WAEC SSCE - General Mathematics - 2005 (Essay)
Question 1 Report
(a) Without using mathematical table or calculator, evaluate : \(\sqrt{\frac{0.18 \times 12.5}{0.05 \times 0.2}}\).
(b) Simplify : \(\frac{8 - 4\sqrt{18}}{\sqrt{50}}\).
(c) x, y and z are related such that x varies directly as the cube of y and inversely as the square of z. If x = 108 when y = 3 and z = 4, find z when x = 4000 and y = 10.
(a) Simplify inside the root first:
\[\frac{0.18\times12.5}{0.05\times0.2} = \frac{2.25}{0.01} = 225,\qquad \sqrt{225} = 15.\]
(b) Write each surd in simplest form: \(\sqrt{18} = 3\sqrt2\) and \(\sqrt{50} = 5\sqrt2\).
\[\frac{8 - 4\sqrt{18}}{\sqrt{50}} = \frac{8 - 12\sqrt2}{5\sqrt2} = \frac{8}{5\sqrt2} - \frac{12\sqrt2}{5\sqrt2} = \frac{4\sqrt2}{5} - \frac{12}{5} = \frac{4\sqrt2 - 12}{5}.\]
(c) The relation is \(x = \dfrac{k y^3}{z^2}\). Use \(x=108,\,y=3,\,z=4\):
\[108 = \frac{k(27)}{16}\Rightarrow k = \frac{108\times16}{27} = 64.\]
Now find z when \(x=4000,\,y=10\):
\[4000 = \frac{64(1000)}{z^2}\Rightarrow z^2 = \frac{64000}{4000} = 16\Rightarrow z = 4.\]
Answer Details
(a) Simplify inside the root first:
\[\frac{0.18\times12.5}{0.05\times0.2} = \frac{2.25}{0.01} = 225,\qquad \sqrt{225} = 15.\]
(b) Write each surd in simplest form: \(\sqrt{18} = 3\sqrt2\) and \(\sqrt{50} = 5\sqrt2\).
\[\frac{8 - 4\sqrt{18}}{\sqrt{50}} = \frac{8 - 12\sqrt2}{5\sqrt2} = \frac{8}{5\sqrt2} - \frac{12\sqrt2}{5\sqrt2} = \frac{4\sqrt2}{5} - \frac{12}{5} = \frac{4\sqrt2 - 12}{5}.\]
(c) The relation is \(x = \dfrac{k y^3}{z^2}\). Use \(x=108,\,y=3,\,z=4\):
\[108 = \frac{k(27)}{16}\Rightarrow k = \frac{108\times16}{27} = 64.\]
Now find z when \(x=4000,\,y=10\):
\[4000 = \frac{64(1000)}{z^2}\Rightarrow z^2 = \frac{64000}{4000} = 16\Rightarrow z = 4.\]
Question 2 Report
(a) Evaluate without using the mathematical table or calculator, \(\log_{10} \sqrt{30} - \log_{10} \sqrt{6} + \log_{10} \sqrt{2}\).
(b) 
\(U = {1, 2, 3, ..., 10} ; A = {1, 2, 3, 4, 5} ; B = {2, 3, 5}\) and \(C = {6, 8, 10}\). (i) Given that the Venn diagram represents the sets above, copy and fill in the elements.
(ii) Find \(A \cap C\) ; (iii) Find \(A \cap B'\).
(a) Evaluate \(\log_{10}\sqrt{30} - \log_{10}\sqrt{6} + \log_{10}\sqrt{2}\).
Write each surd as a power and combine using the laws of logarithms. Since \(\sqrt{n}=n^{1/2}\),
\[\tfrac{1}{2}\log_{10}30 - \tfrac{1}{2}\log_{10}6 + \tfrac{1}{2}\log_{10}2 = \tfrac{1}{2}\left(\log_{10}30 - \log_{10}6 + \log_{10}2\right).\]Using \(\log a - \log b + \log c = \log\dfrac{ac}{b}\):
\[\tfrac{1}{2}\log_{10}\!\left(\frac{30\times 2}{6}\right) = \tfrac{1}{2}\log_{10}\!\left(\frac{60}{6}\right) = \tfrac{1}{2}\log_{10}10.\]Since \(\log_{10}10 = 1\),
\[= \tfrac{1}{2}\times 1 = \boxed{\tfrac{1}{2}}.\](b) \(U=\{1,2,3,\dots,10\}\), \(A=\{1,2,3,4,5\}\), \(B=\{2,3,5\}\), \(C=\{6,8,10\}\).
(i) Filling the Venn diagram. Every element of \(B\) is also in \(A\) (\(2,3,5\in A\)), so the circle \(B\) lies completely inside circle \(A\), exactly as shown. \(C\) has no element in common with \(A\), so it is drawn separately. The regions are filled as:
- Inside \(B\) (which is inside \(A\)): \(2,\;3,\;5\)
- Inside \(A\) but outside \(B\): \(1,\;4\)
- Inside \(C\): \(6,\;8,\;10\)
- Outside every circle: \(7,\;9\)
(The diagram already shows \(1\) in the \(A\)-only region, \(5\) in \(B\), \(6,10\) in \(C\) and \(9\) outside; the remaining elements \(2,3\) join \(B\), \(4\) joins the \(A\)-only region, \(8\) joins \(C\) and \(7\) joins the outside.)
(ii) \(A\cap C\). \(A=\{1,2,3,4,5\}\) and \(C=\{6,8,10\}\) share no element, so
\[A\cap C = \varnothing\;\;(\text{the empty set}).\](iii) \(A\cap B'\). First \(B' = U - B = \{1,4,6,7,8,9,10\}\). Then take the elements common to \(A\):
\[A\cap B' = \{1,2,3,4,5\}\cap\{1,4,6,7,8,9,10\} = \{1,\;4\}.\]Answer Details
(a) Evaluate \(\log_{10}\sqrt{30} - \log_{10}\sqrt{6} + \log_{10}\sqrt{2}\).
Write each surd as a power and combine using the laws of logarithms. Since \(\sqrt{n}=n^{1/2}\),
\[\tfrac{1}{2}\log_{10}30 - \tfrac{1}{2}\log_{10}6 + \tfrac{1}{2}\log_{10}2 = \tfrac{1}{2}\left(\log_{10}30 - \log_{10}6 + \log_{10}2\right).\]Using \(\log a - \log b + \log c = \log\dfrac{ac}{b}\):
\[\tfrac{1}{2}\log_{10}\!\left(\frac{30\times 2}{6}\right) = \tfrac{1}{2}\log_{10}\!\left(\frac{60}{6}\right) = \tfrac{1}{2}\log_{10}10.\]Since \(\log_{10}10 = 1\),
\[= \tfrac{1}{2}\times 1 = \boxed{\tfrac{1}{2}}.\](b) \(U=\{1,2,3,\dots,10\}\), \(A=\{1,2,3,4,5\}\), \(B=\{2,3,5\}\), \(C=\{6,8,10\}\).
(i) Filling the Venn diagram. Every element of \(B\) is also in \(A\) (\(2,3,5\in A\)), so the circle \(B\) lies completely inside circle \(A\), exactly as shown. \(C\) has no element in common with \(A\), so it is drawn separately. The regions are filled as:
- Inside \(B\) (which is inside \(A\)): \(2,\;3,\;5\)
- Inside \(A\) but outside \(B\): \(1,\;4\)
- Inside \(C\): \(6,\;8,\;10\)
- Outside every circle: \(7,\;9\)
(The diagram already shows \(1\) in the \(A\)-only region, \(5\) in \(B\), \(6,10\) in \(C\) and \(9\) outside; the remaining elements \(2,3\) join \(B\), \(4\) joins the \(A\)-only region, \(8\) joins \(C\) and \(7\) joins the outside.)
(ii) \(A\cap C\). \(A=\{1,2,3,4,5\}\) and \(C=\{6,8,10\}\) share no element, so
\[A\cap C = \varnothing\;\;(\text{the empty set}).\](iii) \(A\cap B'\). First \(B' = U - B = \{1,4,6,7,8,9,10\}\). Then take the elements common to \(A\):
\[A\cap B' = \{1,2,3,4,5\}\cap\{1,4,6,7,8,9,10\} = \{1,\;4\}.\]Question 3 Report
(a) Find the smallest integer that satisfies the inequality \(x + 8 < 4x - 15\).
(b) A sales girl is paid a monthly salary of N2,500 in addition to a commission of 5 kobo in the naira on all sales made by her during the month. If her sales for a month amounts to N200,000.00, calculate her income for that month.
(c) 
The diagram shows a window consisting of a rectangular and semi- circular parts. The radius of the semi- circular part is 35 cm and the height of the rectangular part is 50 cm. Find the area of the window. [Take \(\pi = \frac{22}{7}\)].
(a) Smallest integer satisfying the inequality
\[x + 8 < 4x - 15\]
Collect like terms:
\[8 + 15 < 4x - x\]
\[23 < 3x\]
\[x > \frac{23}{3} = 7\tfrac{2}{3} \approx 7.67\]
The smallest integer greater than \(7.67\) is \(\mathbf{8}\).
(b) Monthly income of the sales girl
Fixed salary \(= \text{N}2{,}500\).
Commission \(= 5\text{ kobo in the naira} = \dfrac{5}{100}\text{ naira per naira} = \text{N}0.05\) per naira of sales.
Commission on N200,000 sales:
\[0.05 \times 200{,}000 = \text{N}10{,}000\]
Total income:
\[2{,}500 + 10{,}000 = \text{N}12{,}500\]
Her income for the month is \(\mathbf{\text{N}12{,}500.00}\).
(c) Area of the window
The window is a rectangle with a semicircle on top. From the diagram, the semicircular part has radius \(35\text{ cm}\), so the width of the rectangle equals the diameter:
\[\text{width} = 2 \times 35 = 70\text{ cm}, \qquad \text{height} = 50\text{ cm}\]
Area of rectangle:
\[A_1 = 70 \times 50 = 3{,}500\text{ cm}^2\]
Area of semicircle: (with \(\pi = \tfrac{22}{7}\))
\[A_2 = \frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7} \times 35^2 = \frac{1}{2}\times \frac{22}{7} \times 1225\]
\[A_2 = \frac{1}{2}\times 22 \times 175 = \frac{1}{2}\times 3{,}850 = 1{,}925\text{ cm}^2\]
Total area of window:
\[A = A_1 + A_2 = 3{,}500 + 1{,}925 = \mathbf{5{,}425\text{ cm}^2}\]
Answer Details
(a) Smallest integer satisfying the inequality
\[x + 8 < 4x - 15\]
Collect like terms:
\[8 + 15 < 4x - x\]
\[23 < 3x\]
\[x > \frac{23}{3} = 7\tfrac{2}{3} \approx 7.67\]
The smallest integer greater than \(7.67\) is \(\mathbf{8}\).
(b) Monthly income of the sales girl
Fixed salary \(= \text{N}2{,}500\).
Commission \(= 5\text{ kobo in the naira} = \dfrac{5}{100}\text{ naira per naira} = \text{N}0.05\) per naira of sales.
Commission on N200,000 sales:
\[0.05 \times 200{,}000 = \text{N}10{,}000\]
Total income:
\[2{,}500 + 10{,}000 = \text{N}12{,}500\]
Her income for the month is \(\mathbf{\text{N}12{,}500.00}\).
(c) Area of the window
The window is a rectangle with a semicircle on top. From the diagram, the semicircular part has radius \(35\text{ cm}\), so the width of the rectangle equals the diameter:
\[\text{width} = 2 \times 35 = 70\text{ cm}, \qquad \text{height} = 50\text{ cm}\]
Area of rectangle:
\[A_1 = 70 \times 50 = 3{,}500\text{ cm}^2\]
Area of semicircle: (with \(\pi = \tfrac{22}{7}\))
\[A_2 = \frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7} \times 35^2 = \frac{1}{2}\times \frac{22}{7} \times 1225\]
\[A_2 = \frac{1}{2}\times 22 \times 175 = \frac{1}{2}\times 3{,}850 = 1{,}925\text{ cm}^2\]
Total area of window:
\[A = A_1 + A_2 = 3{,}500 + 1{,}925 = \mathbf{5{,}425\text{ cm}^2}\]
Question 4 Report
(a) A plane flies due East from A(lat. 53°N, long. 25°E) to a point B(lat. 53°N, long. 85°E) at an average speed of 400 km/h. The plane then flies South from B to a point C 2000km away. Calculate, correct to the nearest whole number :
(a) the distance between A and B.
(b) the time the plane takes to reach point B ;
(c) the latitude of C.
[Take radius of the earth = 6400km; \(\pi = \frac{22}{7}\)].
(a) Distance A to B along the parallel of latitude 53°N, through a longitude difference of \(85^\circ - 25^\circ = 60^\circ\):
\[AB = \frac{60}{360}\times 2\pi R\cos53^\circ = \frac{60}{360}\times 2\times\frac{22}{7}\times6400\times\cos53^\circ.\]
\[AB = \frac{1}{6}\times 40228.6\times0.6018 = 4035\text{ km (to the nearest km)}.\]
(b) Time to reach B at 400 km/h:
\[t = \frac{4035}{400} = 10.09 \approx 10\text{ hours}.\]
(c) Latitude of C. Flying south along a meridian, 1° corresponds to
\[\frac{2\pi R}{360} = \frac{1}{360}\times2\times\frac{22}{7}\times6400 = 111.75\text{ km}.\]
The 2000 km southward change in latitude is
\[\frac{2000}{111.75} = 17.9^\circ.\]
\[\text{Latitude of C} = 53^\circ - 17.9^\circ = 35.1^\circ \approx 35^\circ\text{N}.\]
Answer Details
(a) Distance A to B along the parallel of latitude 53°N, through a longitude difference of \(85^\circ - 25^\circ = 60^\circ\):
\[AB = \frac{60}{360}\times 2\pi R\cos53^\circ = \frac{60}{360}\times 2\times\frac{22}{7}\times6400\times\cos53^\circ.\]
\[AB = \frac{1}{6}\times 40228.6\times0.6018 = 4035\text{ km (to the nearest km)}.\]
(b) Time to reach B at 400 km/h:
\[t = \frac{4035}{400} = 10.09 \approx 10\text{ hours}.\]
(c) Latitude of C. Flying south along a meridian, 1° corresponds to
\[\frac{2\pi R}{360} = \frac{1}{360}\times2\times\frac{22}{7}\times6400 = 111.75\text{ km}.\]
The 2000 km southward change in latitude is
\[\frac{2000}{111.75} = 17.9^\circ.\]
\[\text{Latitude of C} = 53^\circ - 17.9^\circ = 35.1^\circ \approx 35^\circ\text{N}.\]
Question 5 Report
(a) Simplify : \(\frac{5}{8} of 2\frac{1}{2} - \frac{3}{4} \div \frac{3}{5}\).
(b) A cone and a right pyramid have equal heights and volumes. If the area of the base of the pyramid is \(154 cm^{2}\), find the base radius of the cone. [Take \(\pi = \frac{22}{7}\)].
(a) Work "of" and \(\div\) before subtraction:
\[\tfrac{5}{8}\text{ of }2\tfrac{1}{2} = \tfrac{5}{8}\times\tfrac{5}{2} = \tfrac{25}{16},\qquad \tfrac{3}{4}\div\tfrac{3}{5} = \tfrac{3}{4}\times\tfrac{5}{3} = \tfrac{5}{4} = \tfrac{20}{16}.\]
\[\tfrac{25}{16} - \tfrac{20}{16} = \tfrac{5}{16}.\]
(b) The cone and the pyramid have equal heights \(h\) and equal volumes.
\[V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h,\qquad V_{\text{pyramid}} = \tfrac{1}{3}(\text{base area})h.\]
Since the volumes and heights are equal, the base areas are equal:
\[\pi r^2 = 154 \Rightarrow \tfrac{22}{7}r^2 = 154 \Rightarrow r^2 = 154\times\tfrac{7}{22} = 49.\]
\[r = 7\text{ cm}.\]
Answer Details
(a) Work "of" and \(\div\) before subtraction:
\[\tfrac{5}{8}\text{ of }2\tfrac{1}{2} = \tfrac{5}{8}\times\tfrac{5}{2} = \tfrac{25}{16},\qquad \tfrac{3}{4}\div\tfrac{3}{5} = \tfrac{3}{4}\times\tfrac{5}{3} = \tfrac{5}{4} = \tfrac{20}{16}.\]
\[\tfrac{25}{16} - \tfrac{20}{16} = \tfrac{5}{16}.\]
(b) The cone and the pyramid have equal heights \(h\) and equal volumes.
\[V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h,\qquad V_{\text{pyramid}} = \tfrac{1}{3}(\text{base area})h.\]
Since the volumes and heights are equal, the base areas are equal:
\[\pi r^2 = 154 \Rightarrow \tfrac{22}{7}r^2 = 154 \Rightarrow r^2 = 154\times\tfrac{7}{22} = 49.\]
\[r = 7\text{ cm}.\]
Question 6 Report
(a) Two fair die are thrown once. Find the probabitlity of getting : (i) the same digit ; (ii) a total score greater than 5.
(b) Given that \(x = \cos 30°\) and \(y = \sin 30°\), evaluate without using a mathematical table or calculator : \(\frac{x^{2} + y^{2}}{y^{2} - x^{2}}\).
(a) Two dice give \(6\times6 = 36\) equally likely outcomes.
(i) Same digit: the pairs are (1,1),(2,2),...,(6,6), that is 6 outcomes.
\[P(\text{same digit}) = \tfrac{6}{36} = \tfrac{1}{6}.\]
(ii) Total greater than 5: count the totals that are 5 or less: total 2 (1 way), 3 (2), 4 (3), 5 (4), giving \(1+2+3+4 = 10\) outcomes.
\[P(\text{total} > 5) = \tfrac{36-10}{36} = \tfrac{26}{36} = \tfrac{13}{18}.\]
(b) \(x = \cos30^\circ = \tfrac{\sqrt3}{2}\Rightarrow x^2 = \tfrac{3}{4};\quad y = \sin30^\circ = \tfrac{1}{2}\Rightarrow y^2 = \tfrac{1}{4}.\)
\[\frac{x^2 + y^2}{y^2 - x^2} = \frac{\tfrac{3}{4}+\tfrac{1}{4}}{\tfrac{1}{4}-\tfrac{3}{4}} = \frac{1}{-\tfrac{1}{2}} = -2.\]
Answer Details
(a) Two dice give \(6\times6 = 36\) equally likely outcomes.
(i) Same digit: the pairs are (1,1),(2,2),...,(6,6), that is 6 outcomes.
\[P(\text{same digit}) = \tfrac{6}{36} = \tfrac{1}{6}.\]
(ii) Total greater than 5: count the totals that are 5 or less: total 2 (1 way), 3 (2), 4 (3), 5 (4), giving \(1+2+3+4 = 10\) outcomes.
\[P(\text{total} > 5) = \tfrac{36-10}{36} = \tfrac{26}{36} = \tfrac{13}{18}.\]
(b) \(x = \cos30^\circ = \tfrac{\sqrt3}{2}\Rightarrow x^2 = \tfrac{3}{4};\quad y = \sin30^\circ = \tfrac{1}{2}\Rightarrow y^2 = \tfrac{1}{4}.\)
\[\frac{x^2 + y^2}{y^2 - x^2} = \frac{\tfrac{3}{4}+\tfrac{1}{4}}{\tfrac{1}{4}-\tfrac{3}{4}} = \frac{1}{-\tfrac{1}{2}} = -2.\]
Question 7 Report
(a) 
In the diagram, A, B, C and D are points on the circumference of a circle. XY is a tangent at A. Find : (i) < CAX ; (ii) < ABY.
(b) If (m + 1) and (m - 3) are factors of \(m^{2} - km + c\), find the values of k and c.
(a) Circle through A, B, C, D with tangent XY at A.
From the diagram: \(\angle ADB=20^\circ\) (at D), and the tangent-secant angle at Y (between tangent \(YA\) and the secant through \(B\) and \(C\)) is \(\angle AYC=69^\circ\), with \(C,\,B,\,Y\) in a straight line.
(i) \(\angle CAX\).
\(\angle ADB=20^\circ\) is an inscribed angle on chord \(AB\), so
\[\text{arc } AB=2\times20^\circ=40^\circ.\]
The tangent-chord angle \(\angle BAY\) equals the angle in the alternate segment \(\angle ADB\):
\[\angle BAY=20^\circ.\]
In triangle \(ABY\), \(\angle AYB=69^\circ\), so
\[\angle ABY=180^\circ-69^\circ-20^\circ=91^\circ,\]
and since \(C,B,Y\) are collinear,
\[\angle ABC=180^\circ-\angle ABY=180^\circ-91^\circ=89^\circ.\]
By the alternate segment theorem, the tangent-chord angle \(\angle CAX\) equals the inscribed angle \(\angle ABC\) in the alternate segment:
\[\angle CAX=\angle ABC=\boxed{89^\circ}.\]
(ii) \(\angle ABY\).
From the working above,
\[\angle ABY=180^\circ-69^\circ-20^\circ=\boxed{91^\circ}.\]
(b) Factors of \(m^{2}-km+c\).
If \((m+1)\) and \((m-3)\) are factors, then
\[m^{2}-km+c=(m+1)(m-3)=m^{2}-3m+m-3=m^{2}-2m-3.\]
Comparing coefficients:
\[-k=-2\ \Rightarrow\ \boxed{k=2},\qquad c=\boxed{-3}.\]
Answer Details
(a) Circle through A, B, C, D with tangent XY at A.
From the diagram: \(\angle ADB=20^\circ\) (at D), and the tangent-secant angle at Y (between tangent \(YA\) and the secant through \(B\) and \(C\)) is \(\angle AYC=69^\circ\), with \(C,\,B,\,Y\) in a straight line.
(i) \(\angle CAX\).
\(\angle ADB=20^\circ\) is an inscribed angle on chord \(AB\), so
\[\text{arc } AB=2\times20^\circ=40^\circ.\]
The tangent-chord angle \(\angle BAY\) equals the angle in the alternate segment \(\angle ADB\):
\[\angle BAY=20^\circ.\]
In triangle \(ABY\), \(\angle AYB=69^\circ\), so
\[\angle ABY=180^\circ-69^\circ-20^\circ=91^\circ,\]
and since \(C,B,Y\) are collinear,
\[\angle ABC=180^\circ-\angle ABY=180^\circ-91^\circ=89^\circ.\]
By the alternate segment theorem, the tangent-chord angle \(\angle CAX\) equals the inscribed angle \(\angle ABC\) in the alternate segment:
\[\angle CAX=\angle ABC=\boxed{89^\circ}.\]
(ii) \(\angle ABY\).
From the working above,
\[\angle ABY=180^\circ-69^\circ-20^\circ=\boxed{91^\circ}.\]
(b) Factors of \(m^{2}-km+c\).
If \((m+1)\) and \((m-3)\) are factors, then
\[m^{2}-km+c=(m+1)(m-3)=m^{2}-3m+m-3=m^{2}-2m-3.\]
Comparing coefficients:
\[-k=-2\ \Rightarrow\ \boxed{k=2},\qquad c=\boxed{-3}.\]
Question 8 Report
(a) Copy and complete the table.
\(y = x^{2} - 2x - 2\) for \(-4 \leq x \leq 4\)
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 22 | -2 | 1 | 6 |
(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = x^{2} - 2x - 2\).
(c) Use your graph to find : (i) the roots of the equation \(x^{2} - 2x - 2 = 0\) ; (ii) the values of x for which \(x^{2} - 2x - 4\frac{1}{2} = 0\) ; (iii) the equation of the line of symmetry of the curve.
(a) Completing the table. For each value of \(x\) we evaluate \(y = x^{2} - 2x - 2\) by building it up in rows:
| \(x\) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| \(x^{2}\) | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
| \(-2x\) | 8 | 6 | 4 | 2 | 0 | -2 | -4 | -6 | -8 |
| \(-2\) | -2 | -2 | -2 | -2 | -2 | -2 | -2 | -2 | -2 |
| \(y\) | 22 | 13 | 6 | 1 | -2 | -3 | -2 | 1 | 6 |
The three missing entries are \(y=13\) (at \(x=-3\)), \(y=1\) (at \(x=-1\)) and \(y=-3\) (at \(x=1\)).
(b) Graph of \(y = x^{2} - 2x - 2\). Plotting the nine points \((x,y)\) and joining them with a smooth curve gives the parabola below. The dashed horizontal line \(y = 2.5\) is used in part (c)(ii) and the vertical dashed line \(x = 1\) is the line of symmetry from part (c)(iii).
(c) Using the graph.
(i) Roots of \(x^{2} - 2x - 2 = 0\). These are the values of \(x\) where the curve cuts the \(x\)-axis, i.e. where \(y = 0\). The curve crosses the \(x\)-axis at \[ x \approx -0.7 \quad \text{and} \quad x \approx 2.7. \]
(ii) Values of \(x\) for which \(x^{2} - 2x - 4\tfrac{1}{2} = 0\). Rearranging so that the left side becomes our plotted expression: \[ x^{2} - 2x - 4\tfrac{1}{2} = 0 \;\Rightarrow\; x^{2} - 2x - 2 = 2\tfrac{1}{2} = 2.5. \] So we draw the horizontal line \(y = 2.5\) and read off where it meets the curve: \[ x \approx -1.3 \quad \text{and} \quad x \approx 3.3. \]
(iii) Equation of the line of symmetry. The lowest point (vertex) of the curve occurs at \(x = \dfrac{-(-2)}{2(1)} = 1\), where \(y = -3\). The curve is symmetrical about the vertical line through this point, so the line of symmetry is \[ x = 1. \]
Answer Details
(a) Completing the table. For each value of \(x\) we evaluate \(y = x^{2} - 2x - 2\) by building it up in rows:
| \(x\) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| \(x^{2}\) | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
| \(-2x\) | 8 | 6 | 4 | 2 | 0 | -2 | -4 | -6 | -8 |
| \(-2\) | -2 | -2 | -2 | -2 | -2 | -2 | -2 | -2 | -2 |
| \(y\) | 22 | 13 | 6 | 1 | -2 | -3 | -2 | 1 | 6 |
The three missing entries are \(y=13\) (at \(x=-3\)), \(y=1\) (at \(x=-1\)) and \(y=-3\) (at \(x=1\)).
(b) Graph of \(y = x^{2} - 2x - 2\). Plotting the nine points \((x,y)\) and joining them with a smooth curve gives the parabola below. The dashed horizontal line \(y = 2.5\) is used in part (c)(ii) and the vertical dashed line \(x = 1\) is the line of symmetry from part (c)(iii).
(c) Using the graph.
(i) Roots of \(x^{2} - 2x - 2 = 0\). These are the values of \(x\) where the curve cuts the \(x\)-axis, i.e. where \(y = 0\). The curve crosses the \(x\)-axis at \[ x \approx -0.7 \quad \text{and} \quad x \approx 2.7. \]
(ii) Values of \(x\) for which \(x^{2} - 2x - 4\tfrac{1}{2} = 0\). Rearranging so that the left side becomes our plotted expression: \[ x^{2} - 2x - 4\tfrac{1}{2} = 0 \;\Rightarrow\; x^{2} - 2x - 2 = 2\tfrac{1}{2} = 2.5. \] So we draw the horizontal line \(y = 2.5\) and read off where it meets the curve: \[ x \approx -1.3 \quad \text{and} \quad x \approx 3.3. \]
(iii) Equation of the line of symmetry. The lowest point (vertex) of the curve occurs at \(x = \dfrac{-(-2)}{2(1)} = 1\), where \(y = -3\). The curve is symmetrical about the vertical line through this point, so the line of symmetry is \[ x = 1. \]
Question 9 Report
(a) Evaluate : \(2 \div (\frac{64}{125})^{-\frac{2}{3}}\)
(b) The lines \(y = 3x + 5\) and \(y = - 4x - 1\) intersect at a point k. Find the coordinates of k.
(a) A negative index inverts the base:
\[\left(\tfrac{64}{125}\right)^{-\frac{2}{3}} = \left(\tfrac{125}{64}\right)^{\frac{2}{3}} = \left(\sqrt[3]{\tfrac{125}{64}}\right)^{2} = \left(\tfrac{5}{4}\right)^{2} = \tfrac{25}{16}.\]
\[2 \div \tfrac{25}{16} = 2 \times \tfrac{16}{25} = \tfrac{32}{25} = 1\tfrac{7}{25}.\]
(b) At the point of intersection k the two y-values are equal:
\[3x + 5 = -4x - 1 \Rightarrow 7x = -6 \Rightarrow x = -\tfrac{6}{7}.\]
\[y = 3\left(-\tfrac{6}{7}\right) + 5 = -\tfrac{18}{7} + \tfrac{35}{7} = \tfrac{17}{7}.\]
Coordinates of k: \(\left(-\tfrac{6}{7},\ \tfrac{17}{7}\right)\).
Answer Details
(a) A negative index inverts the base:
\[\left(\tfrac{64}{125}\right)^{-\frac{2}{3}} = \left(\tfrac{125}{64}\right)^{\frac{2}{3}} = \left(\sqrt[3]{\tfrac{125}{64}}\right)^{2} = \left(\tfrac{5}{4}\right)^{2} = \tfrac{25}{16}.\]
\[2 \div \tfrac{25}{16} = 2 \times \tfrac{16}{25} = \tfrac{32}{25} = 1\tfrac{7}{25}.\]
(b) At the point of intersection k the two y-values are equal:
\[3x + 5 = -4x - 1 \Rightarrow 7x = -6 \Rightarrow x = -\tfrac{6}{7}.\]
\[y = 3\left(-\tfrac{6}{7}\right) + 5 = -\tfrac{18}{7} + \tfrac{35}{7} = \tfrac{17}{7}.\]
Coordinates of k: \(\left(-\tfrac{6}{7},\ \tfrac{17}{7}\right)\).
Question 10 Report
(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find :
(i) the value of n ; (ii) the sum of all the interior angles.
(b) A boy walks 6km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13km, on a bearing of 146°.
(i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometre, the distance PR.
(a) The regular polygon
(i) Value of n. At any vertex the interior and exterior angles are supplementary, and here the interior angle is 120° greater than the exterior angle. Let the exterior angle be \(x\):
\[x + (x + 120^\circ) = 180^\circ\] \[2x + 120^\circ = 180^\circ \implies 2x = 60^\circ \implies x = 30^\circ.\]For any polygon the exterior angles sum to \(360^\circ\), so
\[n = \frac{360^\circ}{\text{exterior angle}} = \frac{360^\circ}{30^\circ} = 12.\]The polygon has 12 sides.
(ii) Sum of all the interior angles.
\[\text{Sum} = (n-2)\times 180^\circ = (12-2)\times 180^\circ = 10 \times 180^\circ = 1800^\circ.\]The interior angles sum to 1800°.
(b) The boy's journey
(i) Sketch of the movement. He walks 6 km from \(P\) to \(Q\) on a bearing of 065°, then 13 km from \(Q\) to \(R\) on a bearing of 146°. A North line is drawn at each point so the bearings can be measured clockwise from North.
(ii) Distance PR. First find the interior angle of the triangle at \(Q\). The bearing of \(P\) from \(Q\) is \(065^\circ + 180^\circ = 245^\circ\), and the bearing of \(R\) from \(Q\) is \(146^\circ\). Working from the North line at \(Q\), the angle between \(QP\) and the North line is \(180^\circ - 65^\circ = 115^\circ\) on the western side, giving the angle inside the triangle:
\[\angle PQR = (180^\circ - 146^\circ) + 65^\circ = 34^\circ + 65^\circ = 99^\circ.\]Apply the cosine rule to triangle \(PQR\):
\[|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2\,|PQ|\,|QR|\cos(\angle PQR)\] \[|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13)\cos 99^\circ\] \[|PR|^{2} = 36 + 169 - 156\cos 99^\circ\]Since \(\cos 99^\circ = -0.1564\):
\[|PR|^{2} = 205 - 156(-0.1564) = 205 + 24.40 = 229.40\] \[|PR| = \sqrt{229.40} = 15.15\text{ km}.\]Correct to the nearest kilometre, \(|PR| \approx \mathbf{15\text{ km}}\).
Answer Details
(a) The regular polygon
(i) Value of n. At any vertex the interior and exterior angles are supplementary, and here the interior angle is 120° greater than the exterior angle. Let the exterior angle be \(x\):
\[x + (x + 120^\circ) = 180^\circ\] \[2x + 120^\circ = 180^\circ \implies 2x = 60^\circ \implies x = 30^\circ.\]For any polygon the exterior angles sum to \(360^\circ\), so
\[n = \frac{360^\circ}{\text{exterior angle}} = \frac{360^\circ}{30^\circ} = 12.\]The polygon has 12 sides.
(ii) Sum of all the interior angles.
\[\text{Sum} = (n-2)\times 180^\circ = (12-2)\times 180^\circ = 10 \times 180^\circ = 1800^\circ.\]The interior angles sum to 1800°.
(b) The boy's journey
(i) Sketch of the movement. He walks 6 km from \(P\) to \(Q\) on a bearing of 065°, then 13 km from \(Q\) to \(R\) on a bearing of 146°. A North line is drawn at each point so the bearings can be measured clockwise from North.
(ii) Distance PR. First find the interior angle of the triangle at \(Q\). The bearing of \(P\) from \(Q\) is \(065^\circ + 180^\circ = 245^\circ\), and the bearing of \(R\) from \(Q\) is \(146^\circ\). Working from the North line at \(Q\), the angle between \(QP\) and the North line is \(180^\circ - 65^\circ = 115^\circ\) on the western side, giving the angle inside the triangle:
\[\angle PQR = (180^\circ - 146^\circ) + 65^\circ = 34^\circ + 65^\circ = 99^\circ.\]Apply the cosine rule to triangle \(PQR\):
\[|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2\,|PQ|\,|QR|\cos(\angle PQR)\] \[|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13)\cos 99^\circ\] \[|PR|^{2} = 36 + 169 - 156\cos 99^\circ\]Since \(\cos 99^\circ = -0.1564\):
\[|PR|^{2} = 205 - 156(-0.1564) = 205 + 24.40 = 229.40\] \[|PR| = \sqrt{229.40} = 15.15\text{ km}.\]Correct to the nearest kilometre, \(|PR| \approx \mathbf{15\text{ km}}\).
Question 11 Report
The table shows the age distributions of the members of a club.
| Age (years) | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 |
| Frequency | 7 | 18 | 25 | 17 | 9 | 4 |
(a) Calculate, correct to one decimal place, the mean age.
(b) (i) Draw a histogram to illustrate the information.
(ii) Use the histogram to estimate the modal age .
(c) If a member is selected at random, what is the probability that he/she is in the modal class?
(a) Mean age
Take the class mid-value (class-mark) \(x\) of each interval and form \(fx\).
| Age (years) | Mid-value \(x\) | Frequency \(f\) | \(fx\) |
| 10 - 14 | 12 | 7 | 84 |
| 15 - 19 | 17 | 18 | 306 |
| 20 - 24 | 22 | 25 | 550 |
| 25 - 29 | 27 | 17 | 459 |
| 30 - 34 | 32 | 9 | 288 |
| 35 - 39 | 37 | 4 | 148 |
| Total | 80 | 1835 |
\[ \bar{x}=\frac{\sum fx}{\sum f}=\frac{1835}{80}=22.9375 \]
Mean age \( \approx \mathbf{22.9\ years}\) (to 1 d.p.).
(b)(i) Histogram
First convert each class to its continuous class boundaries, then draw bars whose heights equal the frequencies (the bars touch, since the boundaries are continuous).
| Age (years) | Class boundaries | Frequency \(f\) |
| 10 - 14 | 9.5 - 14.5 | 7 |
| 15 - 19 | 14.5 - 19.5 | 18 |
| 20 - 24 | 19.5 - 24.5 | 25 |
| 25 - 29 | 24.5 - 29.5 | 17 |
| 30 - 34 | 29.5 - 34.5 | 9 |
| 35 - 39 | 34.5 - 39.5 | 4 |
(b)(ii) Modal age from the histogram
The tallest bar is the modal class \(20\text{-}24\) (boundaries \(19.5\text{-}24.5\)). To read the mode, join the top-left corner of the modal bar to the top-left corner of the bar after it, and the top-right corner of the modal bar to the top-right corner of the bar before it. From the point where these two lines cross, drop a vertical line to the age axis.
The vertical line meets the axis at approximately \(\mathbf{21.8\ years}\).
This agrees with the calculation \[ \text{Mode}=L+\frac{\Delta_1}{\Delta_1+\Delta_2}\times c =19.5+\frac{25-18}{(25-18)+(25-17)}\times 5 =19.5+\frac{7}{15}\times 5 =19.5+2.3=21.8\ \text{years}. \]
(c) Probability of being in the modal class
The modal class \(20\text{-}24\) contains \(25\) members out of a total of \(80\):
\[ P(\text{modal class})=\frac{25}{80}=\frac{5}{16}=0.3125 \]
Answer Details
(a) Mean age
Take the class mid-value (class-mark) \(x\) of each interval and form \(fx\).
| Age (years) | Mid-value \(x\) | Frequency \(f\) | \(fx\) |
| 10 - 14 | 12 | 7 | 84 |
| 15 - 19 | 17 | 18 | 306 |
| 20 - 24 | 22 | 25 | 550 |
| 25 - 29 | 27 | 17 | 459 |
| 30 - 34 | 32 | 9 | 288 |
| 35 - 39 | 37 | 4 | 148 |
| Total | 80 | 1835 |
\[ \bar{x}=\frac{\sum fx}{\sum f}=\frac{1835}{80}=22.9375 \]
Mean age \( \approx \mathbf{22.9\ years}\) (to 1 d.p.).
(b)(i) Histogram
First convert each class to its continuous class boundaries, then draw bars whose heights equal the frequencies (the bars touch, since the boundaries are continuous).
| Age (years) | Class boundaries | Frequency \(f\) |
| 10 - 14 | 9.5 - 14.5 | 7 |
| 15 - 19 | 14.5 - 19.5 | 18 |
| 20 - 24 | 19.5 - 24.5 | 25 |
| 25 - 29 | 24.5 - 29.5 | 17 |
| 30 - 34 | 29.5 - 34.5 | 9 |
| 35 - 39 | 34.5 - 39.5 | 4 |
(b)(ii) Modal age from the histogram
The tallest bar is the modal class \(20\text{-}24\) (boundaries \(19.5\text{-}24.5\)). To read the mode, join the top-left corner of the modal bar to the top-left corner of the bar after it, and the top-right corner of the modal bar to the top-right corner of the bar before it. From the point where these two lines cross, drop a vertical line to the age axis.
The vertical line meets the axis at approximately \(\mathbf{21.8\ years}\).
This agrees with the calculation \[ \text{Mode}=L+\frac{\Delta_1}{\Delta_1+\Delta_2}\times c =19.5+\frac{25-18}{(25-18)+(25-17)}\times 5 =19.5+\frac{7}{15}\times 5 =19.5+2.3=21.8\ \text{years}. \]
(c) Probability of being in the modal class
The modal class \(20\text{-}24\) contains \(25\) members out of a total of \(80\):
\[ P(\text{modal class})=\frac{25}{80}=\frac{5}{16}=0.3125 \]
Question 12 Report
The sketch shows a plot of land .
(a) Using a scale of 1 cm to 10m, draw an accurate diagram of the plot ;
(b) Construct : (i) The locus \(l_{1}\) of points equidistant from AC and BC ; (ii) the locus \(l_{2}\) of points 60m from A.
(c) A tree T inside the plot is on both \(l_{1}\) and \(l_{2}\). Locate T and find |TC| in metres.
(d) A flagpole, P is to be placed such that it it is nearer AC than BC and more than 60m from A. Shade the regions where P can be located.
The supplied reference value \(|TC|=60\text{ m}\) is not consistent with the stated plot dimensions. The \(6\text{ cm}\) measurement is the radius of the circle centred at \(A\), so it represents \(|AT|=60\text{ m}\), not \(|TC|\). Using \(\angle A=40^\circ\), \(\angle B=95^\circ\), and \(|AB|=85\text{ m}\), \(|TC|\) is approximately \(72\text{ m}\).
(a) Scale drawing
With scale \(1\text{ cm}:10\text{ m}\), draw \(|AB|=8.5\text{ cm}\). Construct a \(40^\circ\) angle at \(A\) and a \(95^\circ\) angle at \(B\); their rays meet at \(C\). The remaining angle is
\[ \angle C=180^\circ-40^\circ-95^\circ=45^\circ. \]
(b), (c) and (d): completed construction
Locus \(l_1\). Points equidistant from the two lines \(AC\) and \(BC\) lie on an angle bisector of \(\angle ACB\). Since only points inside the plot are relevant, draw the internal angle bisector of the \(45^\circ\) angle at \(C\). It makes angles of \(22.5^\circ\) with both \(AC\) and \(BC\).
Locus \(l_2\). Points \(60\text{ m}\) from \(A\) lie on a circle with centre \(A\) and radius \(60\text{ m}\). At the given scale, this radius is \(6\text{ cm}\). Only the arc inside the triangular plot is needed.
Locating \(T\) and finding \(|TC|\). The tree is the intersection of the internal angle bisector and the \(60\text{ m}\) arc. To check the measurement numerically, first find \(AC\):
\[ AC=\frac{85\sin95^\circ}{\sin45^\circ}\approx119.8\text{ m}. \]
In triangle \(ACT\), \(|AT|=60\text{ m}\) and \(\angle ACT=22.5^\circ\). Applying the cosine rule:
\[ 60^2=119.8^2+TC^2-2(119.8)(TC)\cos22.5^\circ. \]
The intersection that is inside the plot gives
\[ |TC|\approx71.8\text{ m}\approx\boxed{72\text{ m}}. \]
Flagpole region. “Nearer \(AC\) than \(BC\)” means the region on the \(AC\) side of \(l_1\). “More than \(60\text{ m}\) from \(A\)” means outside \(l_2\). Therefore, shade the part of the triangle that is on the \(AC\) side of the angle bisector and outside the \(60\text{ m}\) arc, as shown in green.
Examination reminder: A circle of radius \(60\text{ m}\) tells you that \(AT=60\text{ m}\). Do not automatically assign that radius to another line such as \(TC\); \(TC\) must be measured or calculated from the intersection.
Answer Details
The supplied reference value \(|TC|=60\text{ m}\) is not consistent with the stated plot dimensions. The \(6\text{ cm}\) measurement is the radius of the circle centred at \(A\), so it represents \(|AT|=60\text{ m}\), not \(|TC|\). Using \(\angle A=40^\circ\), \(\angle B=95^\circ\), and \(|AB|=85\text{ m}\), \(|TC|\) is approximately \(72\text{ m}\).
(a) Scale drawing
With scale \(1\text{ cm}:10\text{ m}\), draw \(|AB|=8.5\text{ cm}\). Construct a \(40^\circ\) angle at \(A\) and a \(95^\circ\) angle at \(B\); their rays meet at \(C\). The remaining angle is
\[ \angle C=180^\circ-40^\circ-95^\circ=45^\circ. \]
(b), (c) and (d): completed construction
Locus \(l_1\). Points equidistant from the two lines \(AC\) and \(BC\) lie on an angle bisector of \(\angle ACB\). Since only points inside the plot are relevant, draw the internal angle bisector of the \(45^\circ\) angle at \(C\). It makes angles of \(22.5^\circ\) with both \(AC\) and \(BC\).
Locus \(l_2\). Points \(60\text{ m}\) from \(A\) lie on a circle with centre \(A\) and radius \(60\text{ m}\). At the given scale, this radius is \(6\text{ cm}\). Only the arc inside the triangular plot is needed.
Locating \(T\) and finding \(|TC|\). The tree is the intersection of the internal angle bisector and the \(60\text{ m}\) arc. To check the measurement numerically, first find \(AC\):
\[ AC=\frac{85\sin95^\circ}{\sin45^\circ}\approx119.8\text{ m}. \]
In triangle \(ACT\), \(|AT|=60\text{ m}\) and \(\angle ACT=22.5^\circ\). Applying the cosine rule:
\[ 60^2=119.8^2+TC^2-2(119.8)(TC)\cos22.5^\circ. \]
The intersection that is inside the plot gives
\[ |TC|\approx71.8\text{ m}\approx\boxed{72\text{ m}}. \]
Flagpole region. “Nearer \(AC\) than \(BC\)” means the region on the \(AC\) side of \(l_1\). “More than \(60\text{ m}\) from \(A\)” means outside \(l_2\). Therefore, shade the part of the triangle that is on the \(AC\) side of the angle bisector and outside the \(60\text{ m}\) arc, as shown in green.
Examination reminder: A circle of radius \(60\text{ m}\) tells you that \(AT=60\text{ m}\). Do not automatically assign that radius to another line such as \(TC\); \(TC\) must be measured or calculated from the intersection.
Question 13 Report
(a) 
In the diagram, \(\Delta\) ABD is right-angled at B. |AB| = 3 cm, |AD| = 5 cm, \(\stackrel\frown{ACB}\) = 61° and \(\stackrel\frown{DAC}\) = x°. Calculate, correct to one decimal place, the value of x.
(b) 
In the diagram, OABCD is a pyramid with a square base of side 2cm and a slant height of 4 cm. Calculate, correct to three significant figures : (i) the vertical height of the pyramid ; (ii) the volume of the pyramid.
(a) Value of x
In the diagram \(\triangle ABD\) is right-angled at \(B\), with \(|AB| = 3\ \text{cm}\) and hypotenuse \(|AD| = 5\ \text{cm}\). Point \(C\) lies on \(BD\), \(\angle ACB = 61^\circ\) and \(\angle DAC = x^\circ\).
Step 1 - the whole angle at A. In right triangle \(ABD\), \(AB\) is adjacent to \(\angle BAD\) and \(AD\) is the hypotenuse, so
\[\cos(\angle BAD)=\frac{|AB|}{|AD|}=\frac{3}{5}=0.6,\qquad \angle BAD=\cos^{-1}(0.6)=53.13^\circ.\]
Step 2 - the lower part of that angle. Triangle \(ACB\) is also right-angled at \(B\), so its three angles give
\[\angle BAC=180^\circ-90^\circ-61^\circ=29^\circ.\]
Step 3 - subtract. Since \(C\) lies on \(BD\), the angle \(\angle DAC\) is the difference between the whole angle at \(A\) and \(\angle BAC\):
\[x=\angle BAD-\angle BAC=53.13^\circ-29^\circ=24.13^\circ.\]
Correct to one decimal place, \(x = \mathbf{24.1}\).
(b) The pyramid
\(OABCD\) is a pyramid on a square base of side \(2\ \text{cm}\) with a slant height of \(4\ \text{cm}\). In a pyramid the slant height is the distance from the apex \(O\) down the middle of a triangular face to the midpoint of a base edge (call that midpoint \(M\)). It is not the sloping corner edge \(OA\); mixing the two up is the usual error here. The foot \(N\) of the vertical height is the centre of the square base.
(i) Vertical height
\(N\) is the centre of the base and \(M\) is the midpoint of a base edge, so \(NM\) is half the side:
\[NM=\tfrac{1}{2}\times 2=1\ \text{cm}.\]
The vertical height \(h=ON\), the slant height \(l=OM=4\ \text{cm}\) and \(NM=1\ \text{cm}\) form a right triangle at \(N\). By Pythagoras' theorem,
\[l^{2}=h^{2}+NM^{2}\ \Rightarrow\ h^{2}=4^{2}-1^{2}=16-1=15,\]
\[h=\sqrt{15}=3.873\ \text{cm}.\]
Correct to three significant figures, \(h = \mathbf{3.87\ \text{cm}}\).
(ii) Volume
\[V=\frac{1}{3}\times(\text{base area})\times h=\frac{1}{3}\times(2\times 2)\times\sqrt{15}=\frac{4\sqrt{15}}{3}.\]
\[V=\frac{4\times 3.873}{3}=\frac{15.49}{3}=5.164\ \text{cm}^{3}.\]
Correct to three significant figures, \(V = \mathbf{5.16\ \text{cm}^{3}}\).
Answer Details
(a) Value of x
In the diagram \(\triangle ABD\) is right-angled at \(B\), with \(|AB| = 3\ \text{cm}\) and hypotenuse \(|AD| = 5\ \text{cm}\). Point \(C\) lies on \(BD\), \(\angle ACB = 61^\circ\) and \(\angle DAC = x^\circ\).
Step 1 - the whole angle at A. In right triangle \(ABD\), \(AB\) is adjacent to \(\angle BAD\) and \(AD\) is the hypotenuse, so
\[\cos(\angle BAD)=\frac{|AB|}{|AD|}=\frac{3}{5}=0.6,\qquad \angle BAD=\cos^{-1}(0.6)=53.13^\circ.\]
Step 2 - the lower part of that angle. Triangle \(ACB\) is also right-angled at \(B\), so its three angles give
\[\angle BAC=180^\circ-90^\circ-61^\circ=29^\circ.\]
Step 3 - subtract. Since \(C\) lies on \(BD\), the angle \(\angle DAC\) is the difference between the whole angle at \(A\) and \(\angle BAC\):
\[x=\angle BAD-\angle BAC=53.13^\circ-29^\circ=24.13^\circ.\]
Correct to one decimal place, \(x = \mathbf{24.1}\).
(b) The pyramid
\(OABCD\) is a pyramid on a square base of side \(2\ \text{cm}\) with a slant height of \(4\ \text{cm}\). In a pyramid the slant height is the distance from the apex \(O\) down the middle of a triangular face to the midpoint of a base edge (call that midpoint \(M\)). It is not the sloping corner edge \(OA\); mixing the two up is the usual error here. The foot \(N\) of the vertical height is the centre of the square base.
(i) Vertical height
\(N\) is the centre of the base and \(M\) is the midpoint of a base edge, so \(NM\) is half the side:
\[NM=\tfrac{1}{2}\times 2=1\ \text{cm}.\]
The vertical height \(h=ON\), the slant height \(l=OM=4\ \text{cm}\) and \(NM=1\ \text{cm}\) form a right triangle at \(N\). By Pythagoras' theorem,
\[l^{2}=h^{2}+NM^{2}\ \Rightarrow\ h^{2}=4^{2}-1^{2}=16-1=15,\]
\[h=\sqrt{15}=3.873\ \text{cm}.\]
Correct to three significant figures, \(h = \mathbf{3.87\ \text{cm}}\).
(ii) Volume
\[V=\frac{1}{3}\times(\text{base area})\times h=\frac{1}{3}\times(2\times 2)\times\sqrt{15}=\frac{4\sqrt{15}}{3}.\]
\[V=\frac{4\times 3.873}{3}=\frac{15.49}{3}=5.164\ \text{cm}^{3}.\]
Correct to three significant figures, \(V = \mathbf{5.16\ \text{cm}^{3}}\).
