WAEC SSCE - General Mathematics - 2004 (Essay)
Tambaya 1 Rahoto
(a) Copy and complete the following table of values for the relation \(y = 2x^{2} - 7x - 3\).
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 19 | -3 | -9 |
(b) Using 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = 2x^{2} - 7x - 3\) for \(-2 \leq x \leq 5\).
(c) From your graph, find the : (i) minimum value of y ;
(ii) gradient of the curve at x = 1.
(d) By drawing a suitable straight line, find the values of x for which \(2x^{2} - 7x - 5 = x + 4\).
(a) For \(y=2x^2-7x-3\), the completed table is:
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(19\) | \(6\) | \(-3\) | \(-8\) | \(-9\) | \(-6\) | \(1\) | \(12\) |
(b) The points were plotted using the stated scales, and joined with a smooth curve. The straight line \(y=x+6\) and the tangent at \(x=1\) are included for parts (c) and (d).
(c)
(i) The lowest point of the curve gives the minimum value
\[\boxed{y\approx -9}\]
(ii) The tangent at \(x=1\) passes through \((1,-8)\). Using two points on the tangent, for example \((0,-5)\) and \((1,-8)\),
\[\text{gradient}=\frac{-8-(-5)}{1-0}=\boxed{-3}.\]
(d) Rearrange the equation in terms of the curve already drawn:
\[2x^2-7x-5=x+4\]
\[2x^2-7x-3=x+6.\]
Thus, draw the straight line \(y=x+6\). From its points of intersection with the parabola on the graph,
\[\boxed{x\approx -0.9\ \text{ or }\ x\approx 4.9}.\]
Bayanin Amsa
(a) For \(y=2x^2-7x-3\), the completed table is:
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(19\) | \(6\) | \(-3\) | \(-8\) | \(-9\) | \(-6\) | \(1\) | \(12\) |
(b) The points were plotted using the stated scales, and joined with a smooth curve. The straight line \(y=x+6\) and the tangent at \(x=1\) are included for parts (c) and (d).
(c)
(i) The lowest point of the curve gives the minimum value
\[\boxed{y\approx -9}\]
(ii) The tangent at \(x=1\) passes through \((1,-8)\). Using two points on the tangent, for example \((0,-5)\) and \((1,-8)\),
\[\text{gradient}=\frac{-8-(-5)}{1-0}=\boxed{-3}.\]
(d) Rearrange the equation in terms of the curve already drawn:
\[2x^2-7x-5=x+4\]
\[2x^2-7x-3=x+6.\]
Thus, draw the straight line \(y=x+6\). From its points of intersection with the parabola on the graph,
\[\boxed{x\approx -0.9\ \text{ or }\ x\approx 4.9}.\]
Tambaya 2 Rahoto
(a) The sides of an isosceles triangle triangle are in the ratio \(7 : 5 : 7\). Calculate, correct to the nearest degree, the angle included between the equal sides.
(b) The sum of the interior angles of a regular polygon is 1440°. Calculate : (i) the number of sides ; (ii) the size of one exterior angle of the polygon.
(a) Sides \(7:5:7\). The angle included between the two equal sides (each \(7k\)) is opposite the side \(5k\). By the cosine rule:
\[\cos\theta=\frac{7^{2}+7^{2}-5^{2}}{2(7)(7)}=\frac{49+49-25}{98}=\frac{73}{98}=0.7449.\]
\[\theta=\cos^{-1}(0.7449)=41.9^{\circ}\approx\mathbf{42^{\circ}}\ (\text{nearest degree}).\]
(b) Sum of interior angles \(=(n-2)\times180^{\circ}=1440^{\circ}\):
\[n-2=\frac{1440}{180}=8\Rightarrow \mathbf{n=10\text{ sides}}.\]
(ii) Each exterior angle \(=\dfrac{360^{\circ}}{n}=\dfrac{360^{\circ}}{10}=\mathbf{36^{\circ}}.\)
Bayanin Amsa
(a) Sides \(7:5:7\). The angle included between the two equal sides (each \(7k\)) is opposite the side \(5k\). By the cosine rule:
\[\cos\theta=\frac{7^{2}+7^{2}-5^{2}}{2(7)(7)}=\frac{49+49-25}{98}=\frac{73}{98}=0.7449.\]
\[\theta=\cos^{-1}(0.7449)=41.9^{\circ}\approx\mathbf{42^{\circ}}\ (\text{nearest degree}).\]
(b) Sum of interior angles \(=(n-2)\times180^{\circ}=1440^{\circ}\):
\[n-2=\frac{1440}{180}=8\Rightarrow \mathbf{n=10\text{ sides}}.\]
(ii) Each exterior angle \(=\dfrac{360^{\circ}}{n}=\dfrac{360^{\circ}}{10}=\mathbf{36^{\circ}}.\)
Tambaya 3 Rahoto
K(lat. 60°N, long. 50°W) is a point on the eart's surface. L is another point due East of K and the third point N is due North of K. The distance KL is 3520km and KN is 10951km.
(a) Calculate: (i) The longitude of L ; (ii) The latitude of N. (Take \(\pi = \frac{22}{7}\) and the radius of the earth = 6400km).
(b) A man was allowed 20% of his income as tax free. He then paid 25 kobo in the naira on the remainder. If he paid N1,200.00 as tax, calculate his total income.
\(R=6400\) km, \(\pi=\tfrac{22}{7}\). \(K(60^{\circ}\N,50^{\circ}\W)\).
(a)(i) Longitude of L (due east of \(K\), along latitude \(60^{\circ}\N\)). Radius of that parallel uses \(\cos60^{\circ}=\tfrac12\):
\[KL=\frac{\theta}{360}\times2\pi R\cos60^{\circ}\Rightarrow 3520=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\frac12.\]
Full parallel \(=\dfrac{44}{7}\times3200=20114.3\) km, so \(\theta=\dfrac{3520\times360}{20114.3}=63^{\circ}\) (eastward).
Starting at \(50^{\circ}\W\) and moving \(63^{\circ}\) east: \(-50^{\circ}+63^{\circ}=+13^{\circ}\). Longitude of \(L=13^{\circ}\E\).
(a)(ii) Latitude of N (due north of \(K\), along a meridian). Full meridian circle \(=2\pi R=\dfrac{44}{7}\times6400=40228.6\) km.
\[KN=\frac{\phi}{360}\times40228.6=10951\Rightarrow\phi=\frac{10951\times360}{40228.6}=98^{\circ}.\]
From \(60^{\circ}\N\), going \(98^{\circ}\) north first reaches the North Pole after \(30^{\circ}\), then continues \(68^{\circ}\) down the opposite meridian. Latitude of \(N=90^{\circ}-68^{\circ}=22^{\circ}\N\) (on the \(130^{\circ}\E\) meridian).
(b) Let total income be \(I\). Tax-free \(=20\%\), so taxable \(=0.8I\); tax \(=25\) kobo per naira \(=25\%\) of taxable:
\[0.25\times0.8I=1200\Rightarrow 0.2I=1200\Rightarrow I=\mathbf{N6000}.\]
Bayanin Amsa
\(R=6400\) km, \(\pi=\tfrac{22}{7}\). \(K(60^{\circ}\N,50^{\circ}\W)\).
(a)(i) Longitude of L (due east of \(K\), along latitude \(60^{\circ}\N\)). Radius of that parallel uses \(\cos60^{\circ}=\tfrac12\):
\[KL=\frac{\theta}{360}\times2\pi R\cos60^{\circ}\Rightarrow 3520=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\frac12.\]
Full parallel \(=\dfrac{44}{7}\times3200=20114.3\) km, so \(\theta=\dfrac{3520\times360}{20114.3}=63^{\circ}\) (eastward).
Starting at \(50^{\circ}\W\) and moving \(63^{\circ}\) east: \(-50^{\circ}+63^{\circ}=+13^{\circ}\). Longitude of \(L=13^{\circ}\E\).
(a)(ii) Latitude of N (due north of \(K\), along a meridian). Full meridian circle \(=2\pi R=\dfrac{44}{7}\times6400=40228.6\) km.
\[KN=\frac{\phi}{360}\times40228.6=10951\Rightarrow\phi=\frac{10951\times360}{40228.6}=98^{\circ}.\]
From \(60^{\circ}\N\), going \(98^{\circ}\) north first reaches the North Pole after \(30^{\circ}\), then continues \(68^{\circ}\) down the opposite meridian. Latitude of \(N=90^{\circ}-68^{\circ}=22^{\circ}\N\) (on the \(130^{\circ}\E\) meridian).
(b) Let total income be \(I\). Tax-free \(=20\%\), so taxable \(=0.8I\); tax \(=25\) kobo per naira \(=25\%\) of taxable:
\[0.25\times0.8I=1200\Rightarrow 0.2I=1200\Rightarrow I=\mathbf{N6000}.\]
Tambaya 4 Rahoto
(a) 
In the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96° at the centre. Calculate, correct to three significant figures, the area of the minor segment cut-off. (Take \(\pi = \frac{22}{7}\)).
(b) 
The figure shows a circle inscribed in a square. If a portion of the circle is shaded with some portions of the square, calculate the total area of the shaded portions. [Take \(\pi = \frac{22}{7}\)].
(a) Area of the minor segment
The chord \(XY\) subtends \(\theta = 96^\circ\) at the centre and the radius is \(r = 5\text{ cm}\). The minor segment is the minor sector minus triangle \(OXY\).
Area of minor sector (with \(\pi = \tfrac{22}{7}\)):
\[\frac{\theta}{360}\pi r^2 = \frac{96}{360}\times\frac{22}{7}\times 5^2 = 20.95\text{ cm}^2\]
Area of triangle OXY:
\[\frac{1}{2}r^2\sin\theta = \frac{1}{2}\times 5^2\times\sin 96^\circ = 12.43\text{ cm}^2\]
Area of minor segment:
\[20.95 - 12.43 = 8.52\text{ cm}^2\]
Correct to three significant figures, the area of the minor segment is \(\mathbf{8.52\text{ cm}^2}\).
(b) Total area of the shaded portions
The square has side 14 cm, so the inscribed circle has radius \(r = 7\text{ cm}\).
Shaded sector of the circle (angle \(80^\circ\)):
\[\frac{80}{360}\times\frac{22}{7}\times 7^2 = 34.22\text{ cm}^2\]
Shaded portion of the square (the four corners between the square and the circle):
\[\text{Area of square} = 14^2 = 196\text{ cm}^2\]
\[\text{Area of circle} = \frac{22}{7}\times 7^2 = 154\text{ cm}^2\]
\[196 - 154 = 42\text{ cm}^2\]
Total shaded area:
\[42 + 34.22 = 76.22 \approx 76.2\text{ cm}^2\]
Bayanin Amsa
(a) Area of the minor segment
The chord \(XY\) subtends \(\theta = 96^\circ\) at the centre and the radius is \(r = 5\text{ cm}\). The minor segment is the minor sector minus triangle \(OXY\).
Area of minor sector (with \(\pi = \tfrac{22}{7}\)):
\[\frac{\theta}{360}\pi r^2 = \frac{96}{360}\times\frac{22}{7}\times 5^2 = 20.95\text{ cm}^2\]
Area of triangle OXY:
\[\frac{1}{2}r^2\sin\theta = \frac{1}{2}\times 5^2\times\sin 96^\circ = 12.43\text{ cm}^2\]
Area of minor segment:
\[20.95 - 12.43 = 8.52\text{ cm}^2\]
Correct to three significant figures, the area of the minor segment is \(\mathbf{8.52\text{ cm}^2}\).
(b) Total area of the shaded portions
The square has side 14 cm, so the inscribed circle has radius \(r = 7\text{ cm}\).
Shaded sector of the circle (angle \(80^\circ\)):
\[\frac{80}{360}\times\frac{22}{7}\times 7^2 = 34.22\text{ cm}^2\]
Shaded portion of the square (the four corners between the square and the circle):
\[\text{Area of square} = 14^2 = 196\text{ cm}^2\]
\[\text{Area of circle} = \frac{22}{7}\times 7^2 = 154\text{ cm}^2\]
\[196 - 154 = 42\text{ cm}^2\]
Total shaded area:
\[42 + 34.22 = 76.22 \approx 76.2\text{ cm}^2\]
Tambaya 5 Rahoto
The table shows the marks scored by a group of students in a class test.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 1 | 4 | 9 | 8 | 5 | 3 |
(a)(i) Calculate the mean mark ; (ii) Find the median.
(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?
(a) Prepare the frequency table:
| Mark, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | Total |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 4 | 9 | 8 | 5 | 3 | 30 |
| \(fx\) | 0 | 4 | 18 | 24 | 20 | 15 | 81 |
| Cumulative frequency | 1 | 5 | 14 | 22 | 27 | 30 |
(i) Mean mark
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{81}{30}=2.7\]
(ii) Median
There are \(30\) observations. The median lies between the 15th and 16th observations. From the cumulative frequencies, both the 15th and 16th observations have mark \(3\).
\[\text{Median}=3\]
(b) Pie chart
The sector angle for each mark is \(\dfrac{f}{30}\times360^\circ\). Thus, for mark \(2\):
\[\frac{9}{30}\times360^\circ=108^\circ\]
Therefore, the sectorial angle for mark \(2\) is \(108^\circ\).
The complete pie chart is shown below. Its sector angles are \(12^\circ,48^\circ,108^\circ,96^\circ,60^\circ\), and \(36^\circ\) for marks \(0,1,2,3,4\), and \(5\) respectively.
Bayanin Amsa
(a) Prepare the frequency table:
| Mark, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | Total |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 4 | 9 | 8 | 5 | 3 | 30 |
| \(fx\) | 0 | 4 | 18 | 24 | 20 | 15 | 81 |
| Cumulative frequency | 1 | 5 | 14 | 22 | 27 | 30 |
(i) Mean mark
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{81}{30}=2.7\]
(ii) Median
There are \(30\) observations. The median lies between the 15th and 16th observations. From the cumulative frequencies, both the 15th and 16th observations have mark \(3\).
\[\text{Median}=3\]
(b) Pie chart
The sector angle for each mark is \(\dfrac{f}{30}\times360^\circ\). Thus, for mark \(2\):
\[\frac{9}{30}\times360^\circ=108^\circ\]
Therefore, the sectorial angle for mark \(2\) is \(108^\circ\).
The complete pie chart is shown below. Its sector angles are \(12^\circ,48^\circ,108^\circ,96^\circ,60^\circ\), and \(36^\circ\) for marks \(0,1,2,3,4\), and \(5\) respectively.
Tambaya 6 Rahoto
(a) Solve \(\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\)
(b) Simplify \(\frac{5}{\sqrt{7} - \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{3}}\), leaving your answer in surd form.
(a) Write both sides in base 3: \(81=3^{4},\ 27=3^{3}\).
\[\frac{1}{81^{x-2}}=3^{-4(x-2)},\qquad 27^{1-x}=3^{3(1-x)}.\]
Equating indices:
\[-4(x-2)=3(1-x)\Rightarrow -4x+8=3-3x\Rightarrow -x=-5\Rightarrow x=5.\]
(b) Rationalise each term:
\[\frac{5}{\sqrt7-\sqrt3}=\frac{5(\sqrt7+\sqrt3)}{7-3}=\frac{5(\sqrt7+\sqrt3)}{4},\qquad \frac{1}{\sqrt7+\sqrt3}=\frac{\sqrt7-\sqrt3}{4}.\]
\[\text{Sum}=\frac{5\sqrt7+5\sqrt3+\sqrt7-\sqrt3}{4}=\frac{6\sqrt7+4\sqrt3}{4}=\frac{3\sqrt7+2\sqrt3}{2}.\]
Bayanin Amsa
(a) Write both sides in base 3: \(81=3^{4},\ 27=3^{3}\).
\[\frac{1}{81^{x-2}}=3^{-4(x-2)},\qquad 27^{1-x}=3^{3(1-x)}.\]
Equating indices:
\[-4(x-2)=3(1-x)\Rightarrow -4x+8=3-3x\Rightarrow -x=-5\Rightarrow x=5.\]
(b) Rationalise each term:
\[\frac{5}{\sqrt7-\sqrt3}=\frac{5(\sqrt7+\sqrt3)}{7-3}=\frac{5(\sqrt7+\sqrt3)}{4},\qquad \frac{1}{\sqrt7+\sqrt3}=\frac{\sqrt7-\sqrt3}{4}.\]
\[\text{Sum}=\frac{5\sqrt7+5\sqrt3+\sqrt7-\sqrt3}{4}=\frac{6\sqrt7+4\sqrt3}{4}=\frac{3\sqrt7+2\sqrt3}{2}.\]
Tambaya 7 Rahoto
(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \(\Delta\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.
(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?
(a) \(O\) is the centre, so \(|OA|=|OB|=r\). Perimeter of \(\triangle AOB\):
\[2r+|AB|=52.2\Rightarrow 2r+24.2=52.2\Rightarrow r=14\text{ cm}.\]
By the cosine rule in \(\triangle AOB\):
\[\cos(\angle AOB)=\frac{r^{2}+r^{2}-|AB|^{2}}{2r^{2}}=\frac{196+196-585.64}{392}=\frac{-193.64}{392}=-0.4939.\]
\[\angle AOB=\cos^{-1}(-0.4939)=119.6^{\circ}\approx\mathbf{120^{\circ}}\ (\text{nearest degree}).\]
(b) Volume of tank \(=60\times80\times100=480{,}000\text{ cm}^{3}\). Half filled:
\[\tfrac12\times480{,}000=240{,}000\text{ cm}^{3}=\frac{240{,}000}{1000}=\mathbf{240\text{ litres}}\quad(1000\text{ cm}^{3}=1\text{ litre}).\]
Bayanin Amsa
(a) \(O\) is the centre, so \(|OA|=|OB|=r\). Perimeter of \(\triangle AOB\):
\[2r+|AB|=52.2\Rightarrow 2r+24.2=52.2\Rightarrow r=14\text{ cm}.\]
By the cosine rule in \(\triangle AOB\):
\[\cos(\angle AOB)=\frac{r^{2}+r^{2}-|AB|^{2}}{2r^{2}}=\frac{196+196-585.64}{392}=\frac{-193.64}{392}=-0.4939.\]
\[\angle AOB=\cos^{-1}(-0.4939)=119.6^{\circ}\approx\mathbf{120^{\circ}}\ (\text{nearest degree}).\]
(b) Volume of tank \(=60\times80\times100=480{,}000\text{ cm}^{3}\). Half filled:
\[\tfrac12\times480{,}000=240{,}000\text{ cm}^{3}=\frac{240{,}000}{1000}=\mathbf{240\text{ litres}}\quad(1000\text{ cm}^{3}=1\text{ litre}).\]
Tambaya 8 Rahoto
(a) Simplify : \(\sqrt{1001_{two}}\), leaving your answer in base two.
(b) 
In the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90°. Find the value of z in terms of x and y.
(c) 
In the diagram, P, Q, R and S are points of the circle centre O. \(\stackrel\frown{POQ} = 160°\), \(\stackrel\frown{QSR} = 45°\) and \(\stackrel\frown{PQS} = 40°\). Calculate, (i) < QPS ; (ii) < RQS.
(a) Simplify \(\sqrt{1001_{two}}\), leaving the answer in base two.
First convert \(1001_{two}\) to base ten:
\[1001_{two}=1(2^3)+0(2^2)+0(2^1)+1(2^0)=8+0+0+1=9_{ten}.\]Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:
\[3_{ten}=1(2^1)+1(2^0)=11_{two}.\]Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).
(b) Find \(z\) in terms of \(x\) and \(y\).
In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):
\[|KP|=\tfrac{1}{2}|PQ|=\tfrac{z}{2}.\]Applying Pythagoras' theorem to right-angled triangle \(OKP\):
\[|OP|^2=|OK|^2+|KP|^2\]\[x^2=y^2+\left(\frac{z}{2}\right)^2\]\[\left(\frac{z}{2}\right)^2=x^2-y^2\]\[\frac{z}{2}=\sqrt{x^2-y^2}\]Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).
(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).
(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:
\[\angle PSQ=\tfrac{1}{2}\times 160^\circ=80^\circ.\]In triangle \(PQS\), the three angles sum to \(180^\circ\):
\[\angle QPS=180^\circ-\angle PQS-\angle PSQ=180^\circ-40^\circ-80^\circ=\mathbf{60^\circ}.\](ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.
- Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
- \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
- \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).
The four arcs around the circle sum to \(360^\circ\):
\[\text{arc }PQ+\text{arc }QR+\text{arc }RS+\text{arc }SP=360^\circ\]\[160^\circ+90^\circ+\text{arc }RS+80^\circ=360^\circ\]\[\text{arc }RS=30^\circ.\]\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):
\[\angle RQS=\tfrac{1}{2}\times30^\circ=\mathbf{15^\circ}.\]Bayanin Amsa
(a) Simplify \(\sqrt{1001_{two}}\), leaving the answer in base two.
First convert \(1001_{two}\) to base ten:
\[1001_{two}=1(2^3)+0(2^2)+0(2^1)+1(2^0)=8+0+0+1=9_{ten}.\]Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:
\[3_{ten}=1(2^1)+1(2^0)=11_{two}.\]Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).
(b) Find \(z\) in terms of \(x\) and \(y\).
In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):
\[|KP|=\tfrac{1}{2}|PQ|=\tfrac{z}{2}.\]Applying Pythagoras' theorem to right-angled triangle \(OKP\):
\[|OP|^2=|OK|^2+|KP|^2\]\[x^2=y^2+\left(\frac{z}{2}\right)^2\]\[\left(\frac{z}{2}\right)^2=x^2-y^2\]\[\frac{z}{2}=\sqrt{x^2-y^2}\]Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).
(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).
(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:
\[\angle PSQ=\tfrac{1}{2}\times 160^\circ=80^\circ.\]In triangle \(PQS\), the three angles sum to \(180^\circ\):
\[\angle QPS=180^\circ-\angle PQS-\angle PSQ=180^\circ-40^\circ-80^\circ=\mathbf{60^\circ}.\](ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.
- Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
- \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
- \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).
The four arcs around the circle sum to \(360^\circ\):
\[\text{arc }PQ+\text{arc }QR+\text{arc }RS+\text{arc }SP=360^\circ\]\[160^\circ+90^\circ+\text{arc }RS+80^\circ=360^\circ\]\[\text{arc }RS=30^\circ.\]\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):
\[\angle RQS=\tfrac{1}{2}\times30^\circ=\mathbf{15^\circ}.\]Tambaya 9 Rahoto
(a) The angles of depression of the top and bottom of a building are 51° and 62° respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.
(b) 
In the diagram, PR is a chord of the circle centre O and radius 30cm, < POR = 120°. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \(\pi = 3.142\)).
(a) Height of the building.
Let \(d\) be the horizontal distance between the tower and the building. The observer is at the top of the tower, height \(72\text{ m}\).
Bottom of the building (same level as the foot of the tower) has an angle of depression of \(62^{\circ}\), so the vertical drop is the full \(72\text{ m}\):
\[\tan 62^{\circ}=\frac{72}{d}\Rightarrow d=\frac{72}{\tan 62^{\circ}}=\frac{72}{1.8807}=38.28\text{ m}\]
Top of the building has an angle of depression of \(51^{\circ}\), so the drop from the tower top down to the building top is:
\[\text{drop}=d\tan 51^{\circ}=38.28\times 1.2349=47.27\text{ m}\]
The height of the building is the tower height minus this drop:
\[h=72-47.27=24.73\text{ m}\]
Correct to 2 significant figures, the height of the building is \(25\text{ m}\).
(b) Circle, centre O, radius 30 cm, \(\angle POR=120^{\circ}\), \(\pi=3.142\).
(i) Length of chord PR.
Using the cosine rule in triangle \(POR\) with \(|OP|=|OR|=30\text{ cm}\):
\[|PR|^{2}=30^{2}+30^{2}-2(30)(30)\cos 120^{\circ}\]
\[=900+900-1800(-0.5)=1800+900=2700\]
\[|PR|=\sqrt{2700}=51.96\text{ cm}\approx 52.0\text{ cm}\]
(ii) Length of arc PQR.
From the diagram, \(Q\) lies on the major arc, so arc \(PQR\) uses the reflex angle:
\[\text{Reflex }\angle POR=360^{\circ}-120^{\circ}=240^{\circ}\]
\[\text{Arc }PQR=\frac{240}{360}\times 2\pi r=\frac{240}{360}\times 2\times 3.142\times 30\]
\[=\frac{240}{360}\times 188.52=0.6667\times 188.52=125.7\text{ cm}\]
(iii) Perimeter of the shaded portion.
The shaded region is the major segment, bounded by the chord \(PR\) and the major arc \(PQR\):
\[\text{Perimeter}=|PR|+\text{arc }PQR=51.96+125.7=177.7\text{ cm}\]
Correct to 3 significant figures, the perimeter of the shaded portion is \(178\text{ cm}\).
Bayanin Amsa
(a) Height of the building.
Let \(d\) be the horizontal distance between the tower and the building. The observer is at the top of the tower, height \(72\text{ m}\).
Bottom of the building (same level as the foot of the tower) has an angle of depression of \(62^{\circ}\), so the vertical drop is the full \(72\text{ m}\):
\[\tan 62^{\circ}=\frac{72}{d}\Rightarrow d=\frac{72}{\tan 62^{\circ}}=\frac{72}{1.8807}=38.28\text{ m}\]
Top of the building has an angle of depression of \(51^{\circ}\), so the drop from the tower top down to the building top is:
\[\text{drop}=d\tan 51^{\circ}=38.28\times 1.2349=47.27\text{ m}\]
The height of the building is the tower height minus this drop:
\[h=72-47.27=24.73\text{ m}\]
Correct to 2 significant figures, the height of the building is \(25\text{ m}\).
(b) Circle, centre O, radius 30 cm, \(\angle POR=120^{\circ}\), \(\pi=3.142\).
(i) Length of chord PR.
Using the cosine rule in triangle \(POR\) with \(|OP|=|OR|=30\text{ cm}\):
\[|PR|^{2}=30^{2}+30^{2}-2(30)(30)\cos 120^{\circ}\]
\[=900+900-1800(-0.5)=1800+900=2700\]
\[|PR|=\sqrt{2700}=51.96\text{ cm}\approx 52.0\text{ cm}\]
(ii) Length of arc PQR.
From the diagram, \(Q\) lies on the major arc, so arc \(PQR\) uses the reflex angle:
\[\text{Reflex }\angle POR=360^{\circ}-120^{\circ}=240^{\circ}\]
\[\text{Arc }PQR=\frac{240}{360}\times 2\pi r=\frac{240}{360}\times 2\times 3.142\times 30\]
\[=\frac{240}{360}\times 188.52=0.6667\times 188.52=125.7\text{ cm}\]
(iii) Perimeter of the shaded portion.
The shaded region is the major segment, bounded by the chord \(PR\) and the major arc \(PQR\):
\[\text{Perimeter}=|PR|+\text{arc }PQR=51.96+125.7=177.7\text{ cm}\]
Correct to 3 significant figures, the perimeter of the shaded portion is \(178\text{ cm}\).
Tambaya 10 Rahoto
(a) In the simultaneous equations : \(px + qy = 5 ; qx + py = -10\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.
(b) Solve : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
(a) Substitute \(x=1,\ y=-2\):
\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]
\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]
From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).
Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).
(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):
\[(4r-3)(3r+4)=(2r-1)(6r+1)\]
\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]
Bayanin Amsa
(a) Substitute \(x=1,\ y=-2\):
\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]
\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]
From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).
Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).
(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):
\[(4r-3)(3r+4)=(2r-1)(6r+1)\]
\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]
Tambaya 11 Rahoto
Using a ruler and a pair of compasses only,
(a) Construct : (i) \(\Delta PQR\) such that /PQ/ = 8cm, /PR/ = 7cm and < QPR = 105°. (ii) locus \(L_{1}\) of points equidistant from P and Q. (iii) locus \(l_{2}\) of points equidistant Q and R.
(b)(i) Label the point T where \(l_{1}\) and \(l_{2}\) intersect ; (ii) With centre T and radius /TQ/, construct a circle \(l_{3}\). (iii) Complete quadrilateral PQSR such that /RS/ = /QS/ and /RQ/ = /TS/.
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Construction of Triangle PQR and the Required Loci
Given: Construct a triangle PQR such that
- PQ = 8 cm,
- PR = 7 cm,
- ∠QPR = 105°.
(a)(i) Construction of ΔPQR
- Using a ruler, draw a line segment PQ = 8 cm.
- At point P, construct an angle of 105°:
- Construct a 90° angle at P by drawing a perpendicular to PQ.
- Construct a 120° ray at P using equal arcs with the compasses.
- Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
- On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
- Join R to Q. This gives the required triangle ΔPQR.
(a)(ii) Locus L1 of points equidistant from P and Q
The locus of points equidistant from P and Q is the perpendicular bisector of PQ.
- With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
- Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
- Join the two arc-intersection points with a ruler.
This straight line is labelled L1.
(a)(iii) Locus l2 of points equidistant from Q and R
The locus of points equidistant from Q and R is the perpendicular bisector of QR.
- Draw equal arcs centred at Q and R, using a radius greater than half of QR.
- Join the two points where the arcs intersect.
This straight line is labelled l2.
(b)(i) Point of Intersection
Label the point where L1 and l2 intersect as T.
Since T lies on the perpendicular bisector of QR, TQ = TR.
Therefore, TP = TQ = TR.
Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.
```Bayanin Amsa
```html
Construction of Triangle PQR and the Required Loci
Given: Construct a triangle PQR such that
- PQ = 8 cm,
- PR = 7 cm,
- ∠QPR = 105°.
(a)(i) Construction of ΔPQR
- Using a ruler, draw a line segment PQ = 8 cm.
- At point P, construct an angle of 105°:
- Construct a 90° angle at P by drawing a perpendicular to PQ.
- Construct a 120° ray at P using equal arcs with the compasses.
- Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
- On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
- Join R to Q. This gives the required triangle ΔPQR.
(a)(ii) Locus L1 of points equidistant from P and Q
The locus of points equidistant from P and Q is the perpendicular bisector of PQ.
- With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
- Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
- Join the two arc-intersection points with a ruler.
This straight line is labelled L1.
(a)(iii) Locus l2 of points equidistant from Q and R
The locus of points equidistant from Q and R is the perpendicular bisector of QR.
- Draw equal arcs centred at Q and R, using a radius greater than half of QR.
- Join the two points where the arcs intersect.
This straight line is labelled l2.
(b)(i) Point of Intersection
Label the point where L1 and l2 intersect as T.
Since T lies on the perpendicular bisector of QR, TQ = TR.
Therefore, TP = TQ = TR.
Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.
```Tambaya 12 Rahoto
(a) Simplify : \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\)
(b) A point P is 40km from Q on a bearing 061°. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.
(c) A man left N5,720 to be shared among his son and three daughters. Each daughter's share was \(\frac{3}{4}\) of the son's share. How much did the son receive?
(a) Factor the first bracket:
\[\frac{x^{2}}{2}-x+\frac12=\frac12\left(x^{2}-2x+1\right)=\frac12(x-1)^{2}.\]
\[\therefore\ \frac12(x-1)^{2}\cdot\frac{1}{x-1}=\frac{x-1}{2}.\]
(b) \(P\) is 40 km from \(Q\) on bearing \(061^{\circ}\) (measured from north).
(i) North of Q: \(40\cos61^{\circ}=40(0.4848)=19.4\text{ km}\ (1\text{ d.p.}).\)
(ii) East of Q: \(40\sin61^{\circ}=40(0.8746)=35.0\text{ km}\ (1\text{ d.p.}).\)
(c) Let the son's share be \(s\). Each daughter gets \(\tfrac34 s\); three daughters get \(3\times\tfrac34 s=\tfrac94 s\).
\[s+\frac94 s=5720\Rightarrow\frac{13}{4}s=5720\Rightarrow s=5720\times\frac{4}{13}=\mathbf{N1760}.\]
Bayanin Amsa
(a) Factor the first bracket:
\[\frac{x^{2}}{2}-x+\frac12=\frac12\left(x^{2}-2x+1\right)=\frac12(x-1)^{2}.\]
\[\therefore\ \frac12(x-1)^{2}\cdot\frac{1}{x-1}=\frac{x-1}{2}.\]
(b) \(P\) is 40 km from \(Q\) on bearing \(061^{\circ}\) (measured from north).
(i) North of Q: \(40\cos61^{\circ}=40(0.4848)=19.4\text{ km}\ (1\text{ d.p.}).\)
(ii) East of Q: \(40\sin61^{\circ}=40(0.8746)=35.0\text{ km}\ (1\text{ d.p.}).\)
(c) Let the son's share be \(s\). Each daughter gets \(\tfrac34 s\); three daughters get \(3\times\tfrac34 s=\tfrac94 s\).
\[s+\frac94 s=5720\Rightarrow\frac{13}{4}s=5720\Rightarrow s=5720\times\frac{4}{13}=\mathbf{N1760}.\]
Tambaya 13 Rahoto
The frequency distribution shows tha marks of 100 students in a Mathematics test.
| Marks | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| No. of Students |
2 | 4 | 9 | 13 | 18 | 32 | 13 | 5 | 3 | 1 |
(a) Draw cumulative frequency curve for the distribution .
(b) Use your curve to estimate : (i) the median ; (ii) the lower quartile ; (iii) the 60th percentile.
(a) Less-than cumulative frequency table
| Marks | Class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 1–10 | 10.5 | 2 | 2 |
| 11–20 | 20.5 | 4 | 6 |
| 21–30 | 30.5 | 9 | 15 |
| 31–40 | 40.5 | 13 | 28 |
| 41–50 | 50.5 | 18 | 46 |
| 51–60 | 60.5 | 32 | 78 |
| 61–70 | 70.5 | 13 | 91 |
| 71–80 | 80.5 | 5 | 96 |
| 81–90 | 90.5 | 3 | 99 |
| 91–100 | 100.5 | 1 | 100 |
Plot cumulative frequency against the upper class boundaries, beginning with 0.5,0 . Draw a smooth increasing curve through the plotted points.
(b) Estimates from the ogive
The total frequency is 0.5+99.5=100 students.
- Median position = rac{100}{2}=50 . From cumulative frequency 50, the corresponding mark is approximately 52 marks.
- Lower quartile position = rac{100}{4}=25 . From cumulative frequency 25, the corresponding mark is approximately 39 marks.
- 60th percentile position = rac{60}{100} imes100=60 . From cumulative frequency 60, the corresponding mark is approximately 55 marks.
Bayanin Amsa
(a) Less-than cumulative frequency table
| Marks | Class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 1–10 | 10.5 | 2 | 2 |
| 11–20 | 20.5 | 4 | 6 |
| 21–30 | 30.5 | 9 | 15 |
| 31–40 | 40.5 | 13 | 28 |
| 41–50 | 50.5 | 18 | 46 |
| 51–60 | 60.5 | 32 | 78 |
| 61–70 | 70.5 | 13 | 91 |
| 71–80 | 80.5 | 5 | 96 |
| 81–90 | 90.5 | 3 | 99 |
| 91–100 | 100.5 | 1 | 100 |
Plot cumulative frequency against the upper class boundaries, beginning with 0.5,0 . Draw a smooth increasing curve through the plotted points.
(b) Estimates from the ogive
The total frequency is 0.5+99.5=100 students.
- Median position = rac{100}{2}=50 . From cumulative frequency 50, the corresponding mark is approximately 52 marks.
- Lower quartile position = rac{100}{4}=25 . From cumulative frequency 25, the corresponding mark is approximately 39 marks.
- 60th percentile position = rac{60}{100} imes100=60 . From cumulative frequency 60, the corresponding mark is approximately 55 marks.
