Question Library

Length = \(\frac{\theta}{360}\) x 2\(\pi\)r = \(\frac{72}{360} \times 2 \frac{22}{7} \times \frac{0.5}{3.5}\) = 4.4

L = \(\frac{\theta}{360} \times 2 \pi r\) 22 = \(\frac{\theta}{360} \times 2 \times \frac{22}{7} \times 15\) \(\theta = \frac{22 \times 360 \times 7}{2 \times 22 \times 15}\) \(\theta\) = 84\(^o\)

New are = (2 x 15) x (2 x \(x\)) = 60cm\(^2\)

S\(O\)T = 180\(^o\) - (50 + 50) = 80\(^o\) P\(Q\)R = 80\(^o\) = \(\frac{360^o - 160^o}{2}\) = 100\(^o\)

3\(x^o\) + 2\(x^o\) + 4\(x^o\) + 3\(x^o\) + 6\(x^o\) = 540\(^o\) \(\frac{18x^o}{18} = \frac{540^o}{18}\) \(x^o\) = 30\(^o\) Smallest angle = 2 x 30\(^o\) = 60\(^o\)

y\(^2\) = 3\(^2\) + 4\(^2\) y\(^2\) = 9 + 16 y = \(\sqrt{25}\) = 5 \(\frac{\cos x}{2 \sin x} = \frac{4}{5} + 2 \times \frac{3}{5}\) = \(\frac{4}{5} \times \frac{5}{6}\) = \(\frac{2}{3}\)

Sin\(\theta = \frac{20}{75}\) \(\theta = sin^{-1} (\frac{20}{75})\) = 15\(^o\)

Z\({O}\)T = 2 x 70\(^o\) 140\(^o\) Length = \(\frac{\theta}{360}\) x 2\(\pi\)r \(\frac{140^o}{360} \times 2 \times \frac{22}{7}\) x 18 = 44cm

∠QRT = ∠SRT = 52 (opposite angles) ∠SRT = y = 52 (alternate angles) y = 52\(^o\)

50\(^o\) + m = 105\(^o\) (exterior angle of \(\triangle\) m = 105\(^o\) - 50\(^o\) m = 55\(^o\)

In \(\triangle\)PQR |PR|\(^2\) = |PQ|\(^2\) + |QR|\(^2\) = 3\(^2\) + 4\(^2\) |PR|\(^2\) = 9 + 16 |PR| = \(\sqrt{25}\) = 5cm In \(\triangle\)PRS |PS|\(^2\) = |PR|\(^2\) + |RS|\(^2\) 13\(^2\) = 5\(^2\) + |RS|\(^2\) 169 =25 + |RS|\(^2\) |RS|\(^2\) = 169 - 25 = 144 |RS| = \(\sqrt{144}\) = 12cm Area of quadrilateral PQRS = Area of \(\triangle\)PQR + Area of \(\triangle\)PQR + Area of \(\triangle\)PRS = (\(\frac{1}{2} \times 3 \times 4\)) + (\(\frac{1}{2} \times 12 \times 5\)) = 6cm\(^2\) + 30cm\(^2\) = 36cm\(^2\)

F \(\infty\) \(\frac{m_1m_ 2}{d^2}\) F = \(\frac{km_1m_2}{d^2}\) 20 = \(\frac{k(25)(10)}{(25)}\) \(\frac{10k}{10} = \frac{20}{10}\) k = 2 (i) F = \(\frac{2m_1m_2}{d_2}\) (ii) Since F = \(\frac{2m_1m_2}{d_2}\) 30 = \(\frac{2(7.5)(4)}{d^2}\) 30 \(\frac{60}{d^2}\) d\(^2\) = \(\frac{60}{30}\) d\(^2\) = 2 d = \(\sqrt{2}\)m (b) Sum of angles = (n - 2) 180 = (5 - 2) 180 = 540\(^o\) x + x + 20 + x + 60 + x + 40 + x + 80 = 540 5x + 200 = 540 5x = 540 - 200 \(\frac{5x}{5} = \frac{340}{5}\) x = 68\(^o\)

(a) When x = 60\(^o\) y = 2cos 60\(^o\) + 3sin 60\(^o\) y = 3.6 When x = 120\(^o\) y = 2 cos 120\(^o\) + 3 sin 180\(^o\) y = 1.6 When x = 180\(^o\) y = 2 cos 180\(^o\) + 3 sin 180\(^o\) y = 2 When x = 300\(^o\) y = 2 cos 300\(^o\) + 3 sin 300\(^o\) y = 1.6 when x = 360\(^o\) y = 2 cos 360\(^o\) + 3 sin 360\(^o\) y = 2 (b) (c)(i) 1555.5\(^o\), 318\(^o\) (ii) When x = 3.42\(^o\), y = 0.9\(^o\)

T\(O\)S = 68\(^o\) x 2 = 136\(^o\) (Angle at centre is twice angle at circumference) T\(R\) = 180\(^o\) - (90 + 68\(^o\)) = 180\(^o\) - 158\(^o\) = 22\(^o\) x = 2 x T\(R\)O = 2 x 22\(^o\) = 44\(^o\) (b) Let the capacity of Tank B = x litres Capacity of Tank A = x + 600 litres If 100 litres is pumoed out of each tank then, (x + 600) - 100 = 3(x - 100) x + 500 = 3x - 300 500 + 300 = 3x - x \(\frac{800}{2} = \frac{2x}{x}\) x = 400 litres Capacity of Tank A = 400 + 600 = 1,000 litres Capacity of Tank B = 400 litres

In the diagram, Thus, m = 2 x 55\(^o\) (is a bisector of obtuse m = 110\(^o\) n = \(\frac{1}{2}\) x 110\(^o\) (angle at centre = 2 x angle at circum) n = 55\(^o\) m + n = 110 + 55 = 165\(^o\)