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\(\frac {\sin A}{a} = \frac {\sin B}{b} = \frac {\sin C}{c}\) \(\implies \frac {\sin 82^0}{43.2} = \frac {\sin 56^0}{AB}\) \(\implies AB \times \sin 82^0 = 43.2 \times \sin 56^0\) \(\therefore AB = \frac {43.2 \times \sin 56^0}{\sin 82^0}\) = 36.2cm (to 3 s.f)

Consider ∆XOB and using Pythagoras theorem 13\(^2\) = 12\(^2\) + h\(^2\) ⇒ 169 = 144 + h\(^2\) ⇒ 169 - 144 =h\(^2\) ⇒ 25 = h\(^2\) ⇒ h = \(\sqrt25\) = 5cm tan \theta = \(\frac {opp}{adj}\) ⇒ tan \theta = \(\frac{12}{5}\) = 2.4 ⇒ \theta = tan\(^{-1}\)(2.4) ⇒ \theta = 67.38\(^0\) ∠AOB = 2\theta = 2 x 67.38\(^o\) = 134.76\(^o\) L = \(\frac{\theta}{360^o} \times 2\pi r\) ⇒ L = \(\frac {134.76}{360} \times 2 \times \frac {22}{7} \times 13 = \frac {77082.72}{2520}\) ∴ L = 30.6cm (to 3 s.f)

An exterior angle of a n-sided regular polygon = \(\frac{360}{n}\) For (n - 1) sided regular polygon = \(\frac{360}{n - 1}\) For (n + 2) sided regular polygon = \(\frac{360}{n + 1}\) ⇒ \(\frac{360}{n - 1} - \frac{360}{n + 2}\) = 6 9Given) ⇒ \(\frac{360(n + 2) - 360(n - 1)}{(n - 1)(n + 2)}\) ⇒ \(\frac{360n + 720 - 360n + 360}{(n - 1)(n + 2)}\) ⇒ \(\frac{1080}{(n - 1)(n + 2)} = \frac{6}{1}\) ⇒ 1080 = 6 (n - 1)(n + 2) ⇒ 180 = (n - 1)(n + 2) ⇒ 180 = n\(^2\)+ 2n - n - 2 ⇒ 180 = n\(^2\) + n - 2 ⇒ n\(^2\)b+ n - 2 - 180 = 0 ⇒ n\(^2\) + n - 182 = 0 ⇒ n\(^2\) + 14n - 13n - 182 = 0 ⇒ n (n + 14) - 13 (n + 14) = 0 ⇒ (n - 13) (n + 14) = 0 ⇒ n - 13 = 0 or n + 14 = 0 ⇒ n = 13 or n = -14 ∴ n = 13 (We can't have a negative number of side)

\(PR^2 = PQ^2 + RQ^2 - 2(PQ)(RQ)cos Q\) \(\implies cos Q = \frac{PQ^2 + RQ^2 - PR^2}{2(PQ)(RQ)}\) \(\implies cos Q = \frac{8^2 + 5^2 - 7^2}{2\times8\times5}\) \(\implies cos Q = \frac{64 + 25 - 49}{80}\) \(\implies cos Q = \frac{40}{80} = 0.5\) \(\implies Q = cos^{-1} (0.5) = 60^0\) \(\therefore x = 60^0\)

Intersecting Chords Theorem states that If two chords intersect in a circle, then the products of the measures of the segments of the chords are equal. ⇒ AE * EB = CE * ED ⇒ 6 * \(x\) = 4 * (\(x\) + 5) ⇒ 6\(x\) = 4\(x\) + 20 ⇒ 6\(x\) - 4\(x\) = 20 ⇒ 2\(x\) = 20 ∴ \(x = \frac{20}{2}\) = 10 units

∠PQR + ∠PSR = 180o (opp. angles of cyclic quad. are supplementary) ⇒ 5\(x\) - \(y\) + 10 + (-2\(x\) + 3\(y\) + 145) = 180 ⇒ 5\(x\) - \(y\) + 10 - 2\(x\) + 3\(y\) + 145 = 180 ⇒ 3\(x\) + 2\(y\) + 155 = 180 ⇒ 3\(x\) + 2\(y\) = 180 - 155 ⇒ 3\(x\) + 2\(y\) = 25 ----- (i) ∠QPS + ∠QRS = 180o (opp. angles of cyclic quad. are supplementary) ⇒ -4\(x\) - 7\(y\) + 150 + (2\(x\) + 8\(y\) + 105) = 180 ⇒ -4\(x\) - 7\(y\) + 75 + 2\(x\) + 8\(y\) + 180 = 180 ⇒ -2\(x\) + \(y\) + 255 = 180 ⇒ -2\(x\) + y = 180 - 255 ⇒ -2\(x\) + \(y\) = -75 ------- (ii) ⇒ \(y\) = -75 + 2\(x\) -------- (iii) Substitute (-75 + 2\(x\)) for \(y\) in equation (i) ⇒ 3\(x\) + 2(-75 + 2\(x\)) = 25 ⇒ 3\(x\) - 150 + 4\(x\) = 25 ⇒ 7\(x\) = 25 + 150 ⇒ 7\(x\) = 175 ⇒ \(x = \frac{175}{7} = 25\) From equation (iii) ⇒ \(y\) = -75 + 2(25) = -75 + 50 ⇒ \(y\) = -25 ∴ \(x\) + \(y\) = 25 + (-25) = 0

On the \(y\)-axis, each box is \(\frac{1 - 0}{5} = \frac{1}{5}\) = 0.2unit On the \(x\)-axis, each box is \(\frac{90 - 0}{6} = \frac{90}{6} = 15^o\) ⇒ \(\theta_1 = 180^o + (2.5\times15^o) = 180^o + 37.5^o = 217.5^o ≃ 218^o \)(2 and half boxes were counted to the right of 180\(^o\)) ⇒ \(\theta_2 = 270^o + (3.5\times15^o) = 270^o + 52.5^o = 322.5^o ≃ 323^o \)(3 and half boxes were counted to the right of 270\(^o\)) ∴ \(\theta = 218^o, 323^o\)

|AP| = |PB| = \(x\) (The perpendicular to a chord bisects the chord if drawn from the center of the circle.) From ∆OPB Using Pythagoras theorem ⇒ \(13^2 = 12^2 + x^2\) ⇒ \(169 = 144 + x^2\) ⇒ \(169 - 144 = x^2\) ⇒ \(x^2 = 25\) ⇒ \(x = \sqrt25 = 5 cm\) ∴ Length of the chord |AB| = \(x + x = 5 + 5 = 10 cm\)

∠SOR = 2 \times 28^\circ = 56^\circ (angle at the centre is twice the angle at the circumference) From ∆SOR |OS| = |OR| (radii) So, ∆SOR is isosceles. ∠ORS = \(\frac{180^0 - 56^0}{2} = \frac{124^0}{2}\) ( base angles of isosceles triangle are equal) ∴ ∠ORS = 62^\circ

Reflex ∠MOR = 2 \times 124^\circ = 248^\circ (angle at the centre is twice the angle at the circumference) ∠MOR = 360^\circ - 248^\circ = 112^\circ (sum of angle at a point is 360^\circ) ∠OMN = 360^\circ - (124^\circ+ 48^\circ + 112^\circ) (sum of angles in a quadrilateral is 360^\circ) = 360^\circ - 284^\circ ∴ ∠OMN = 76^\circ

∠RMN = 90^\circ (angles in a semicircle is a right angle) ∠SRN = ∠RMN + ∠MNR (ext. angle of a ∆ is equal to the sum of two opp. int. angles) ⇒ 5x + 20 = 90 + x ⇒ 5x - x = 90 - 20 ⇒ 4x = 70 x = \(\frac{70}{4}\) = 17.4^\circ therefore, 2x = 2\(\times17.5\) = 35^\circ

t = ∠NTS (the angle between a tangent and a chord is equal to the angle in the alternate segment). Therefore, t = 36^\circ

a. 2m = 2 \times n (the angle at the centre is twice the angle at the circumference) ∴ m = n ∠SOR = ∠QRO = 54^\circ (alternate angles are equal) From ∆ROQ Since |OQ| = |OR| (radii), therefore, ∆ROQ is an isosceles triangle 2m = 180^\circ - 2(54^\circ) = 180^\circ - 108^\circ (sum of angles in a ∆ is 180^\circ) 2m = 72^\circ m = \(\frac{72}{2}\) = 36^\circ Recall: m = n ∴ m = n = 36^\circ b. Let the length of the rectangle be L and the width be W L = W + 4 (given) Perimeter = 2(L + W) 40 = 2(L + W) 40 = 2((W + 4) + W) (Recall: L = W + 4) 40 = 2(2W + 4) 40 = 4W + 8 40 - 8 = 4W 32 = 4W W = \(\frac{32}{4}\) W = 8cm L = W + 4 = 8 + 4 L = 12cm Area of a rectangle = L x W = 12 \times 8 = \(96 cm^2\)

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\(402_y = 102_ten\) ⇒ 4 x y\(^2\) + 0 x y\(^1\) + 2 x y\(^0\) = 102 ⇒ 4y\(^2\) + 0 + 2 x 1 = 102 ⇒ 4y\(^2\) + 2 = 102 ⇒ 4y\(^2\) = 102 - 2 ⇒ 4y\(^2\) = 100 ⇒ y\(^2 = \frac {100}{4}\) ⇒ y\(^2\) = 25 ∴ y = \sqrt25 = 5