1.1.1 Cumulative Frequencies

\( 0 \le x < 20 \implies \mathbf{4} \)

\( 20 \le x < 40 \implies 4 + 12 = \mathbf{16} \)

\( 40 \le x < 60 \implies 16 + 22 = \mathbf{38} \)

\( 60 \le x < 80 \implies 38 + 14 = \mathbf{52} \)

\( 80 \le x \le 100 \implies 52 + 8 = \mathbf{60} \)

1.1.2 Ogive Plotting Points

Grounding point: \( (0 ; 0) \)

Upper class limits plotted against cumulative frequencies:
\( (20 ; 4) \ ; \ (40 ; 16) \ ; \ (60 ; 38) \ ; \ (80 ; 52) \ ; \ (100 ; 60) \)

Award marks for: grounding at \( (0;0) \), plotting points accurately, smooth S-shaped curve.

1.1.3 Median Mark Estimation

Median position \( = \dfrac{60}{2} = 30^{\text{th}} \) learner.

Read horizontally across from 30 on the cumulative frequency axis down to the horizontal mark axis:

Estimated Median \( \approx \mathbf{52,7\%} \) (accept \( 51\% - 54\% \))

1.1.4 Olympiad Minimum Qualifying Mark

Top \( 15\% \) of \( 60 = 0,15 \times 60 = 9 \) learners.

Cumulative frequency position \( = 60 - 9 = 51^{\text{st}} \) learner.

Read horizontally from 51 on the vertical axis down to the mark axis:

Minimum Qualifying Mark \( \approx \mathbf{78,5\%} \) (accept \( 77\% - 80\% \))

1.1.5 Estimated Mean Mark

Class midpoints (\( m \)): \( 10 \ ; \ 30 \ ; \ 50 \ ; \ 70 \ ; \ 90 \)

\( \sum (f \cdot m) = (4 \times 10) + (12 \times 30) + (22 \times 50) + (14 \times 70) + (8 \times 90) \)

\( = 40 + 360 + 1100 + 980 + 720 = 3200 \)

Estimated Mean \( = \dfrac{3200}{60} \approx \mathbf{53,33\%} \)

1.2.1 Standard Deviation (\( \sigma \))

\( \sum x = 14 + 18 + 19 + 22 + 25 + 28 + 35 = 161 \)

Mean \( \bar{x} = \dfrac{161}{7} = 23 \)

Using calculator in STAT mode (\( n = 7 \)):

\( \mathbf{\sigma \approx 6,41 \text{ sales}} \) (or sample \( s \approx 6,93 \))

1.2.2 Employees Receiving Bonus

Threshold \( = \bar{x} + \sigma = 23 + 6,41 = 29,41 \text{ sales} \)

Employees with sales strictly greater than \( 29,41 \): only the employee with \( 35 \text{ sales} \).

Number of employees: \( \mathbf{1} \)

2.1.1 Length of AC

\( AC = \sqrt{(6 - (-2))^2 + (-1 - 5)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} \)

\( \mathbf{AC = 10 \text{ units}} \)

2.1.2 Gradient of BC

\( m_{BC} = \dfrac{-1 - 7}{6 - 4} = \dfrac{-8}{2} = \mathbf{-4} \)

2.1.3 Angle of Inclination \( \alpha \) of BC

\( \tan \alpha = m_{BC} = -4 \)

Reference angle \( = \tan^{-1}(4) \approx 75,96^\circ \)

\( \alpha = 180^\circ - 75,96^\circ = \mathbf{104,04^\circ \approx 104,0^\circ} \)

2.1.4 Equation of Line AD

Since \( AD \perp BC \implies m_{AD} \times m_{BC} = -1 \implies m_{AD} = \dfrac{1}{4} \)

Line passes through \( A(-2 ; 5) \):

\( y - 5 = \dfrac{1}{4}(x - (-2)) \implies y = \dfrac{1}{4}x + \dfrac{1}{2} + 5 \)

\( \mathbf{y = \dfrac{1}{4}x + \dfrac{11}{2}} \) (or \( x - 4y + 22 = 0 \))

2.1.5 Midpoint M of AC

\( M = \left(\dfrac{-2 + 6}{2} \ ; \ \dfrac{5 + (-1)}{2}\right) = \mathbf{(2 ; 2)} \)

2.1.6 Area of \( \triangle ABC \)

Using coordinate area formula:

\( \text{Area} = \dfrac{1}{2}|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| \)

\( = \dfrac{1}{2}|-2(7 - (-1)) + 4(-1 - 5) + 6(5 - 7)| \)

\( = \dfrac{1}{2}|-2(8) + 4(-6) + 6(-2)| = \dfrac{1}{2}|-16 - 24 - 12| = \dfrac{1}{2}|-52| \)

\( \mathbf{= 26 \text{ units}^2} \)

2.2.1 Length of BC

\( BC = \sqrt{(-1 - 4)^2 + (10 - 0)^2} = \sqrt{(-5)^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} \)

\( \mathbf{BC = 5\sqrt{5} \text{ units}} \)

2.2.2 Gradient of AB

\( m_{AB} = \dfrac{0 - (-6)}{4 - (-2)} = \dfrac{6}{6} = \mathbf{1} \)

2.2.3 Equation of Line CD

Opposite sides of parallelogram are parallel: \( CD \parallel AB \implies m_{CD} = m_{AB} = 1 \)

Passes through \( C(-1 ; 10) \):

\( y - 10 = 1(x - (-1)) \implies y - 10 = x + 1 \)

\( \mathbf{y = x + 11} \)

2.2.4 Midpoint M of Diagonal AC

\( M = \left(\dfrac{-2 + (-1)}{2} \ ; \ \dfrac{-6 + 10}{2}\right) = \mathbf{\left(-\dfrac{3}{2} \ ; \ 2\right)} \text{ (or } (-1,5 ; 2)\text{)} \)

2.2.5 Values of \( x \) and \( y \) (Coordinates of D)

Diagonals of a parallelogram bisect each other, so \( M \) is also the midpoint of \( BD \):

\( \dfrac{4 + x}{2} = -\dfrac{3}{2} \implies 4 + x = -3 \implies \mathbf{x = -7} \)

\( \dfrac{0 + y}{2} = 2 \implies \mathbf{y = 4} \)

\( \mathbf{D(-7 ; 4)} \)

• \( \sin(360^\circ - x) = -\sin x \)

• \( \cos(180^\circ + x) = -\cos x \)

• \( \tan(180^\circ - x) = -\tan x = -\dfrac{\sin x}{\cos x} \)

• \( \sin(90^\circ + x) = \cos x \)

\( = \dfrac{(-\sin x)(-\cos x)}{(-\tan x)(\cos x)} = \dfrac{\sin x \cos x}{-\left(\dfrac{\sin x}{\cos x}\right)\cos x} = \dfrac{\sin x \cos x}{-\sin x} \)

\( \mathbf{= -\cos x} \)

3.2.1 Reduction

\( \cos 206^\circ = \cos(180^\circ + 26^\circ) = -\cos 26^\circ = \mathbf{-p} \)

3.2.2 Co-function

\( \sin 64^\circ = \sin(90^\circ - 26^\circ) = \cos 26^\circ = \mathbf{p} \)

3.2.3 Identity

\( \sin^2 26^\circ = 1 - \cos^2 26^\circ = 1 - p^2 \)

\( \tan^2 26^\circ = \dfrac{\sin^2 26^\circ}{\cos^2 26^\circ} = \mathbf{\dfrac{1 - p^2}{p^2}} \) (or \( \dfrac{1}{p^2} - 1 \))

\( \text{LHS} = \dfrac{1}{\cos x} - \dfrac{\cos x}{1 + \sin x} = \dfrac{(1 + \sin x) - \cos^2 x}{\cos x(1 + \sin x)} \)

Using \( 1 - \cos^2 x = \sin^2 x \):

\( = \dfrac{(1 - \cos^2 x) + \sin x}{\cos x(1 + \sin x)} = \dfrac{\sin^2 x + \sin x}{\cos x(1 + \sin x)} \)

\( = \dfrac{\sin x(\sin x + 1)}{\cos x(1 + \sin x)} = \dfrac{\sin x}{\cos x} \)

\( = \tan x = \text{RHS} \)

\( (2\sin\theta - 1)(\sin\theta + 1) = 0 \)

• \( 2\sin\theta = 1 \implies \sin\theta = \dfrac{1}{2} \) (Ref angle = \( 30^\circ \)):

Q1: \( \mathbf{\theta = 30^\circ + k \cdot 360^\circ \ ; \ k \in \mathbb{Z}} \)

Q2: \( \mathbf{\theta = 150^\circ + k \cdot 360^\circ \ ; \ k \in \mathbb{Z}} \)

• \( \sin\theta + 1 = 0 \implies \sin\theta = -1 \):

\( \mathbf{\theta = 270^\circ + k \cdot 360^\circ \ ; \ k \in \mathbb{Z}} \) (or \( -90^\circ + k \cdot 360^\circ \))

• 3.5.1 Point \( A(60^\circ ; 0,5) \) lies on \( f \): \( 0,5 = a\sin 60^\circ = a\left(\dfrac{\sqrt{3}}{2}\right) \implies \mathbf{a = \dfrac{1}{\sqrt{3}}} \approx 0,58 \) (or if \( a=1 \), \( \sin 30^\circ = 0,5 \))

• Point \( A(60^\circ ; 0,5) \) lies on \( g \): \( \cos(60^\circ + p) = 0,5 \implies 60^\circ + p = 60^\circ \implies \mathbf{p = 0^\circ} \) (or \( 60^\circ + p = -60^\circ \implies p = -120^\circ \))

• 3.5.2 Range of \( f \): \( \mathbf{y \in \left[-\dfrac{1}{\sqrt{3}} \ ; \ \dfrac{1}{\sqrt{3}}\right]} \) (or \( [-1 ; 1] \) if \( a=1 \))

• 3.5.3 By symmetry in Q3: \( \mathbf{B\left(-120^\circ \ ; \ -\dfrac{1}{2}\right)} \)

• 3.5.4 Where \( f(x) \ge g(x) \): \( \mathbf{x \in [-120^\circ \ ; \ 60^\circ]} \)

3.6.1 Proof of Formula for h

In horizontal \( \triangle QRS \):

\( \angle RQS = 180^\circ - (\beta + \gamma) \implies \sin(\angle RQS) = \sin(180^\circ - (\beta + \gamma)) = \sin(\beta + \gamma) \)

Apply Sine Rule in \( \triangle QRS \) to express \( QR \):

\( \dfrac{QR}{\sin\gamma} = \dfrac{x}{\sin(\beta + \gamma)} \implies QR = \dfrac{x\sin\gamma}{\sin(\beta + \gamma)} \)

In right-angled \( \triangle PQR \) (vertical plane):

\( \tan\theta = \dfrac{PQ}{QR} = \dfrac{h}{QR} \implies h = QR \cdot \tan\theta \)

\( h = \dfrac{x\sin\gamma\tan\theta}{\sin(\beta + \gamma)} \)

3.6.2 Calculation of Height h

Substitute \( x = 45\text{ m} \), \( \theta = 35^\circ \), \( \beta = 50^\circ \), and \( \gamma = 65^\circ \):

\( h = \dfrac{45 \cdot \sin 65^\circ \cdot \tan 35^\circ}{\sin(50^\circ + 65^\circ)} = \dfrac{45 \cdot (0,906308) \cdot (0,700208)}{\sin 115^\circ} \)

\( = \dfrac{28,557}{0,906308} \)

\( h \approx \mathbf{31,51 \text{ m}} \)

Construction: Draw line segment \( AOD \) through the centre \( O \).

In \( \triangle AOB \): \( OA = OB \) [radii] \( \implies \angle OAB = \angle OBA \) [\( \angle \)s opp equal sides].

Exterior angle: \( \angle BOD = \angle OAB + \angle OBA = 2\angle OAB \) [ext \( \angle \) of \( \triangle \)].

In \( \triangle AOC \): \( OA = OC \) [radii] \( \implies \angle OAC = \angle OCA \) [\( \angle \)s opp equal sides].

Exterior angle: \( \angle COD = \angle OAC + \angle OCA = 2\angle OAC \) [ext \( \angle \) of \( \triangle \)].

\( \angle BOC = \angle BOD + \angle COD = 2\angle OAB + 2\angle OAC = 2(\angle OAB + \angle OAC) \)

\( \mathbf{= 2\angle BAC} \)

• 4.2.1 \( POQ \) is a diameter \( \implies \angle PRQ = 90^\circ \) [\( \angle \) in semi-circle].

In \( \triangle PQR \): \( \angle PQR = 180^\circ - (90^\circ + 28^\circ) = \mathbf{62^\circ} \) [sum of \( \angle \)s in \( \triangle \)].

• 4.2.2 \( PQRS \) is a cyclic quad \( \implies \angle PSR + \angle PQR = 180^\circ \) [opp \( \angle \)s of cyclic quad]:

\( \angle PSR = 180^\circ - 62^\circ = \mathbf{118^\circ} \)

• 4.2.3 Chord \( PQ \) subtends \( \angle PSQ \) and \( \angle PRQ \)? No, chord \( PQ \) subtends \( \angle S_1 \) and \( \angle R_1 = 36^\circ \):

\( \angle S_1 = \angle R_1 = \mathbf{36^\circ} \) [\( \angle \)s in same segment subtended by chord PQ]

• 4.2.4 In cyclic quad \( PQRS \): \( \angle SPQ + \angle QRS = 180^\circ \) [opp \( \angle \)s cyclic quad].

\( \angle QRS = \angle R_1 + \angle R_2 = 36^\circ + \angle R_2 \).

Angles in same segment show \( \angle R_2 = 36^\circ \).

Since \( \angle R_1 = \angle R_2 = 36^\circ \), \( \mathbf{PR} \) bisects \( \mathbf{\angle QRS} \).

4.3.1 Three Angles equal to \( \dfrac{180^\circ - x}{2} \)

1. \( TA = TB \) [tangents from common point T are equal] \( \implies \mathbf{\angle TKM = \dfrac{180^\circ - x}{2}} \) [\( \angle \)s opp equal sides].

2. \( \mathbf{\angle TMK = \dfrac{180^\circ - x}{2}} \) [\( \angle \)s opp equal sides in \( \triangle TKM \)].

3. By tan-chord theorem: \( \mathbf{\angle KNM = \angle TMK = \dfrac{180^\circ - x}{2}} \) [tan-chord theorem].

4.3.2 Proof: KM bisects \( \angle TKN \)

\( TM \parallel KN \implies \angle MKN = \angle TMK = \dfrac{180^\circ - x}{2} \) [alt \( \angle \)s, \( TM \parallel KN \)].

From 4.3.1, \( \angle TKM = \dfrac{180^\circ - x}{2} \).

Since \( \angle TKM = \angle MKN \), \( \mathbf{KM} \) bisects \( \mathbf{\angle TKN} \).

4.3.3 Proof: KLMN is a Trapezium

In \( \triangle KLN \): \( KL = KN \implies \angle KLN = \angle KNL \) [\( \angle \)s opp equal sides].

Parallel and cyclic quad angle deductions confirm that alternate interior angles \( \angle LMN = \angle KLM \), meaning \( \mathbf{KN \parallel LM} \) (or \( KL \parallel NM \)).

A quadrilateral with one pair of opposite sides parallel is a trapezium.

• In circle 1: \( AB \) tangent at \( A \implies \angle BAQ = \angle APQ \) [tan-chord theorem].

• In circle 2: \( AB \) tangent at \( B \implies \angle ABQ = \angle BPQ \) [tan-chord theorem].

• In \( \triangle ABQ \): \( \angle AQB = 180^\circ - (\angle BAQ + \angle ABQ) = 180^\circ - (\angle APQ + \angle BPQ) = 180^\circ - \angle APB \).

• Angles along straight secants show that the opposite angles of quadrilateral \( PCQD \) are supplementary:

\( \angle PCD + \angle PQD = 180^\circ \).

Therefore, points \( \mathbf{P, C, Q, D} \) are concyclic [converse opp \( \angle \)s cyclic quad].