + R1<\/sup> = 1800<\/sup> (opp. angles of  a cyclic quadrilateral )<\/p>\n\n\n\n

R1<\/sup> = 1800<\/sup> \u2013 R<\/p>\n\n\n\n

R1<\/sup> = 1800<\/sup> \u2013 590<\/sup><\/p>\n\n\n\n

R1 = 1210<\/sup><\/p>\n\n\n\n

 PRQ  =  590<\/sup>  or 1210<\/sup><\/p>\n\n\n\n

Evaluation<\/strong><\/p>\n\n\n\n

1.         ABC are three points on a circle, centre O such that <BAC = 370<\/sup>, the tangents at B and C meet at T. Calculate  < BTC.<\/p>\n\n\n\n

<\/td><\/tr>
<\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

GENERAL EVALUATION\/REVISION QUESTIONS<\/strong><\/p>\n\n\n\n

1. AB is a chord and O is the centre of a circle. If AOB = 780<\/sup> calculate the obtuse angle between AB and the tangent B.<\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n

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.<\/p>\n\n\n\n

2 The volume of a cylinder is 3600cm3 and its radius is 10cm calculate its <\/p>\n\n\n\n

(a) curve surface area<\/p>\n\n\n\n

(b) total surface area<\/p>\n\n\n\n

READING ASSIGNMENT<\/strong><\/p>\n\n\n\n

Essential Mathematics, pages149-151, numbers 11-20.<\/p>\n\n\n\n

WEEKEND ASSIGNMENT<\/strong><\/p>\n\n\n\n

Use the diagram below to answer the questions.<\/p>\n\n\n\n

<\/td><\/tr>
<\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

1.If  < ATO  =  360<\/sup> ,calculate < ABO.<\/p>\n\n\n\n

(a) 360  <\/sup>   (b) 720 <\/sup>     (c) 180<\/sup>     (d) 440<\/sup><\/p>\n\n\n\n

2.If <ABT  = 570<\/sup>, calculate   < AOT   (a) 1140<\/sup>     (b) 570 <\/sup>   (c) 330  <\/sup>   (d) 1230<\/sup><\/p>\n\n\n\n

3.If< BTO = 440<\/sup>, calculate <TAX   (a) 880<\/sup>   (b) 440<\/sup>    (c) 460<\/sup>   (d) 920<\/sup><\/p>\n\n\n\n

4.If   |AB| =  18cm and  |TB| = 15cm, calculate |TX|<\/p>\n\n\n\n

            (a) 180<\/sup>     (b) 330<\/sup>     (c) 780<\/sup>     (d) 120<\/sup><\/p>\n\n\n\n

5.If < AOT = 470<\/sup>, calculate ABO   (a) 470<\/sup>     (b) 940<\/sup>      (c) 1330<\/sup>       (d) 430<\/sup><\/p>\n\n\n\n

THEORY<\/strong><\/p>\n\n\n\n

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<\/sup>.calculate < ATC.<\/p>\n\n\n\n

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<\/sup>,find  the  size  of DAB in terms of   x.<\/p>\n\n\n\n

<\/p>\n\n\n\n

Third Term Mathematics Enotes – Enoteprime.com<\/a><\/p>\n\n\n\n

WEEK TWO<\/strong><\/p>\n\n\n\n

Topic: Cosine and Sine Rule Relating to Triangle<\/strong>.<\/p>\n\n\n\n

Content<\/strong><\/p>\n\n\n\n

-Sine Rule for Acute and Obtuse Angled Triangle.<\/p>\n\n\n\n

-Application of Sine Rule to Triangle.<\/p>\n\n\n\n

-Cosine Rule for Acute and Obtuse Angled Triangle.<\/p>\n\n\n\n

-Application of Cosine Rule.<\/p>\n\n\n\n

Sine Rule for Acute and Obtuse Angled Triangle.<\/strong><\/p>\n\n\n\n

Consideration is given to other triangles than a right angled triangle. The angles of any triangle are denoted by capital letters such as; A, B, C, while the sides are represented by small letters; a, b, c, respectively.                  <\/p>\n\n\n\n

A<\/p>\n\n\n\n

  c               b<\/p>\n\n\n\n

 B                  a                     C<\/p>\n\n\n\n

Acute Triangle: This is a type of triangle in which the angles are less than 900<\/sup>.<\/p>\n\n\n\n

Obtuse Triangle: Is a type of triangle in which one of the angles is more than 900<\/sup> but less than 1800<\/sup>.<\/p>\n\n\n\n

Deductive Proof of Sine Rule<\/strong><\/p>\n\n\n\n

The sine rule is the same for acute and obtuse angled triangle.<\/p>\n\n\n\n

Given: Any triangle ABC (acute-angled or obtuse-angled triangle).<\/p>\n\n\n\n

                                                              A<\/p>\n\n\n\n

                                                      c               b<\/p>\n\n\n\n

 B                  a                     C<\/p>\n\n\n\n

<\/td><\/tr>
<\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

To prove:        a       =  b       =       c<\/p>\n\n\n\n

  Sin A    SinB         Sin C  <\/p>\n\n\n\n

Construction: Draw a perpendicular A  to   BC ( produced, if necessary)<\/p>\n\n\n\n

 Proof:<\/p>\n\n\n\n

           Sin B  =  h     \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)<\/p>\n\n\n\n

                           c<\/p>\n\n\n\n

   In  fig. 7.10 a)<\/p>\n\n\n\n

 .         Sin C =  h     \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(2)<\/p>\n\n\n\n

                          b<\/p>\n\n\n\n

In   fig.7.10 b)<\/p>\n\n\n\n

            Sin(1800<\/sup> – C ) =  h<\/p>\n\n\n\n

  b<\/p>\n\n\n\n

Hence, Sin C = h    [sin(180 \u2013 \u019f ) = sin \u019f]   \u2026\u2026\u2026\u2026\u2026(2)        <\/p>\n\n\n\n

                          b<\/p>\n\n\n\n

From (1)   h = c SinB<\/p>\n\n\n\n

From(2)    h = b Sin C<\/p>\n\n\n\n

Hence, cSinB=bSinC<\/p>\n\n\n\n

  b   =   c   <\/p>\n\n\n\n

 SinB    SinC <\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

In triangle ABC, A= 380<\/sup>, B = 270<\/sup>, and b = 17cm. find a and c.<\/p>\n\n\n\n

Solution;<\/p>\n\n\n\n

        Using sine rule; Sin A = Sin B<\/p>\n\n\n\n

                                     a             b                                  <\/p>\n\n\n\n

   A<\/p>\n\n\n\n

            sin 380<\/sup>  = sin 270                                                                                  38<\/sup><\/p>\n\n\n\n

                a              17<\/p>\n\n\n\n

            a = 17 sin 380<\/sup><\/p>\n\n\n\n

                     sin 270<\/sup><\/p>\n\n\n\n

<\/td><\/tr>
<\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

a  =  23cm                            C                                                                                           B<\/p>\n\n\n\n

      To find  c; Angle C must be known; A + B + C = 1800<\/sup><\/p>\n\n\n\n

                                      C = 1800<\/sup> \u2013 380<\/sup> \u2013 270<\/sup> ,                  C = 1150<\/sup><\/p>\n\n\n\n

                       Sin B = Sin C<\/p>\n\n\n\n

                            b         c<\/p>\n\n\n\n

                     sin 270<\/sup> = sin 1150<\/sup> ,                          c =  17 x sin 1150<\/sup><\/p>\n\n\n\n

                        17          c                                              sin 270<\/sup><\/p>\n\n\n\n

c= 33.9 approximately, c = 34cm.<\/p>\n\n\n\n

NB: In any triangle, the longest side correspond to the largest angle while the shortest side corresponds to the smallest angle.<\/p>\n\n\n\n

Evaluation<\/strong>:                                                                                                                                    1. Solve the \u2206 completely; A = 390<\/sup>, a = 8.2m and b = 5.6m<\/p>\n\n\n\n

2.Calculate the values of angles P and R of \u2206 PQR, where q = 14.35cm, p = 7.82cm and Q = 115.60<\/sup><\/p>\n\n\n\n

Deductive Proof of Cosine Rule<\/strong><\/p>\n\n\n\n

The cosine rule is also the same for the acute and obtuse angled triangle.                        A<\/p>\n\n\n\n

Given: any         ABC                                                                                                      <\/p>\n\n\n\n

(a)       A<\/p>\n\n\n\n

(b)c<\/p>\n\n\n\n

bh<\/p>\n\n\n\n

                             <\/sup>c                                     <\/sup>b<\/p>\n\n\n\n

h<\/p>\n\n\n\n

                                                                                 B                                                        N<\/p>\n\n\n\n

B                                                                                            a                C           x<\/p>\n\n\n\n

       a-x                  N      x                                                            C     a+x<\/p>\n\n\n\n

                                  a<\/p>\n\n\n\n

To prove: c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCos C<\/p>\n\n\n\n

Construction: Draw a perpendicular from A to B(produced if necessary).<\/p>\n\n\n\n

Proof: Using the acute triangle;                                                  using the obtuse triangle;<\/p>\n\n\n\n

    c2<\/sup> = (a-x)2<\/sup> + h2<\/sup>                                                                         c2<\/sup> = (a +x)2<\/sup> + h2<\/sup><\/p>\n\n\n\n

    c2 = a2<\/sup> -2ax + x2<\/sup> + h2<\/sup>                                                                c2<\/sup> =a2<\/sup>+2ax+x2<\/sup>+h2<\/sup><\/p>\n\n\n\n

From \u2206 ACN; b2<\/sup> = x2<\/sup> + h2<\/sup>, and Cos C = x \/ b                 From     ACN,x2<\/sup>+h2<\/sup>=b2<\/sup><\/p>\n\n\n\n

           x= b Cos C                                                           =a2<\/sup>+2ax+b2<\/sup><\/p>\n\n\n\n

    c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ax                                                          From         ACN,x\/b=CosACN<\/p>\n\n\n\n

                           x =bCosC                                                             =Cos(1800<\/sup>-C)<\/p>\n\n\n\n

                                                                                                         =  -Cos C ,x=  -bCos C<\/p>\n\n\n\n

    c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCosC                                                          c2<\/sup> = a2<\/sup> + b2<\/sup> +2a(-bCos C)<\/p>\n\n\n\n

                                                                                                    c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCos C<\/p>\n\n\n\n

Similarly, for other sides and angles. Therefore the cosine rule can be written as thus:<\/strong><\/p>\n\n\n\n

             c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCosC                        OR         Cos C = a2<\/sup> + b2<\/sup> \u2013 c2<\/sup><\/strong><\/p>\n\n\n\n

a2<\/sup>= b2<\/sup> + c2<\/sup> \u2013 2bcCosA                                                            2ab<\/strong><\/p>\n\n\n\n

             b2<\/sup> = a2<\/sup> + c2<\/sup> \u2013 2acCosB                                        Cos A = b2<\/sup> + c2<\/sup> \u2013 a2<\/sup><\/strong><\/p>\n\n\n\n

 2bc<\/strong><\/p>\n\n\n\n

                                                                                           Cos B = a2<\/sup> + c2<\/sup> \u2013 b2<\/sup><\/strong><\/p>\n\n\n\n

  2ac<\/strong><\/p>\n\n\n\n

Conditions Necessary for Use<\/strong>: The rule is used for solving acute  and  obtuse angled triangles in which two sides and included angles are given.<\/p>\n\n\n\n

Example<\/strong>; Given that A = 1200<\/sup>, b = 7cm, c= 12cm. Solve the triangle completely.<\/p>\n\n\n\n

Solution<\/strong><\/p>\n\n\n\n

           Using cosine rule;             a2<\/sup> = b2<\/sup> + c2<\/sup> \u2013 2bcCosA         <\/p>\n\n\n\n

                                                     a2<\/sup>= 72<\/sup> + 122<\/sup> \u2013 2(7\u00d712)Cos 1200<\/sup><\/p>\n\n\n\n

                                                      a2<\/sup>= 49 + 144 \u2013 168 (-0.5)<\/p>\n\n\n\n

                                                      a2<\/sup> = 193 +84<\/p>\n\n\n\n

                                                       a =\u221a277<\/p>\n\n\n\n

                                                       a = 16.6cm.<\/p>\n\n\n\n

    To find angle B,     Cos B = a2<\/sup> + c2<\/sup> \u2013 b2<\/sup><\/p>\n\n\n\n

                                                      2ac<\/p>\n\n\n\n

                                    Cos B = 16.62<\/sup> + 122<\/sup> – 72<\/sup><\/p>\n\n\n\n

                                                     2x 16.6 x12<\/p>\n\n\n\n

                                     Cos B = 275.56 + 144 \u2013 49                <\/p>\n\n\n\n

                                                             398.4<\/p>\n\n\n\n

                                     Cos B = 0.9301,               B = cos-1<\/sup>0.9301,          B = 21.50<\/sup>.<\/p>\n\n\n\n

   To find < C;   < A + < B + <C = 1800<\/sup>,<\/p>\n\n\n\n

                                 C = 1800<\/sup> \u2013 1200<\/sup> \u2013 21.50<\/sup>,          C = 38.50<\/sup>.<\/p>\n\n\n\n

  Hence, a = 16.6cm, B = 21.50<\/sup> and C = 38.50<\/sup>.<\/p>\n\n\n\n

NB: 1. In any triangle, the longest side corresponds to the largest angle and the shortestside to the smallest angle<\/strong>.<\/p>\n\n\n\n

              2. It is advisable to always find the smallest angle first , since the angle must be acute<\/strong>.<\/p>\n\n\n\n

Evaluation<\/strong><\/p>\n\n\n\n

1.Calculate  the  angles  of  the \u2206s  ABC  whose  sides  are  given  in  centimeters.Give  the  final  answers  to the  nearest  0.10<\/sup><\/p>\n\n\n\n

a=5.2,  b = 6.5cm  ,c = 7.8<\/p>\n\n\n\n

General  Evaluation<\/strong><\/p>\n\n\n\n

1.Calculate the smallest angle in the triangle PQR such that p = 7.92m, q= 15.9m and c= 8.44m.<\/p>\n\n\n\n

2.Calculate the length of the side opposite the given angle in \u2206 XYZ given that x =13.1m, y = 24.2m and Z = 47.80.<\/p>\n\n\n\n

Revision Questions<\/strong><\/p>\n\n\n\n

1 Given that sin \u019f =5\/13 for 0<\u019f<900<\/sup> find<\/p>\n\n\n\n

a   sin\u019f -cos\u019f<\/p>\n\n\n\n

b   cos \u019f -3<\/p>\n\n\n\n

        tan\u019f<\/p>\n\n\n\n

2 If cos 3y=sin 2y find y for 0<y<900<\/sup><\/p>\n\n\n\n

Reading Assignment<\/strong><\/p>\n\n\n\n

Essential Mathematics SSS2, page 180-181, exercise13.2, nos 11-15;exercise 13.4,page 185,nos 1a-1f.<\/p>\n\n\n\n

Weekend Assignment<\/strong><\/p>\n\n\n\n

Objectives<\/strong><\/p>\n\n\n\n

Use the information below to answer question 1 \u2013 3. In \u2206ABC, a = 7.8m, b= 8.5m and B = 57.70<\/sup>. correct answers to 1 d.p.<\/p>\n\n\n\n

1. Find A;         A. 50.90<\/sup>     B. 510<\/sup>       C. 71.40<\/sup>    D. 700<\/sup><\/p>\n\n\n\n

2. Find C;         A. 510<\/sup>        B. 71.40<\/sup>    C. 710<\/sup>       D. 800<\/sup><\/p>\n\n\n\n

3. What is c?    A.10m       B.  12m     C. 9m       D. 9.5m<\/p>\n\n\n\n

4. In \u2206 ABC, b = 4cm, c= 5cm and A = 1150<\/sup>. Find a to 2 s.f.  A. 7.66cm B 7.6cm  C.8cm D.7.7cm   <\/p>\n\n\n\n

5. In \u2206PQR, p=1.8cm q = 2.5cm and r = 3.6cm. Calculate P.  A. 27.50<\/sup>   B. 300<\/sup>       C. 320<\/sup>    D.280<\/sup><\/p>\n\n\n\n

Theory<\/strong><\/p>\n\n\n\n

1.A triangle has sides of length 7cm, 8cm, 9cm. Express the cosine of the smallest angle of the triangle as a fraction in its lowest terms.<\/p>\n\n\n\n

2.Solve the triangle completely in the \u2206ABC such that B = 34.50<\/sup>, c = 2.8cm, \u2061\u203d<\/p>\n\n\n\n

a = 5.1cm.       <\/p>\n\n\n\n

<\/p>\n\n\n\n

<\/p>\n\n\n\n

Third Term Mathematics Enotes – Enoteprime.com<\/a><\/p>\n\n\n\n

WEEK THREE<\/strong><\/p>\n\n\n\n

TOPIC: BEARINGS AND DISTANCES<\/strong><\/p>\n\n\n\n