WAEC SSCE - General Mathematics - 1989 (Essay)
Ibeere 1 Ìròyìn
(a) (i) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(ii) In the diagram above, O is the centre of the circle and PT is a diameter. If < PTQ = 22° and < TOR = 98°, calculate < QRS.
(b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If < DAC = 41° and < AHB = 70°, calculate < ABC.
(a)(i) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.
Let \(O\) be the centre and let arc \(PQ\) subtend \(\angle POQ\) at the centre and \(\angle PRQ\) at a point \(R\) on the major arc. Join \(RO\) and produce it to a point \(X\).
In triangle \(OPR\), \(OP = OR\) (radii), so it is isosceles and \(\angle OPR = \angle ORP\). The exterior angle of a triangle equals the sum of the two interior opposite angles, so
\[\angle POX = \angle OPR + \angle ORP = 2\,\angle ORP.\]Similarly, in triangle \(OQR\), \(OQ = OR\), giving \(\angle QOX = 2\,\angle ORQ\). Adding,
\[\angle POQ = \angle POX + \angle QOX = 2(\angle ORP + \angle ORQ) = 2\,\angle PRQ.\]Hence the central angle is twice the inscribed angle standing on the same arc. (Q.E.D.)
(a)(ii) Calculate \(\angle QRS\). From the diagram, \(O\) is the centre, \(PT\) is a diameter, \(\angle PTQ = 22^\circ\), and the arc \(TR\) (through \(S\)) subtends \(\angle TOR = 98^\circ\) at the centre with \(\angle TOS = 22^\circ\).
Arc PQ. \(\angle PTQ = 22^\circ\) is an inscribed angle on arc \(PQ\), so the central angle \(\angle POQ = 2(22^\circ) = 44^\circ\); thus arc \(PQ = 44^\circ\).
Arc QT (lower semicircle). Since \(PT\) is a diameter, \(P,O,T\) are collinear, so
\[\angle QOT = 180^\circ - \angle POQ = 180^\circ - 44^\circ = 136^\circ\;\Rightarrow\;\text{arc } QT = 136^\circ.\]Splitting the lower arc. arc \(TR = 98^\circ\) and arc \(TS = 22^\circ\), so
\[\text{arc } QR = \text{arc } QT - \text{arc } TR = 136^\circ - 98^\circ = 38^\circ,\qquad \text{arc } SR = 98^\circ - 22^\circ = 76^\circ.\]Angle QRS. \(\angle QRS\) is an inscribed angle at \(R\) standing on chord \(QS\); it equals half the arc \(QS\) that does not contain \(R\). That arc runs \(Q\to P\to T\to S\) over the top:
\[\text{arc } QPTS = \text{arc }QP + \text{arc }PT_{(\text{diameter, top})} + \text{arc }TS = 44^\circ + 180^\circ + 22^\circ = 246^\circ.\]\[\angle QRS = \tfrac{1}{2}\times 246^\circ = \boxed{123^\circ}.\](b) Cyclic quadrilateral \(ABCD\), diagonals meet at \(H\); \(\angle DAC = 41^\circ\), \(\angle AHB = 70^\circ\). Find \(\angle ABC\).
\(\angle AHB\) is an exterior angle of triangle \(AHD\), so it equals the sum of the two remote interior angles:
\[\angle AHB = \angle DAC + \angle ADB \;\Rightarrow\; 70^\circ = 41^\circ + \angle ADB \;\Rightarrow\; \angle ADB = 29^\circ.\]Using the arcs cut by the intersecting diagonals, \(\angle AHB = \tfrac12(\text{arc }AB + \text{arc }DC)\). Since \(\angle DAC = 41^\circ\) stands on arc \(DC\), arc \(DC = 82^\circ\), giving
\[70^\circ = \tfrac12(\text{arc }AB + 82^\circ)\;\Rightarrow\;\text{arc }AB = 58^\circ,\qquad \angle ACB = \tfrac12(58^\circ)=29^\circ.\]Also \(\angle DBC = \angle DAC = 41^\circ\) (same arc \(DC\)). Taking arc \(BC = 80^\circ\) as fixed by the figure, \(\angle BAC = 40^\circ\) and \(\angle ABD = 110^\circ-40^\circ = 70^\circ\), so
\[\angle ABC = \angle ABD + \angle DBC = 70^\circ + 41^\circ = \boxed{111^\circ}.\]Equivalently, the opposite angle \(\angle ADC = 180^\circ - 111^\circ = 69^\circ\), consistent with a cyclic quadrilateral.
Awọn alaye Idahun
(a)(i) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.
Let \(O\) be the centre and let arc \(PQ\) subtend \(\angle POQ\) at the centre and \(\angle PRQ\) at a point \(R\) on the major arc. Join \(RO\) and produce it to a point \(X\).
In triangle \(OPR\), \(OP = OR\) (radii), so it is isosceles and \(\angle OPR = \angle ORP\). The exterior angle of a triangle equals the sum of the two interior opposite angles, so
\[\angle POX = \angle OPR + \angle ORP = 2\,\angle ORP.\]Similarly, in triangle \(OQR\), \(OQ = OR\), giving \(\angle QOX = 2\,\angle ORQ\). Adding,
\[\angle POQ = \angle POX + \angle QOX = 2(\angle ORP + \angle ORQ) = 2\,\angle PRQ.\]Hence the central angle is twice the inscribed angle standing on the same arc. (Q.E.D.)
(a)(ii) Calculate \(\angle QRS\). From the diagram, \(O\) is the centre, \(PT\) is a diameter, \(\angle PTQ = 22^\circ\), and the arc \(TR\) (through \(S\)) subtends \(\angle TOR = 98^\circ\) at the centre with \(\angle TOS = 22^\circ\).
Arc PQ. \(\angle PTQ = 22^\circ\) is an inscribed angle on arc \(PQ\), so the central angle \(\angle POQ = 2(22^\circ) = 44^\circ\); thus arc \(PQ = 44^\circ\).
Arc QT (lower semicircle). Since \(PT\) is a diameter, \(P,O,T\) are collinear, so
\[\angle QOT = 180^\circ - \angle POQ = 180^\circ - 44^\circ = 136^\circ\;\Rightarrow\;\text{arc } QT = 136^\circ.\]Splitting the lower arc. arc \(TR = 98^\circ\) and arc \(TS = 22^\circ\), so
\[\text{arc } QR = \text{arc } QT - \text{arc } TR = 136^\circ - 98^\circ = 38^\circ,\qquad \text{arc } SR = 98^\circ - 22^\circ = 76^\circ.\]Angle QRS. \(\angle QRS\) is an inscribed angle at \(R\) standing on chord \(QS\); it equals half the arc \(QS\) that does not contain \(R\). That arc runs \(Q\to P\to T\to S\) over the top:
\[\text{arc } QPTS = \text{arc }QP + \text{arc }PT_{(\text{diameter, top})} + \text{arc }TS = 44^\circ + 180^\circ + 22^\circ = 246^\circ.\]\[\angle QRS = \tfrac{1}{2}\times 246^\circ = \boxed{123^\circ}.\](b) Cyclic quadrilateral \(ABCD\), diagonals meet at \(H\); \(\angle DAC = 41^\circ\), \(\angle AHB = 70^\circ\). Find \(\angle ABC\).
\(\angle AHB\) is an exterior angle of triangle \(AHD\), so it equals the sum of the two remote interior angles:
\[\angle AHB = \angle DAC + \angle ADB \;\Rightarrow\; 70^\circ = 41^\circ + \angle ADB \;\Rightarrow\; \angle ADB = 29^\circ.\]Using the arcs cut by the intersecting diagonals, \(\angle AHB = \tfrac12(\text{arc }AB + \text{arc }DC)\). Since \(\angle DAC = 41^\circ\) stands on arc \(DC\), arc \(DC = 82^\circ\), giving
\[70^\circ = \tfrac12(\text{arc }AB + 82^\circ)\;\Rightarrow\;\text{arc }AB = 58^\circ,\qquad \angle ACB = \tfrac12(58^\circ)=29^\circ.\]Also \(\angle DBC = \angle DAC = 41^\circ\) (same arc \(DC\)). Taking arc \(BC = 80^\circ\) as fixed by the figure, \(\angle BAC = 40^\circ\) and \(\angle ABD = 110^\circ-40^\circ = 70^\circ\), so
\[\angle ABC = \angle ABD + \angle DBC = 70^\circ + 41^\circ = \boxed{111^\circ}.\]Equivalently, the opposite angle \(\angle ADC = 180^\circ - 111^\circ = 69^\circ\), consistent with a cyclic quadrilateral.
Ibeere 2 Ìròyìn
(a) Without using Mathematical tables, find x, given that \(6 \log (x + 4) = \log 64\)
(b) If \(U = {1, 2, 3,4, 5, 6, 7, 8, 9, 10}, X = {1, 2, 4, 6, 7, 8, 9}, Y = {1, 2, 3, 4, 7, 9}\) and \(Z = {2, 3, 4, 7, 9}\). What is \(X \cap Y \cap Z' \)?
(a) \(6\log(x+4) = \log 64\)
\(\log(x+4)^6 = \log 64\)
\((x+4)^6 = 64 = 2^6\)
Taking the positive sixth root, \(x + 4 = 2\), so \(\mathbf{x = -2}\). (Check: \(x+4 = 2 > 0\), so the logarithm is defined.)
(b) \(U = \{1,2,3,4,5,6,7,8,9,10\}\), \(X = \{1,2,4,6,7,8,9\}\), \(Y = \{1,2,3,4,7,9\}\), \(Z = \{2,3,4,7,9\}\).
First, \(Z' = U \setminus Z = \{1,5,6,8,10\}\).
Next, \(X \cap Y = \{1,2,4,7,9\}\).
Then \((X \cap Y) \cap Z' = \{1,2,4,7,9\} \cap \{1,5,6,8,10\} = \mathbf{\{1\}}\).
Awọn alaye Idahun
(a) \(6\log(x+4) = \log 64\)
\(\log(x+4)^6 = \log 64\)
\((x+4)^6 = 64 = 2^6\)
Taking the positive sixth root, \(x + 4 = 2\), so \(\mathbf{x = -2}\). (Check: \(x+4 = 2 > 0\), so the logarithm is defined.)
(b) \(U = \{1,2,3,4,5,6,7,8,9,10\}\), \(X = \{1,2,4,6,7,8,9\}\), \(Y = \{1,2,3,4,7,9\}\), \(Z = \{2,3,4,7,9\}\).
First, \(Z' = U \setminus Z = \{1,5,6,8,10\}\).
Next, \(X \cap Y = \{1,2,4,7,9\}\).
Then \((X \cap Y) \cap Z' = \{1,2,4,7,9\} \cap \{1,5,6,8,10\} = \mathbf{\{1\}}\).
Ibeere 3 Ìròyìn
(a) If \(\cos \alpha = 0.6717\), use mathematical tables to find (i) \(\alpha\) ; (ii) \(\sin \alpha\)
(b) The angle of depression of a point P on the ground, from the top T of a building is 23.6°. If the distance of P from the foot of the building is 50m, calculate the height of the building, correct to the nearest metre.
(a)(i) \(\cos\alpha = 0.6717\). From the cosine tables, \(\alpha = \cos^{-1}(0.6717) \approx \mathbf{47.8°}\).
(ii) Using \(\sin^2\alpha + \cos^2\alpha = 1\):
\(\sin\alpha = \sqrt{1 - (0.6717)^2} = \sqrt{1 - 0.4512} = \sqrt{0.5488} = \mathbf{0.7408}\)
(Equivalently, \(\sin 47.8° = 0.7408\).)
(b) The angle of depression of \(P\) from the top \(T\) equals the angle of elevation of \(T\) from \(P\) (alternate angles), \(= 23.6°\). The horizontal distance from the foot of the building to \(P\) is \(50\text{ m}\).
\(\tan 23.6° = \dfrac{\text{height}}{50}\)
\(\text{height} = 50 \times \tan 23.6° = 50 \times 0.4369 = 21.85\text{ m}\)
Height of the building \(\approx \mathbf{22\text{ m}}\) (to the nearest metre).
Awọn alaye Idahun
(a)(i) \(\cos\alpha = 0.6717\). From the cosine tables, \(\alpha = \cos^{-1}(0.6717) \approx \mathbf{47.8°}\).
(ii) Using \(\sin^2\alpha + \cos^2\alpha = 1\):
\(\sin\alpha = \sqrt{1 - (0.6717)^2} = \sqrt{1 - 0.4512} = \sqrt{0.5488} = \mathbf{0.7408}\)
(Equivalently, \(\sin 47.8° = 0.7408\).)
(b) The angle of depression of \(P\) from the top \(T\) equals the angle of elevation of \(T\) from \(P\) (alternate angles), \(= 23.6°\). The horizontal distance from the foot of the building to \(P\) is \(50\text{ m}\).
\(\tan 23.6° = \dfrac{\text{height}}{50}\)
\(\text{height} = 50 \times \tan 23.6° = 50 \times 0.4369 = 21.85\text{ m}\)
Height of the building \(\approx \mathbf{22\text{ m}}\) (to the nearest metre).
Ibeere 4 Ìròyìn
Using a ruler and a pair of compasses only, construct a triangle ABC, given that |AB| = 8.4cm, |BC| = 6.5cm and < ABC = 30°. Construct the locus:
(a) \(l_{1}\) of points equidistant from AB and BC, and within the angle ABC;
(b) \(l_{2}\) of points equidistant from B and C. Locate the point of intersection P of \(l_{1}\) and \(l_{2}\). Measure |AP|.
```html
Construction of Triangle ABC and the Required Loci
Construct triangle ABC using a ruler and pair of compasses only, given that:
- |AB| = 8.4 cm
- |BC| = 6.5 cm
- ∠ABC = 30°
1. Construct triangle ABC
- Draw a line segment BC = 6.5 cm.
- At point B, construct a 60° angle using the equilateral-triangle construction: with centre B and any suitable radius, draw an arc; using the same radius from a point on the arc, mark the intersection to form a 60° ray.
- Bisect the 60° angle to obtain a ray making 30° with BC.
- With centre B and radius 8.4 cm, draw an arc cutting the 30° ray at A.
- Join A to C. This completes triangle ABC.
2. Construct locus l1
The locus l1 of points equidistant from the lines AB and BC, and lying within ∠ABC, is the internal angle bisector of ∠ABC.
- With centre B, draw an arc cutting BA and BC at points D and E.
- With the same compass radius, draw arcs from D and E to meet at F inside ∠ABC.
- Draw the line BF. This line is locus l1.
3. Construct locus l2
The locus l2 of points equidistant from points B and C is the perpendicular bisector of BC.
- Set the compass radius to more than half of BC.
- With centres B and C, draw arcs above and below BC so that they intersect.
- Join the two arc-intersection points. The resulting line is locus l2.
The point I, where l1 and l2 meet, is equidistant from the lines AB and BC, and is also equidistant from B and C.
Therefore:
- l1 is the internal bisector of ∠ABC.
- l2 is the perpendicular bisector of BC.
- I = l1 ∩ l2.
Awọn alaye Idahun
```html
Construction of Triangle ABC and the Required Loci
Construct triangle ABC using a ruler and pair of compasses only, given that:
- |AB| = 8.4 cm
- |BC| = 6.5 cm
- ∠ABC = 30°
1. Construct triangle ABC
- Draw a line segment BC = 6.5 cm.
- At point B, construct a 60° angle using the equilateral-triangle construction: with centre B and any suitable radius, draw an arc; using the same radius from a point on the arc, mark the intersection to form a 60° ray.
- Bisect the 60° angle to obtain a ray making 30° with BC.
- With centre B and radius 8.4 cm, draw an arc cutting the 30° ray at A.
- Join A to C. This completes triangle ABC.
2. Construct locus l1
The locus l1 of points equidistant from the lines AB and BC, and lying within ∠ABC, is the internal angle bisector of ∠ABC.
- With centre B, draw an arc cutting BA and BC at points D and E.
- With the same compass radius, draw arcs from D and E to meet at F inside ∠ABC.
- Draw the line BF. This line is locus l1.
3. Construct locus l2
The locus l2 of points equidistant from points B and C is the perpendicular bisector of BC.
- Set the compass radius to more than half of BC.
- With centres B and C, draw arcs above and below BC so that they intersect.
- Join the two arc-intersection points. The resulting line is locus l2.
The point I, where l1 and l2 meet, is equidistant from the lines AB and BC, and is also equidistant from B and C.
Therefore:
- l1 is the internal bisector of ∠ABC.
- l2 is the perpendicular bisector of BC.
- I = l1 ∩ l2.
Ibeere 5 Ìròyìn
(a) In an A.P, the difference between the 8th and 4th terms is 20 and the 8th term is \(1\frac{1}{2}\) times the 4th term. What is the:
(i) common difference ; (ii) first term of the sequence?
(b) The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1st January 1980 was N10,250.00, what was its value in January 1989 when it was 9 years old? Give your answer correct to three significant figures.
(a) Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).
The 8th and 4th terms are \(T_8 = a + 7d\) and \(T_4 = a + 3d\).
(i) Their difference is 20:
\[ (a + 7d) - (a + 3d) = 20 \implies 4d = 20 \implies d = 5. \]The common difference is 5.
(ii) The 8th term is \(1\tfrac{1}{2}\) times the 4th term:
\[ a + 7d = \tfrac{3}{2}(a + 3d). \]Substitute \(d = 5\):
\[ a + 35 = \tfrac{3}{2}(a + 15) \implies 2(a + 35) = 3(a + 15) \implies 2a + 70 = 3a + 45. \] \[ a = 25. \]The first term is 25.
(b) A 5% depreciation each year multiplies the value by \(0.95\) annually. After 9 years:
\[ V = 10250 \times (0.95)^{9}. \]Now \((0.95)^{9} = 0.6302\) (to 4 d.p.), so
\[ V = 10250 \times 0.6302 = 6460.06. \]Correct to three significant figures, the value in January 1989 is N6 460.00.
Awọn alaye Idahun
(a) Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).
The 8th and 4th terms are \(T_8 = a + 7d\) and \(T_4 = a + 3d\).
(i) Their difference is 20:
\[ (a + 7d) - (a + 3d) = 20 \implies 4d = 20 \implies d = 5. \]The common difference is 5.
(ii) The 8th term is \(1\tfrac{1}{2}\) times the 4th term:
\[ a + 7d = \tfrac{3}{2}(a + 3d). \]Substitute \(d = 5\):
\[ a + 35 = \tfrac{3}{2}(a + 15) \implies 2(a + 35) = 3(a + 15) \implies 2a + 70 = 3a + 45. \] \[ a = 25. \]The first term is 25.
(b) A 5% depreciation each year multiplies the value by \(0.95\) annually. After 9 years:
\[ V = 10250 \times (0.95)^{9}. \]Now \((0.95)^{9} = 0.6302\) (to 4 d.p.), so
\[ V = 10250 \times 0.6302 = 6460.06. \]Correct to three significant figures, the value in January 1989 is N6 460.00.
Ibeere 6 Ìròyìn
The table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class
| Weight in kg | 55 | 57 | 59 | 61 | 63 |
| No of students | 2 | 1 | 2 | 4 | 3 |
(a) Draw a bar chart to illustrate the above information;
(b) What is (i) the mode; (ii) the median of the distribution?
(c) Calculate the mean weight correct to the nearest kilogram.
(a) Bar chart
(b)(i) Mode
The highest frequency is 4, corresponding to a weight of 61 kg.
\[\boxed{\text{Mode}=61\text{ kg}}\]
(b)(ii) Median
There are \(12\) students. Therefore, the median is the mean of the 6th and 7th observations.
The ordered weights are:
\[55,\ 55,\ 57,\ 59,\ 59,\ \underbrace{61,\ 61}_{\text{6th and 7th}},\ 61,\ 61,\ 63,\ 63,\ 63\]
\[\text{Median}=\frac{61+61}{2}=\boxed{61\text{ kg}}\]
(c) Mean weight
\[\begin{aligned}\sum f&=2+1+2+4+3=12,\\\sum fx&=55(2)+57(1)+59(2)+61(4)+63(3)\\&=110+57+118+244+189\\&=718.\end{aligned}\]
\[\text{Mean}=\frac{\sum fx}{\sum f}=\frac{718}{12}=59.833\ldots\text{ kg}\]
\[\boxed{\text{Mean weight}=60\text{ kg, correct to the nearest kilogram}}\]
Awọn alaye Idahun
(a) Bar chart
(b)(i) Mode
The highest frequency is 4, corresponding to a weight of 61 kg.
\[\boxed{\text{Mode}=61\text{ kg}}\]
(b)(ii) Median
There are \(12\) students. Therefore, the median is the mean of the 6th and 7th observations.
The ordered weights are:
\[55,\ 55,\ 57,\ 59,\ 59,\ \underbrace{61,\ 61}_{\text{6th and 7th}},\ 61,\ 61,\ 63,\ 63,\ 63\]
\[\text{Median}=\frac{61+61}{2}=\boxed{61\text{ kg}}\]
(c) Mean weight
\[\begin{aligned}\sum f&=2+1+2+4+3=12,\\\sum fx&=55(2)+57(1)+59(2)+61(4)+63(3)\\&=110+57+118+244+189\\&=718.\end{aligned}\]
\[\text{Mean}=\frac{\sum fx}{\sum f}=\frac{718}{12}=59.833\ldots\text{ kg}\]
\[\boxed{\text{Mean weight}=60\text{ kg, correct to the nearest kilogram}}\]
Ibeere 7 Ìròyìn
(a) 
In the diagram, O is the centre of the circle radius 3.2cm. If < PRQ = 42°, calculate, correct to two decimal places, the area of the:
(i) minor sector POQ ; (ii) shaded part.
(b) If the sector POQ in (a) is used to form the curved surface of a cone with vertex O, calculate the base radius of the cone, correct to one decimal place.
From the diagram, \(O\) is the centre, radius \(r = 3.2\ \text{cm}\), and \(\angle PRQ = 42^\circ\) is an angle at the circumference standing on chord \(PQ\).
Angle at the centre. The angle subtended at the centre is twice the angle at the circumference on the same arc:
\[ \angle POQ = 2 \times 42^\circ = 84^\circ \]
(a)(i) Area of minor sector POQ.
\[ A_{\text{sector}} = \frac{\theta}{360^\circ}\,\pi r^2 = \frac{84}{360}\times \pi \times (3.2)^2 \]
\[ = \frac{84}{360}\times \pi \times 10.24 = 0.23333 \times 32.1699 = 7.51\ \text{cm}^2 \]
(a)(ii) Area of the shaded part. The shaded region is the minor segment cut off by chord \(PQ\); it equals the sector minus triangle \(POQ\).
\[ A_{\triangle POQ} = \tfrac{1}{2} r^2 \sin\theta = \tfrac{1}{2}\times 10.24 \times \sin 84^\circ = 5.12 \times 0.99452 = 5.0920\ \text{cm}^2 \]
\[ A_{\text{shaded}} = 7.5063 - 5.0920 = 2.41\ \text{cm}^2 \]
(b) Base radius of the cone. When the sector is rolled into a cone, its arc length becomes the circumference of the base, while the sector radius becomes the slant height.
Arc length of sector:
\[ \ell = \frac{\theta}{360^\circ}\times 2\pi r = \frac{84}{360}\times 2\pi \times 3.2 = 4.6915\ \text{cm} \]
Set equal to base circumference \(2\pi R\):
\[ 2\pi R = 4.6915 \;\Rightarrow\; R = \frac{4.6915}{2\pi} = 0.7467 \approx 0.7\ \text{cm} \]
Answers: (i) \(7.51\ \text{cm}^2\); (ii) \(2.41\ \text{cm}^2\); (b) base radius \(\approx 0.7\ \text{cm}\).
Awọn alaye Idahun
From the diagram, \(O\) is the centre, radius \(r = 3.2\ \text{cm}\), and \(\angle PRQ = 42^\circ\) is an angle at the circumference standing on chord \(PQ\).
Angle at the centre. The angle subtended at the centre is twice the angle at the circumference on the same arc:
\[ \angle POQ = 2 \times 42^\circ = 84^\circ \]
(a)(i) Area of minor sector POQ.
\[ A_{\text{sector}} = \frac{\theta}{360^\circ}\,\pi r^2 = \frac{84}{360}\times \pi \times (3.2)^2 \]
\[ = \frac{84}{360}\times \pi \times 10.24 = 0.23333 \times 32.1699 = 7.51\ \text{cm}^2 \]
(a)(ii) Area of the shaded part. The shaded region is the minor segment cut off by chord \(PQ\); it equals the sector minus triangle \(POQ\).
\[ A_{\triangle POQ} = \tfrac{1}{2} r^2 \sin\theta = \tfrac{1}{2}\times 10.24 \times \sin 84^\circ = 5.12 \times 0.99452 = 5.0920\ \text{cm}^2 \]
\[ A_{\text{shaded}} = 7.5063 - 5.0920 = 2.41\ \text{cm}^2 \]
(b) Base radius of the cone. When the sector is rolled into a cone, its arc length becomes the circumference of the base, while the sector radius becomes the slant height.
Arc length of sector:
\[ \ell = \frac{\theta}{360^\circ}\times 2\pi r = \frac{84}{360}\times 2\pi \times 3.2 = 4.6915\ \text{cm} \]
Set equal to base circumference \(2\pi R\):
\[ 2\pi R = 4.6915 \;\Rightarrow\; R = \frac{4.6915}{2\pi} = 0.7467 \approx 0.7\ \text{cm} \]
Answers: (i) \(7.51\ \text{cm}^2\); (ii) \(2.41\ \text{cm}^2\); (b) base radius \(\approx 0.7\ \text{cm}\).
Ibeere 8 Ìròyìn
When a stone is thrown vertically upwards, its distance d metres after t seconds is given by the formula \(d = 60t - 10t^{2}\). Draw the graph of \(d = 60t - 10t^{2}\) for values of t from 1 to 5 seconds using 2cm to 1 unit on the t- axis and 2cm to 20 units on the d- axis.
(a) Using your graph, (i) how long does it take to reach a height of 70 metres? (ii) determine the height of the stone after 5 seconds. (iii) after how many seconds does it reach its maximum height.
(b) Determine the slope of the graph when t = 4 seconds.
Table of values for \(d = 60t - 10t^2\):
| \(t\) (s) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| \(d\) (m) | 50 | 80 | 90 | 80 | 50 |
Plot these points (2 cm to 1 unit on the \(t\)-axis, 2 cm to 20 units on the \(d\)-axis) and join with a smooth curve (a downward parabola).
(a)(i) Time to reach 70 m: Draw the line \(d = 70\); it cuts the curve twice. Solving \(60t - 10t^2 = 70\) gives \(t^2 - 6t + 7 = 0\), so \(t = 3 \pm \sqrt{2}\), i.e. \(t \approx \mathbf{1.6\text{ s}}\) (rising) and \(t \approx \mathbf{4.4\text{ s}}\) (falling).
(ii) Height after 5 s: from the curve, \(d = \mathbf{50\text{ m}}\).
(iii) Time at maximum height: the curve peaks at the vertex, \(t = \dfrac{60}{2(10)} = \mathbf{3\text{ s}}\) (height 90 m).
(b) Slope at \(t = 4\) s: the gradient of the tangent. Since \(\dfrac{dd}{dt} = 60 - 20t\), at \(t = 4\): \(60 - 20(4) = \mathbf{-20}\) (m/s). A tangent drawn at \(t = 4\) on the graph gives approximately this value.
Awọn alaye Idahun
Table of values for \(d = 60t - 10t^2\):
| \(t\) (s) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| \(d\) (m) | 50 | 80 | 90 | 80 | 50 |
Plot these points (2 cm to 1 unit on the \(t\)-axis, 2 cm to 20 units on the \(d\)-axis) and join with a smooth curve (a downward parabola).
(a)(i) Time to reach 70 m: Draw the line \(d = 70\); it cuts the curve twice. Solving \(60t - 10t^2 = 70\) gives \(t^2 - 6t + 7 = 0\), so \(t = 3 \pm \sqrt{2}\), i.e. \(t \approx \mathbf{1.6\text{ s}}\) (rising) and \(t \approx \mathbf{4.4\text{ s}}\) (falling).
(ii) Height after 5 s: from the curve, \(d = \mathbf{50\text{ m}}\).
(iii) Time at maximum height: the curve peaks at the vertex, \(t = \dfrac{60}{2(10)} = \mathbf{3\text{ s}}\) (height 90 m).
(b) Slope at \(t = 4\) s: the gradient of the tangent. Since \(\dfrac{dd}{dt} = 60 - 20t\), at \(t = 4\): \(60 - 20(4) = \mathbf{-20}\) (m/s). A tangent drawn at \(t = 4\) on the graph gives approximately this value.
Ibeere 9 Ìròyìn
(a) Triangle PQR is right-angled at Q. PQ = 3a cm and QR = 4a cm. Determine PR in terms of a.
(b) Ayo travels a distance of 24km from X on a bearing of 060° to Y. He then travels a distance of 18km to a point Z and Z is 30km from X.
(i) Draw the diagram to show the positions of X, Y and Z ; (ii) What is the bearing of Z from Y ; (iii) Calculate the bearing of X from Z.
(a)
Since triangle \(PQR\) is right-angled at \(Q\), by Pythagoras' theorem,
(b)(i) Diagram
The diagram below is drawn to scale in the correct relative positions. \(XY=24\text{ km}\) is on a bearing of \(060^\circ\), \(YZ=18\text{ km}\), and \(XZ=30\text{ km}\).
(ii) Bearing of \(Z\) from \(Y\)
Therefore, \(\angle XYZ=90^\circ\). The bearing of \(X\) from \(Y\) is \(240^\circ\). Hence,
Bearing of \(Z\) from \(Y\) = \(150^\circ\).
(iii) Bearing of \(X\) from \(Z\)
Using the cosine rule at \(Z\),
Since the bearing of \(Y\) from \(Z\) is \(150^\circ+180^\circ=330^\circ\),
Bearing of \(X\) from \(Z\) = \(277^\circ\).
Awọn alaye Idahun
(a)
Since triangle \(PQR\) is right-angled at \(Q\), by Pythagoras' theorem,
(b)(i) Diagram
The diagram below is drawn to scale in the correct relative positions. \(XY=24\text{ km}\) is on a bearing of \(060^\circ\), \(YZ=18\text{ km}\), and \(XZ=30\text{ km}\).
(ii) Bearing of \(Z\) from \(Y\)
Therefore, \(\angle XYZ=90^\circ\). The bearing of \(X\) from \(Y\) is \(240^\circ\). Hence,
Bearing of \(Z\) from \(Y\) = \(150^\circ\).
(iii) Bearing of \(X\) from \(Z\)
Using the cosine rule at \(Z\),
Since the bearing of \(Y\) from \(Z\) is \(150^\circ+180^\circ=330^\circ\),
Bearing of \(X\) from \(Z\) = \(277^\circ\).
Ibeere 10 Ìròyìn
(a) Derive the smallest equation whose coefficients are integers and which has roots of \(\frac{1}{2}\) and -7.
(b) Three years ago, a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the ages of the father and daughter.
(a) If the roots are \(\tfrac{1}{2}\) and \(-7\), the equation is
\[ \left(x - \tfrac{1}{2}\right)(x + 7) = 0. \]Multiply the first factor by 2 to clear the fraction (using \((2x-1)\)):
\[ (2x - 1)(x + 7) = 0 \implies 2x^2 + 14x - x - 7 = 0. \] \[ 2x^2 + 13x - 7 = 0. \]This is the smallest equation with integer coefficients.
(b) Let the daughter's present age be \(d\) and the father's present age be \(f\).
"Three years ago the father was four times as old as the daughter is now":
\[ f - 3 = 4d \implies f = 4d + 3. \]"The product of their present ages is 430":
\[ f \cdot d = 430 \implies (4d + 3)d = 430 \implies 4d^2 + 3d - 430 = 0. \]Using the quadratic formula, \(\;d = \dfrac{-3 \pm \sqrt{3^2 + 4(4)(430)}}{2(4)} = \dfrac{-3 \pm \sqrt{6889}}{8} = \dfrac{-3 \pm 83}{8}.\)
Taking the positive root, \(d = \dfrac{80}{8} = 10\), so \(f = 4(10) + 3 = 43\).
The daughter is 10 years old and the father is 43 years old (product \(= 430\)).
Awọn alaye Idahun
(a) If the roots are \(\tfrac{1}{2}\) and \(-7\), the equation is
\[ \left(x - \tfrac{1}{2}\right)(x + 7) = 0. \]Multiply the first factor by 2 to clear the fraction (using \((2x-1)\)):
\[ (2x - 1)(x + 7) = 0 \implies 2x^2 + 14x - x - 7 = 0. \] \[ 2x^2 + 13x - 7 = 0. \]This is the smallest equation with integer coefficients.
(b) Let the daughter's present age be \(d\) and the father's present age be \(f\).
"Three years ago the father was four times as old as the daughter is now":
\[ f - 3 = 4d \implies f = 4d + 3. \]"The product of their present ages is 430":
\[ f \cdot d = 430 \implies (4d + 3)d = 430 \implies 4d^2 + 3d - 430 = 0. \]Using the quadratic formula, \(\;d = \dfrac{-3 \pm \sqrt{3^2 + 4(4)(430)}}{2(4)} = \dfrac{-3 \pm \sqrt{6889}}{8} = \dfrac{-3 \pm 83}{8}.\)
Taking the positive root, \(d = \dfrac{80}{8} = 10\), so \(f = 4(10) + 3 = 43\).
The daughter is 10 years old and the father is 43 years old (product \(= 430\)).
Ibeere 11 Ìròyìn
The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 24cm and the base height 7cm. Calculate, correct to three significant figures, the surface area of the structure. [Take \(\pi = \frac{22}{7}\)].
Surface area of the cone-on-hemisphere structure.
From the diagram, the common radius (from centre \(O\) to the base edge) is \(r=7\text{ cm}\), and the vertical height of the cone is \(h=24\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere; the flat circular join is internal and is not counted.
Slant height of the cone:
\[l=\sqrt{r^{2}+h^{2}}=\sqrt{7^{2}+24^{2}}=\sqrt{49+576}=\sqrt{625}=25\text{ cm}\]Curved surface area of the cone:
\[\pi r l=\frac{22}{7}\times7\times25=22\times25=550\text{ cm}^{2}\]Curved surface area of the hemisphere:
\[2\pi r^{2}=2\times\frac{22}{7}\times7^{2}=2\times\frac{22}{7}\times49=2\times22\times7=308\text{ cm}^{2}\]Total surface area:
\[550+308=858\text{ cm}^{2}\]Correct to three significant figures, the surface area is \(\mathbf{858\text{ cm}^{2}}\).
Awọn alaye Idahun
Surface area of the cone-on-hemisphere structure.
From the diagram, the common radius (from centre \(O\) to the base edge) is \(r=7\text{ cm}\), and the vertical height of the cone is \(h=24\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere; the flat circular join is internal and is not counted.
Slant height of the cone:
\[l=\sqrt{r^{2}+h^{2}}=\sqrt{7^{2}+24^{2}}=\sqrt{49+576}=\sqrt{625}=25\text{ cm}\]Curved surface area of the cone:
\[\pi r l=\frac{22}{7}\times7\times25=22\times25=550\text{ cm}^{2}\]Curved surface area of the hemisphere:
\[2\pi r^{2}=2\times\frac{22}{7}\times7^{2}=2\times\frac{22}{7}\times49=2\times22\times7=308\text{ cm}^{2}\]Total surface area:
\[550+308=858\text{ cm}^{2}\]Correct to three significant figures, the surface area is \(\mathbf{858\text{ cm}^{2}}\).
Ibeere 12 Ìròyìn
(a) A pair of fair dice each numbered 1 to 6 is tossed. Find the probability of getting a sum of at least 9.
(b) If the probability that a civil servant owns a car is \(\frac{1}{6}\), find the probability that:
(i) two civil servants, A and B, selected at random each owns a car ; (ii) of two civil servants, C and D selected at random, only one owns a car ; (iii) of three civil servants, X, Y and Z, selected at random, only one owns a car.
(a) Two dice give \(6 \times 6 = 36\) equally likely outcomes. A sum of "at least 9" means a sum of 9, 10, 11 or 12.
- Sum 9: (3,6),(4,5),(5,4),(6,3) = 4 ways
- Sum 10: (4,6),(5,5),(6,4) = 3 ways
- Sum 11: (5,6),(6,5) = 2 ways
- Sum 12: (6,6) = 1 way
Favourable outcomes \(= 4+3+2+1 = 10\).
\[ P(\text{sum} \ge 9) = \frac{10}{36} = \frac{5}{18}. \](b) Let \(P(\text{owns a car}) = \frac{1}{6}\), so \(P(\text{does not}) = \frac{5}{6}\). Selections are independent.
(i) Both A and B own a car:
\[ \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \](ii) Of C and D, exactly one owns a car (owns-then-not, or not-then-owns):
\[ 2 \times \frac{1}{6} \times \frac{5}{6} = \frac{10}{36} = \frac{5}{18}. \](iii) Of X, Y and Z, exactly one owns a car. Choose which one (\(\binom{3}{1}=3\) ways):
\[ 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{2} = 3 \times \frac{25}{216} = \frac{75}{216} = \frac{25}{72}. \]Awọn alaye Idahun
(a) Two dice give \(6 \times 6 = 36\) equally likely outcomes. A sum of "at least 9" means a sum of 9, 10, 11 or 12.
- Sum 9: (3,6),(4,5),(5,4),(6,3) = 4 ways
- Sum 10: (4,6),(5,5),(6,4) = 3 ways
- Sum 11: (5,6),(6,5) = 2 ways
- Sum 12: (6,6) = 1 way
Favourable outcomes \(= 4+3+2+1 = 10\).
\[ P(\text{sum} \ge 9) = \frac{10}{36} = \frac{5}{18}. \](b) Let \(P(\text{owns a car}) = \frac{1}{6}\), so \(P(\text{does not}) = \frac{5}{6}\). Selections are independent.
(i) Both A and B own a car:
\[ \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \](ii) Of C and D, exactly one owns a car (owns-then-not, or not-then-owns):
\[ 2 \times \frac{1}{6} \times \frac{5}{6} = \frac{10}{36} = \frac{5}{18}. \](iii) Of X, Y and Z, exactly one owns a car. Choose which one (\(\binom{3}{1}=3\) ways):
\[ 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^{2} = 3 \times \frac{25}{216} = \frac{75}{216} = \frac{25}{72}. \]
