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\(\frac {\sin\theta}{\cos\theta}\) \(\frac{\cos \theta {\frac{d(\sin \theta)}{d \theta}} - \sin \theta {\frac{d(\cos \theta)}{d \theta}}}{\cos^2 \theta}\) \(\frac{\cos \theta. \cos \theta - \sin \theta (-\sin \theta)}{cos^2\theta}\) \(\frac{cos^2\theta + \sin^2 \theta}{cos^2\theta}\) Recall that sin2 \(\theta\) + cos2 \(\theta\) = 1 \(\frac{1}{\cos^2\theta}\) = sec2 \(\theta\)

cos xo = \(\frac{1}{r}\); \(\sqrt{r^2 - 1}\) By Pythagoras r2 = 12 + x2 - 1 x = \(\sqrt{r^2 - 1}\) tan xo = \(\sqrt{r^2 - 1}\) = \(\sqrt{r^2 - 1}\)

Each interior angle = \(\frac{360^o}{n}\) 45o = \(\frac{360^o}{n}\) 45on = 360o n = \(\frac{360^o}{45}\) = 8

The bearing S 20^\circ W is equivalent to a bearing of 200^\circ. Moving clockwise 90^\circ ⇒ 200^\circ + 90^\circ = 290^\circ This is equivalent to N 70^\circ W.

A prism has 3 rectangular faces and 2 triangular faces and 2 rectangular faces = 10(3 + 4 + 5) = 120 Area of triangular faces = \(\sqrt{s(s - a) (s - b) (s - c)}\) where s = \(\frac{a + b + c}{2}\) = \(\frac{3 + 4 + 5}{2}\) = \(\frac{12}{2}\) = 6 Area of \(\bigtriangleup\) = \(\sqrt{6(6 - 30(6 - 4)(6 - 5)}\) = \(\sqrt{6 \times 3 \times 2 \times 1}\) = 6 Area of triangle faces = 2 x 6 = 12cm2 Total surface area = Area of rectangular face + Area of \(\bigtriangleup\) = 120 + 12 = 132cm2

No expanded working has been attached for this question yet. The verified answer is Option C.

Perimeter of sector = 2r + \(\frac{\theta}{360^o} \times 2\pi r\) \(\pi + 8 = 2 \times 4 + \frac{\theta}{3360^o} \times 2 \pi \times 4\) \(\pi + 8 + \frac{\theta}{360^o} \times 8 \pi\) P + 8 - 8 = \(\frac{\theta \pi}{456o}\) \(\pi = \frac{\theta \pi}{45^o}\) \(\theta \pi = 45^o\)

volume of cylinder = \(\pi r^2\)h 0.216\(\pi\) m\(^3\) = \(\pi \times r^2 \times h\) Since r = h, 0.216 = r\(^3\) r\(^3\) = 0.216 r = \(\sqrt[3]{0.216}\) = 0.6

No expanded working has been attached for this question yet. The verified answer is Option C.

A = L x B = (18.6 x 12.5)m2 = 232.5m2 = 233m2

tan \(\theta\) = \(\frac{9}{50}\) = 0.18 \(\theta = tan^{-1} 0.18\) \(\theta\) = 10.20

It is angle PKS, angles in the same segment. Therefore, PKS is congruent with QKR

o Also o ( o x + R = 360 - 240 = 120; Since x = R x + x = 120 2x = 120 Since x = R x + x = 120 2x = 120 x = \(\frac{120}{2}\) = 60o

x + 52o = 90 x = 90 - 52 x = 38o k = opposite angle Z k = 38o y + k = 90o y + 38o = 90o y = 90o - 38o y = 52o y = x = 180o(sum of angles on straight line) 52 + x = 180o x = 180 - 52 x = 128

o = 180o (o - 110o = 70o In \(\bigtriangleup\) XTU, o i.e. o = 180 o = 70o Also o = 180 o in \(\bigtriangleup\) SYT + o(Sum of interior o - 60 2(o TYW = \(\frac{300}{2}\) = 150o < SYT