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(cos 40^\circ - sin 30^\circ) = 0.7660 - 0.5 = 0.2660

tan θ - 7/3.5 = 2 tan θ= 2.0000 θ = 63o

tan21o =x/10 x = 10 tan 21o = 3.3839 = 3.8m

< YXS = 74^\circ (alternate angles, N\(_1\) S\(_1\) // N\(_2\) S\(_2\)) N\(_2\) E\(_2\) S\(_2\) = 180^\circ \(\therefore\) Bearing of Y from X = 180^\circ + 74^\circ = 254^\circ

\(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) Simplifying from the innermost radical and progressing outwards we have the given expression \(\sqrt{160r^2 + \sqrt{71r^4 + \sqrt{100r^8}}}\) = \(\sqrt{160r^2 + \sqrt{71r^4 + 10r^4}}\) = \(\sqrt{160r^2 + \sqrt{81r^4}}\) \(\sqrt{160r^2 + 9r^2}\) = \(\sqrt{169r^2}\) = 13r

9(x + y)2 - 4(x - y)2 Using difference of two squares which says a2 - b2 = (a + b)(a - b) = 9(x + y)2 - 4(x - y)2 = [3(x + y)]2 - [2(x - y)]-2 = [3(x + y) + 2(x - y) + 2(x - y)][3(x + y) - 2(x - y)] = [3x +3y + 2x - 2y][3x + 3y - 2x + 2y] = (5x + y)(x + 5y)

Perimeter. of a rectangle = 2(L + B) = 24 Area of a rectangle = L x B = 35 From above, L + B = 12 L = 12 - B L x B = 35(given) 12 - B)B = 35 12B - B\(^2\) = 35 B\(^2\) - 12B + 35 = 0 B\(^2\) - 7B - 5B + 35 = 0 B(B - 7) - 5(B - 7) = 0 (B-5)(B -7) = 0 B = 5m OR 7m

Let his assistant work for x days ∴ his master worked (x + 10) day. Amount received by master = 40(x + 10), amount got by his assistance = 10x Total amount collected = N2000.00 ∴ 40(x + 10) + 10x = 2000 = 40x + 400 + 10x = 2000 50x + 400 = 2000 50x = 2000 - 400 50x = 1600 x = \(\frac{1600}{50}\) x = 32 days

Cos \(\theta\) = \(\frac{12}{13}\) x\(^2\) + 12\(^2\) = 13\(^2\) x\(^2\) = 169- 144 = 25 x = \(\sqrt{25}\) = 5 Hence, tan\(\theta\) = \(\frac{5}{12}\) and cos\(\theta\) = \(\frac{12}{13}\) If cot\(^2\)\(\theta\) = 1 + \(\frac{1}{tan^2\theta}\) = 1 + \(\frac{1}{\frac{(5)^2}{12^2}}\) = 1 + \(\frac{1}{\frac{25}{144}}\) = 1 + \(\frac{144}{25}\) = \(\frac{25 + 144}{25}\) = \(\frac{169}{25}\)

Since PS = QS = RS S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius QPR \(\pm\) \(\frac{100^o}{2}\) = 50o

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

yzx + 105o + 30o = 180o yzx = 180 - 155 = 45o \(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\) yz = \(\frac{8 \sin 30}{sin 45}\) = 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\) = 4 \(\div\) \(\frac{1}{\sqrt{2}}\) = 4\(\sqrt{2}\)cm

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

(a) Given any triangle ABC To prove that < A + < B + < C = 180^\circ Construction: Produce BC to X and draw CP // BA. Proof: with the lettering \(a_{1} = a_{2}\) (alternate angles); \(b_{1} = b_{2}\) (corresponding angles) \(< BCA + < A + < B = 180^\circ ; < ACB + < A + < B = 180^\circ\) \(\therefore < A + < B + < C = 180^\circ\) (b) From the given triangle and with reference to letterings; \(\hat{a} + \hat{c} + \hat{c} = \hat{b}\) (exterior opposite angles of a triangle) \(a + c = e\) (exterior opposite angles of a triangle) \(a + b = a + a + c + c \implies a + b = 2(a + c)\) = \(2e \text{where e = a + c}\) i.e \(< CAE + < CBD = 2 < CED\) (Proved)

(i) From the figure, \(\sin 30^\circ = \frac{4}{RT} \implies RT = \frac{4}{\sin 30^\circ}\) = \(\frac{4}{0.5} = 8m\) (ii) |RQ| = |XY| \(\therefore |XY|^{2} = 8^{2} - 4^{2} = 64 - 16 = 48\) \(|XY| = \sqrt{48} = 4\sqrt{3} m = 6.928m\) But |PY| = |PX| + |XY| = 20 + 6.928 = 26.928m \(\approxeq\) 26.9m. (iii) Let the angle of elevation = \(\theta\) \(\tan \theta = \frac{TY}{PY} = \frac{7}{26.928}\) \(\tan \theta = 0.26 \implies \theta = \tan^{-1} (0.26)\) \(\theta = 14.57^\circ \approxeq 14.6^\circ\) (to 1 decimal place)