TOPICS:
TRIGONOMETRY (30%)
ANALYTICAL GEOMETRY (30%)
EUCLIDIAN GEOMETRY (40%)
TRIGONOMETRY (30%)
Q1) Consider the identity:
8 sin(180−𝑥)cos⁡(𝑥−360)
𝑠𝑖𝑛2 𝑥−𝑠𝑖𝑛2 (90+𝑥)
= −4𝑡𝑎𝑛2𝑥
1.1)Prove the identity
(6)
Solution:
1.2) For which value(s) of x in the interval 00 < 𝑥 < 1800 will the identity be undefined
(2)
Solution:
[8]
QUESTION 2
In the figure below, ACP and ADP are triangles with 𝐶̂ = 00 ; 𝐶𝑃 = 4√3; 𝐴𝑃 = 8⁡⁡𝑎𝑛𝑑⁡𝐷𝑃 = 4. PA bisects 𝐷𝑃̂𝐶.
Let 𝐶𝐴̂𝑃 = 𝑥⁡⁡𝑎𝑛𝑑⁡𝐷𝐴̂𝑃 = 𝑦
2.1) Show, by calculation, that 𝑥 = 600
(2)
Solution:
2.2) Calculate the length of AD
(4)
Solution:
2.3) Determine y
Solution:
(3)
[9]
QUESTION 3
In the diagram below, the graph of 𝑓(𝑥) = −2𝑐𝑜𝑠𝑥 is drawn for −1200 ≤ 𝑥 ≤ 2400 .
3.1)Draw the graph of 𝑔(𝑥) = 𝑠𝑖𝑛(𝑥 + 600 )⁡⁡𝑓𝑜𝑟⁡ − 1200 ≤ 𝑥 ≤ 2400 on the grid provided in the question paper. (3)
Solution:
3.2) Show that the equation: sin(𝑥 + 600 ) = −2𝑐𝑜𝑠𝑥 can be rewritten as 𝑡𝑎𝑛𝑥 = −4 − √3
(4)
Solution:
3.3) Hence, or otherwise, determine the solutions of the equation: sin(𝑥 + 600 ) = −2𝑐𝑜𝑠𝑥 in the interval
−1800 ≤ 𝑥 ≤ 1800
(3)
Solution:
3.4) Given 𝑠𝑖𝑛160 = 𝑝. Determine the following in terms of p, without using a calculator:
3.4.1) Sin196
Solution:
(2)
3.4.2) cos16
(1)
Solution:
OR
Using SOH-CAH-TOA in the right angled triangle
[13]
ANALYTICAL GEOMETRY (30%)
QUESTION 4
7
In the diagram, 𝐴, 𝐵(−6; −5)⁡𝑎𝑛𝑑⁡𝐶(8; −4) are points in the Cartesian plane. 𝐹 (3; 2) ⁡⁡𝑎𝑛𝑑⁡𝐺 are points on the line AC
such that 𝐴𝐹 = 𝐹𝐺. E is the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡⁡𝑜𝑓⁡𝐴𝐵.
4.1)Calculate:
4.1.1) The equation of AC in the form y = mx+c
Solution:
(4)
OR
4.1.2) The coordinates of G if the equation of BG is 7𝑥 − 10𝑦 = 8
(3)
Solution:
4.2) Show by calculation that the coordinates of A is (2; 5)
Solution:
(2)
4.3)Prove that 𝐸𝐹 || 𝐵𝐺
(4)
Solution:
4.4) Calculate the coordinates of D if ABCD is a parallelogram with D is in the first quadrant
(4)
Solution:
[17]
QUESTION 5
In the diagram, 𝑃(−4; 5)⁡𝑎𝑛𝑑⁡𝐾(0; −3) are the end points of the diameter of a circle with centre M. S and R are
respectively the x-and y- intercept of the tangent to circle at P. 𝜃 is the inclination of PK wit the positive x-axis.
5.1)Determine the coordinates of centre M
(2)
Solution:
−4+0 5+(−3)
; 2 )√
2
𝑀=(
then
𝑀 = (−2; 1) √
5.2) Write the equation of above circle in the form (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2
(4)
Solution:
1
1
1
𝑟 = 𝑃𝐾 = √(−4 − 0)2 + (5 + 3)2 = (4√5) = 2√5 √ and Centre M = (-2;1)
2
2
2
(𝑥 + 2)2 + (𝑦 − 1)2 = (2√5)2 = 20 √√√
5.3) Calculate the inclination angle of the line PK
(3)
Solution:
5.4) Determine the equation of the tangent to circle at K in the form y = mx + c
Solution:
(4)
[13]
EUCLIDIAN GEOMETRY (40%)
QUESTION 6
6.1)Use the information in the figure to show that ABC ||| DAC
(3)
Solution:
𝐵𝐶
𝐴𝐶
8
2
𝐴𝐵
10
2
𝐴𝐶
12
2
= 12 = 3 ;⁡ 𝐴𝐷 = 15 = 3 ;⁡ 𝐶𝐷 = 18 = 3 √ for numerator √ for denominator;
ABC ||| DAC (S.S.S) √
6.2) In the diagram, ABC and ACD are drawn. F and G are points on sides AB and AC respectively such that
AF = 3x; FB = 2x; AG = 12y and GC = 8y. H, E and K are points on side AD such that GH // CK and GE // CD.
Prove that:
6.2.1) FG // BC
Solution:
𝐴𝐹
𝐹𝐵
=
3𝑥
2𝑥
6.2.2)
3
2
𝐴𝐻
𝐴𝐸
= 𝐸𝐷
𝐻𝐾
= ⁡⁡⁡𝑎𝑛𝑑⁡
(2)
𝐴𝐺
𝐺𝐶
=
12𝑦
8𝑦
=
3
2
√ so FG || BC ( converse of proportionality theorem) √
(3)
Solution:
𝐴𝐻
𝐻𝐾
= 𝐺𝐶 ⁡(𝑝𝑟𝑜𝑝. 𝑡ℎ𝑚, 𝐺𝐻 ∕∕ ⁡𝐶K) √
𝐴𝐺
𝐴𝐸
𝐸𝐷
= 𝐺𝐶 ⁡(𝑝𝑟𝑜𝑝. 𝑡ℎ𝑚.,⁡ GE || CD) √;
𝐴𝐺
𝐴𝐻
𝐻𝐾
𝐴𝐸
= 𝐸𝐷 √
[8]
QUESTION 7
In the diagram below, points A; B; D and C lie on a circle. CF || AB with E on AD produced. Chords CB and AD intersect at F.
̂1 = 150 .
̂2 = 500 ⁡⁡⁡𝑎𝑛𝑑⁡𝐶
𝐷
7.1) Calculate, with reasons, the size of:
7.1.1) 𝐴̂
Solution:
(3)
̂2
7.1.2) 𝐶
Solution:
(2)
7.2) Prove, with a reason, that CF is a tangent to circle passing through points C; D and E.
Solution:
(2)
[7]
QUESTION 8
8.1) Complete the statements below by filling in the missing word(s) to make the statement correct.
8.1.1) Opposite interior angles of a cyclic quadrilateral is ………………………………..
(1)
Solution:
Supplementary
8.1.2) If the square of the longest side in a triangle is equal to the sum of the squares of the other two sides then
the triangle is………………………………………………………………
(1)
Solution:
Right-angled
8.2) In the diagram below O is the center of the circle. PQRS is cyclic quadrilateral. Prove the theorem which states that:
𝐴̂ + 𝐶̂ = 1800 .
NOTE: Draw diagram in the answer book if necessary.
(5)
Solution:
OR
8.3) In the figure below, 𝑂𝐴 ⊥ 𝐵𝐶; ⁡⁡⁡⁡𝐴𝐾 = 2⁡𝑐𝑚⁡⁡⁡𝑎𝑛𝑑⁡𝐾𝐶 = 4⁡𝑐𝑚.
Find, with reason, the length of:
8.3.1) BK
(2)
Solution:
BK= 4 √ (line from center ⊥ the chord) √
8.3.2) Radius of circle
(3)
Solution: OC = r; OK = r – 2
OC2 = OK2 + KC2 (pyth. Thm ) √
 r2 = (r – 2)2 +16 √
 r2 = r2 – 4r +4+16 then 4r = 20 and then r = 10 √
[12]
QUESTION 9
In the diagram below, BC is a diameter of the circle. The tangent at point D on the circle meets CB produced at A. CD is
produced to E such that EA 𝐸𝐴 ⊥ 𝐴𝐶. BD is drawn. Let 𝐶̂ = 𝑥.
9.1)Give a reason why:
̂3 = 900
9.1.1) 𝐷
(1)
Solution:
9.1.2) ABDE is a cyclic quadrilateral
Solution:
(1)
̂2 = 𝑥
9.1.3) 𝐷
(1)
Solution:
̂1
9.2) Find two other two angles, with reasons, equal to 𝐵
(3)
Solution:
̂1 = 90 − 𝑥⁡⁡⁡(𝑆𝑢𝑚⁡𝑜𝑓⁡ <′ 𝑠⁡𝑖𝑛⁡) √
𝐵
̂1 = 180 − 𝐷
̂2 − 𝐷
̂3 = 180 − 𝑥 − 90 = 90 − 𝑥⁡⁡(<′ sin 𝑡ℎ𝑒⁡𝑠𝑡𝑟. 𝑙𝑖𝑛𝑒) √
𝐷
𝐸̂ = 180 −A-C= 180-90 – x =90 – x (Sum of <’s in  ) √
̂1 = 𝐷
̂1 = ⁡ 𝐸̂
𝐵
9.3) Hence, give a reason why: AD = AD
(1)
Solution:
AD = AE ( sides opp equal <’s)
√
9.4) Prove that  ADB |||  ACD
(4)
Solution:
9.5)It is further given that BC = 2AB = 2r. Hence, prove that 𝐴𝐷 2 = 3𝑟 2
(2)
Solution:
𝐴𝐷
𝐴𝐶
=
𝐴𝐵
⁡
𝐴𝐷
(III ’s; sides in ratio) √
 AD2 = ACAB = 2rr = 3 r2  AD = 𝑟√3 √
[13]
TRIGONOMETRY LINE CIRCLE EUCLIDIAN GEO
GEO EQN
Q1 Q2 Q3 Q4
Q5
Q6 Q7 Q8 Q9 TOTAL
8
9
13 17
13
8
7
12 13 100
30%
30%
30%
NAME, SURNAME:………………………………………………………………………………………………………………………………………… GRADE: 11…….
For Q3.1)
For Q8.2)