WAEC SSCE - General Mathematics - 1991 (Essay)
Tambaya 1 Rahoto
(a) 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.
(b) 
In the diagram, O is the centre of the circle, < OQR = 32° and < MPQ = 15°. Calculate (i) < QPR ; (ii) < MQO.
(a) Proof: the angle at the centre is twice the angle at the circumference
Given: A circle centre \(O\), an arc \(AB\), with \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining part of the circumference, both standing on the same arc \(AB\).
To prove: \(\angle AOB = 2\,\angle ACB\).
Construction: Join \(C\) to \(O\) and produce it to a point \(D\).
Proof: In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles and \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the two opposite interior angles:
\[ \angle AOD = \angle OAC + \angle OCA = 2\,\angle OCA \]
Similarly, in triangle \(OBC\), \(OB = OC\) (radii), so \(\angle OBC = \angle OCB\) and
\[ \angle BOD = \angle OBC + \angle OCB = 2\,\angle OCB \]
Adding:
\[ \angle AOD + \angle BOD = 2\,\angle OCA + 2\,\angle OCB \]
\[ \angle AOB = 2(\angle OCA + \angle OCB) = 2\,\angle ACB \]
Hence the angle subtended at the centre is twice that subtended at the circumference. (The same argument holds when \(O\) lies outside triangle \(ACB\), using subtraction instead of addition.) \(\blacksquare\)
(b) Calculations
\(O\) is the centre, \(\angle OQR = 32^{\circ}\) and \(\angle MPQ = 15^{\circ}\).
(i) \(\angle QPR\)
In triangle \(OQR\), \(OQ = OR\) (radii), so it is isosceles and \(\angle ORQ = \angle OQR = 32^{\circ}\). Then
\[ \angle QOR = 180^{\circ} - 32^{\circ} - 32^{\circ} = 116^{\circ} \]
\(\angle QOR\) is the angle at the centre and \(\angle QPR\) is the angle at the circumference standing on the same arc \(QR\). By the theorem proved in (a):
\[ \angle QPR = \tfrac{1}{2}\,\angle QOR = \tfrac{1}{2}\times 116^{\circ} = 58^{\circ} \]
(ii) \(\angle MQO\)
\(\angle MPQ = 15^{\circ}\) is the angle at the circumference standing on arc \(MQ\). The angle at the centre on the same arc is
\[ \angle MOQ = 2 \times 15^{\circ} = 30^{\circ} \]
In triangle \(OMQ\), \(OM = OQ\) (radii), so it is isosceles with equal base angles:
\[ \angle MQO = \angle OMQ = \frac{180^{\circ} - 30^{\circ}}{2} = 75^{\circ} \]
\(\angle QPR = 58^{\circ}\) and \(\angle MQO = 75^{\circ}\).
Bayanin Amsa
(a) Proof: the angle at the centre is twice the angle at the circumference
Given: A circle centre \(O\), an arc \(AB\), with \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining part of the circumference, both standing on the same arc \(AB\).
To prove: \(\angle AOB = 2\,\angle ACB\).
Construction: Join \(C\) to \(O\) and produce it to a point \(D\).
Proof: In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles and \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the two opposite interior angles:
\[ \angle AOD = \angle OAC + \angle OCA = 2\,\angle OCA \]
Similarly, in triangle \(OBC\), \(OB = OC\) (radii), so \(\angle OBC = \angle OCB\) and
\[ \angle BOD = \angle OBC + \angle OCB = 2\,\angle OCB \]
Adding:
\[ \angle AOD + \angle BOD = 2\,\angle OCA + 2\,\angle OCB \]
\[ \angle AOB = 2(\angle OCA + \angle OCB) = 2\,\angle ACB \]
Hence the angle subtended at the centre is twice that subtended at the circumference. (The same argument holds when \(O\) lies outside triangle \(ACB\), using subtraction instead of addition.) \(\blacksquare\)
(b) Calculations
\(O\) is the centre, \(\angle OQR = 32^{\circ}\) and \(\angle MPQ = 15^{\circ}\).
(i) \(\angle QPR\)
In triangle \(OQR\), \(OQ = OR\) (radii), so it is isosceles and \(\angle ORQ = \angle OQR = 32^{\circ}\). Then
\[ \angle QOR = 180^{\circ} - 32^{\circ} - 32^{\circ} = 116^{\circ} \]
\(\angle QOR\) is the angle at the centre and \(\angle QPR\) is the angle at the circumference standing on the same arc \(QR\). By the theorem proved in (a):
\[ \angle QPR = \tfrac{1}{2}\,\angle QOR = \tfrac{1}{2}\times 116^{\circ} = 58^{\circ} \]
(ii) \(\angle MQO\)
\(\angle MPQ = 15^{\circ}\) is the angle at the circumference standing on arc \(MQ\). The angle at the centre on the same arc is
\[ \angle MOQ = 2 \times 15^{\circ} = 30^{\circ} \]
In triangle \(OMQ\), \(OM = OQ\) (radii), so it is isosceles with equal base angles:
\[ \angle MQO = \angle OMQ = \frac{180^{\circ} - 30^{\circ}}{2} = 75^{\circ} \]
\(\angle QPR = 58^{\circ}\) and \(\angle MQO = 75^{\circ}\).
Tambaya 2 Rahoto
In a certain class, 22 pupils take one or more of Chemistry, Economics and Government. 12 take Economics (E), 8 take Government (G) and 7 take Chemistry (C). Nobody takes Economics and Chemistry and 4 pupils take Economics and Government.
(a)(i) Using set notation and the letters indicated above, write down the two statements in the last sentence; (ii) Draw a Venn diagram to illustrate the information.
(b) How many pupils take (i) both Chemistry and Government ? (ii) Government only?
(a)(i)
“Nobody takes Economics and Chemistry” is written as
\[n(E\cap C)=0\]
“4 pupils take Economics and Government” is written as
\[n(E\cap G)=4\]
(a)(ii) The Venn diagram is:
(b)(i) Let \(n(C\cap G)=x\). Since \(E\cap C=\varnothing\), there is no triple intersection.
Using the total number of pupils,
\[22=12+8+7-4-x\]
\[22=23-x\]
\[x=1\]
Therefore, the number who take both Chemistry and Government is \(\boxed{1}\).
(b)(ii)
\[\text{Government only}=8-4-1=3\]
Therefore, the number who take Government only is \(\boxed{3}\).
Bayanin Amsa
(a)(i)
“Nobody takes Economics and Chemistry” is written as
\[n(E\cap C)=0\]
“4 pupils take Economics and Government” is written as
\[n(E\cap G)=4\]
(a)(ii) The Venn diagram is:
(b)(i) Let \(n(C\cap G)=x\). Since \(E\cap C=\varnothing\), there is no triple intersection.
Using the total number of pupils,
\[22=12+8+7-4-x\]
\[22=23-x\]
\[x=1\]
Therefore, the number who take both Chemistry and Government is \(\boxed{1}\).
(b)(ii)
\[\text{Government only}=8-4-1=3\]
Therefore, the number who take Government only is \(\boxed{3}\).
Tambaya 3 Rahoto
The table below shows the distribution of the waiting times for some customers in a certain petrol station.
| Waiting time (in mins) | No of customers |
| 1.5 - 1.9 | 3 |
| 2.0 - 2.4 | 10 |
| 2.5 - 2.9 | 18 |
| 3.0 - 3.4 | 10 |
| 3.5 - 3.9 | 7 |
| 4.0 - 4.4 | 2 |
(a) Write down the class boundaries of the distribution.
(b) Construct a cumulative frequency curve for the data;
(c) Using your graph, estimate: (i) the interquartile range of the distribution ; (ii) the proportion of customers who could have waited for more than 3 minutes.
(a) Class boundaries. Subtract 0.05 from each lower limit and add 0.05 to each upper limit.
| Class | Class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 1.5-1.9 | 1.45-1.95 | 3 | 3 |
| 2.0-2.4 | 1.95-2.45 | 10 | 13 |
| 2.5-2.9 | 2.45-2.95 | 18 | 31 |
| 3.0-3.4 | 2.95-3.45 | 10 | 41 |
| 3.5-3.9 | 3.45-3.95 | 7 | 48 |
| 4.0-4.4 | 3.95-4.45 | 2 | 50 |
(b) Plot cumulative frequency against the upper boundaries: \((1.95,3),(2.45,13),(2.95,31),(3.45,41),(3.95,48),(4.45,50)\) and join with a smooth curve. \(N=50\).
(c)(i) Interquartile range. \(Q_1\) at \(\frac{N}{4}=12.5\) (in class 1.95-2.45):
\[ Q_1=1.95+\frac{12.5-3}{10}\times 0.5 = 1.95+0.475 = 2.43 \]\(Q_3\) at \(\frac{3N}{4}=37.5\) (in class 2.95-3.45):
\[ Q_3=2.95+\frac{37.5-31}{10}\times 0.5 = 2.95+0.325 = 3.28 \] \[ \text{IQR}=Q_3-Q_1=3.28-2.43 \approx 0.85 \text{ min} \](ii) Proportion waiting more than 3 minutes. Read the c.f. at \(3.0\): between \(2.95\,(31)\) and \(3.45\,(41)\),
\[ \text{c.f.}(3.0)=31+\frac{3.0-2.95}{0.5}\times 10 = 32 \]Number waiting more than 3 min \(=50-32=18\), so the proportion is
\[ \frac{18}{50}=0.36 \;\;(36\%) \]Bayanin Amsa
(a) Class boundaries. Subtract 0.05 from each lower limit and add 0.05 to each upper limit.
| Class | Class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 1.5-1.9 | 1.45-1.95 | 3 | 3 |
| 2.0-2.4 | 1.95-2.45 | 10 | 13 |
| 2.5-2.9 | 2.45-2.95 | 18 | 31 |
| 3.0-3.4 | 2.95-3.45 | 10 | 41 |
| 3.5-3.9 | 3.45-3.95 | 7 | 48 |
| 4.0-4.4 | 3.95-4.45 | 2 | 50 |
(b) Plot cumulative frequency against the upper boundaries: \((1.95,3),(2.45,13),(2.95,31),(3.45,41),(3.95,48),(4.45,50)\) and join with a smooth curve. \(N=50\).
(c)(i) Interquartile range. \(Q_1\) at \(\frac{N}{4}=12.5\) (in class 1.95-2.45):
\[ Q_1=1.95+\frac{12.5-3}{10}\times 0.5 = 1.95+0.475 = 2.43 \]\(Q_3\) at \(\frac{3N}{4}=37.5\) (in class 2.95-3.45):
\[ Q_3=2.95+\frac{37.5-31}{10}\times 0.5 = 2.95+0.325 = 3.28 \] \[ \text{IQR}=Q_3-Q_1=3.28-2.43 \approx 0.85 \text{ min} \](ii) Proportion waiting more than 3 minutes. Read the c.f. at \(3.0\): between \(2.95\,(31)\) and \(3.45\,(41)\),
\[ \text{c.f.}(3.0)=31+\frac{3.0-2.95}{0.5}\times 10 = 32 \]Number waiting more than 3 min \(=50-32=18\), so the proportion is
\[ \frac{18}{50}=0.36 \;\;(36\%) \]Tambaya 4 Rahoto
A man has 9 identical balls in a bag. Out of these, 3 are black, 2 are blue and the remaining are red.
(a) If a ball is drawn at random, what is the probability that it is (i) not blue? (ii) not red?
(b) If 2 balls are drawn at random, one after the other, what is the probability that both of them will be (i) black, if there is no replacement? (ii) blue, if there is a replacement?
Total balls \(= 9\): black \(= 3\), blue \(= 2\), red \(= 9 - 3 - 2 = 4\).
(a)(i) P(not blue) \(= \dfrac{9 - 2}{9} = \dfrac{7}{9}\).
(a)(ii) P(not red) \(= \dfrac{9 - 4}{9} = \dfrac{5}{9}\).
(b)(i) Both black, no replacement. First black: \(\dfrac{3}{9}\). After removing one black, 2 black remain out of 8: \(\dfrac{2}{8}\).
\[P = \frac{3}{9} \times \frac{2}{8} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}.\]
(b)(ii) Both blue, with replacement. The bag is restored, so each draw is \(\dfrac{2}{9}\).
\[P = \frac{2}{9} \times \frac{2}{9} = \frac{4}{81}.\]
Bayanin Amsa
Total balls \(= 9\): black \(= 3\), blue \(= 2\), red \(= 9 - 3 - 2 = 4\).
(a)(i) P(not blue) \(= \dfrac{9 - 2}{9} = \dfrac{7}{9}\).
(a)(ii) P(not red) \(= \dfrac{9 - 4}{9} = \dfrac{5}{9}\).
(b)(i) Both black, no replacement. First black: \(\dfrac{3}{9}\). After removing one black, 2 black remain out of 8: \(\dfrac{2}{8}\).
\[P = \frac{3}{9} \times \frac{2}{8} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}.\]
(b)(ii) Both blue, with replacement. The bag is restored, so each draw is \(\dfrac{2}{9}\).
\[P = \frac{2}{9} \times \frac{2}{9} = \frac{4}{81}.\]
Tambaya 5 Rahoto
Using a scale of 2cm to 1 unit on the x- axis and 1cm to 1 unit on the y- axis, draw on the same axes the graphs of \(y = 3 + 2x - x^{2}; y = 2x - 3\) for \(-3 \leq x \leq 4\). Using your graph:
(i) solve the equation \(6 - x^{2} = 0\);
(ii) find the maximum value of \(3 + 2x - x^{2}\);
(iii) find the range of x for which \(3 + 2x - x^{2} \leq 1\), expressing all your answers correct to one decimal place.
Table of values
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| \(y=3+2x-x^2\) | -12 | -5 | 0 | 3 | 4 | 3 | 0 | -5 |
| \(y=2x-3\) | -9 | -7 | -5 | -3 | -1 | 1 | 3 | 5 |
Using the stated scales, the required graph is:
(i) At the points of intersection,
\[3+2x-x^2=2x-3\]
\[6-x^2=0.\]
Reading the two intersection abscissae from the graph gives
\[x\approx -2.4\quad\text{or}\quad x\approx 2.4.\]
(ii) The highest point of the parabola is \((1,4)\). Hence, the maximum value of \(3+2x-x^2\) is
\[\boxed{4.0}.\]
(iii) The curve \(y=3+2x-x^2\) meets the line \(y=1\) at approximately \(x=-0.7\) and \(x=2.7\). The part of the parabola on or below \(y=1\) is therefore
\[\boxed{-3.0\leq x\leq -0.7\quad\text{or}\quad 2.7\leq x\leq4.0}.\]
Bayanin Amsa
Table of values
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| \(y=3+2x-x^2\) | -12 | -5 | 0 | 3 | 4 | 3 | 0 | -5 |
| \(y=2x-3\) | -9 | -7 | -5 | -3 | -1 | 1 | 3 | 5 |
Using the stated scales, the required graph is:
(i) At the points of intersection,
\[3+2x-x^2=2x-3\]
\[6-x^2=0.\]
Reading the two intersection abscissae from the graph gives
\[x\approx -2.4\quad\text{or}\quad x\approx 2.4.\]
(ii) The highest point of the parabola is \((1,4)\). Hence, the maximum value of \(3+2x-x^2\) is
\[\boxed{4.0}.\]
(iii) The curve \(y=3+2x-x^2\) meets the line \(y=1\) at approximately \(x=-0.7\) and \(x=2.7\). The part of the parabola on or below \(y=1\) is therefore
\[\boxed{-3.0\leq x\leq -0.7\quad\text{or}\quad 2.7\leq x\leq4.0}.\]
Tambaya 6 Rahoto
A man bought 5 reams of duplicating paper, each of which are supposed to contain 480 sheets. The actual number of sheets in the packets were : 435, 420, 405, 415 and 440.
(a) Calculate, correct to the nearest whole number, the percentage error for the packets of paper;
(b) If the agreed price for a full ream was N35.00, find, correct to the nearest naira, the amount by which the buyer was cheated.
Each ream is supposed to contain 480 sheets, so 5 reams should hold \(5 \times 480 = 2400\) sheets.
Actual sheets: \(435 + 420 + 405 + 415 + 440 = 2115\).
Total shortage \(= 2400 - 2115 = 285\) sheets.
(a) Percentage error.
\[\text{Percentage error} = \frac{\text{shortage}}{\text{supposed number}}\times 100 = \frac{285}{2400}\times 100 = 11.875\%.\]
To the nearest whole number, the percentage error \(= \mathbf{12\%}\).
(b) Amount cheated. A full ream of 480 sheets costs \(\text{N}35.00\), so one sheet is worth \(\dfrac{35}{480}\) naira.
The buyer paid for 2400 sheets but received only 2115, a shortage of 285 sheets:
\[\text{Amount cheated} = 285 \times \frac{35}{480} = \frac{9975}{480} = \text{N}20.78.\]
To the nearest naira, the buyer was cheated by \(\mathbf{\text{N}21}\).
Bayanin Amsa
Each ream is supposed to contain 480 sheets, so 5 reams should hold \(5 \times 480 = 2400\) sheets.
Actual sheets: \(435 + 420 + 405 + 415 + 440 = 2115\).
Total shortage \(= 2400 - 2115 = 285\) sheets.
(a) Percentage error.
\[\text{Percentage error} = \frac{\text{shortage}}{\text{supposed number}}\times 100 = \frac{285}{2400}\times 100 = 11.875\%.\]
To the nearest whole number, the percentage error \(= \mathbf{12\%}\).
(b) Amount cheated. A full ream of 480 sheets costs \(\text{N}35.00\), so one sheet is worth \(\dfrac{35}{480}\) naira.
The buyer paid for 2400 sheets but received only 2115, a shortage of 285 sheets:
\[\text{Amount cheated} = 285 \times \frac{35}{480} = \frac{9975}{480} = \text{N}20.78.\]
To the nearest naira, the buyer was cheated by \(\mathbf{\text{N}21}\).
Tambaya 7 Rahoto
(a) Using logarithm table, evaluate \(\frac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\) correct to three significant figure.
(b) Without using Mathematical tables, find the value of \(\frac{\log 81}{\log \frac{1}{3}}\).
(a) Let \(N=\dfrac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\). Using logarithms:
| Number | log | operation |
|---|---|---|
| \(1.376\) | \(0.1387\) | \(\div3=0.04623\) |
| \(0.007\) | \(\bar{3}.8451=-2.1549\) | \(\div4=-0.53873\) |
\[\log N=0.04623-(-0.53873)=0.58496.\]
\[N=\text{antilog}(0.58496)=3.85\ (3\text{ s.f.}).\]
(b) Express in one base: \(81=3^{4}\) and \(\tfrac13=3^{-1}\).
\[\frac{\log81}{\log\frac13}=\frac{\log3^{4}}{\log3^{-1}}=\frac{4\log3}{-\log3}=-4.\]
Bayanin Amsa
(a) Let \(N=\dfrac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\). Using logarithms:
| Number | log | operation |
|---|---|---|
| \(1.376\) | \(0.1387\) | \(\div3=0.04623\) |
| \(0.007\) | \(\bar{3}.8451=-2.1549\) | \(\div4=-0.53873\) |
\[\log N=0.04623-(-0.53873)=0.58496.\]
\[N=\text{antilog}(0.58496)=3.85\ (3\text{ s.f.}).\]
(b) Express in one base: \(81=3^{4}\) and \(\tfrac13=3^{-1}\).
\[\frac{\log81}{\log\frac13}=\frac{\log3^{4}}{\log3^{-1}}=\frac{4\log3}{-\log3}=-4.\]
Tambaya 8 Rahoto
Simplify :
(i) \(2\frac{2}{3} - (2\frac{1}{2} - 1\frac{4}{5})\)
(ii) \(\frac{3.25 - 1.64}{2.47 - 2.01}\)
(i) Work the bracket first:
\[2\tfrac12-1\tfrac45=\frac52-\frac95=\frac{25-18}{10}=\frac{7}{10}.\]
Then
\[2\tfrac23-\frac{7}{10}=\frac83-\frac{7}{10}=\frac{80-21}{30}=\frac{59}{30}=1\tfrac{29}{30}.\]
(ii)
\[\frac{3.25-1.64}{2.47-2.01}=\frac{1.61}{0.46}=3.5.\]
Bayanin Amsa
(i) Work the bracket first:
\[2\tfrac12-1\tfrac45=\frac52-\frac95=\frac{25-18}{10}=\frac{7}{10}.\]
Then
\[2\tfrac23-\frac{7}{10}=\frac83-\frac{7}{10}=\frac{80-21}{30}=\frac{59}{30}=1\tfrac{29}{30}.\]
(ii)
\[\frac{3.25-1.64}{2.47-2.01}=\frac{1.61}{0.46}=3.5.\]
Tambaya 9 Rahoto
ABCDE is a regular pentagon and a rectangle AXYE is drawn on the side AE such that the vertices X and Y lie on the sides BC and CD respectively. Calculate the size of
(i) an interior angle of the pentagon ;
(ii) < BXA.
(i) Interior angle of the pentagon. The sum of the interior angles of an \(n\)-sided polygon is \((n - 2)\times 180^\circ\). For a regular pentagon \(n = 5\):
\[\text{Each interior angle} = \frac{(5 - 2)\times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ.\]
(ii) < BXA. In the rectangle \(AXYE\), the side \(AE\) is a side of the pentagon and \(AX \perp AE\), so \(< XAE = 90^\circ\).
The interior angle of the pentagon at \(A\) is \(< BAE = 108^\circ\). Since \(X\) lies on \(BC\), the ray \(AX\) splits \(< BAE\):
\[< BAX = < BAE - < XAE = 108^\circ - 90^\circ = 18^\circ.\]
Also \(X\) lies on \(BC\), so in triangle \(ABX\) the angle at \(B\) equals the interior angle of the pentagon:
\[< ABX = 108^\circ.\]
The angles of triangle \(ABX\) sum to \(180^\circ\):
\[< BXA = 180^\circ - < ABX - < BAX = 180^\circ - 108^\circ - 18^\circ = 54^\circ.\]
Therefore \(< BXA = \mathbf{54^\circ}\).
Bayanin Amsa
(i) Interior angle of the pentagon. The sum of the interior angles of an \(n\)-sided polygon is \((n - 2)\times 180^\circ\). For a regular pentagon \(n = 5\):
\[\text{Each interior angle} = \frac{(5 - 2)\times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ.\]
(ii) < BXA. In the rectangle \(AXYE\), the side \(AE\) is a side of the pentagon and \(AX \perp AE\), so \(< XAE = 90^\circ\).
The interior angle of the pentagon at \(A\) is \(< BAE = 108^\circ\). Since \(X\) lies on \(BC\), the ray \(AX\) splits \(< BAE\):
\[< BAX = < BAE - < XAE = 108^\circ - 90^\circ = 18^\circ.\]
Also \(X\) lies on \(BC\), so in triangle \(ABX\) the angle at \(B\) equals the interior angle of the pentagon:
\[< ABX = 108^\circ.\]
The angles of triangle \(ABX\) sum to \(180^\circ\):
\[< BXA = 180^\circ - < ABX - < BAX = 180^\circ - 108^\circ - 18^\circ = 54^\circ.\]
Therefore \(< BXA = \mathbf{54^\circ}\).
Tambaya 10 Rahoto
From a horizontal distance of 8.5 km, a pilot observes that the angles of depression of the top and the base of a control tower are 30° and 33° respectively. Calculate, correct to three significant figures :
(a) the shortest distance between the pilot and the base of the control tower;
(b) the height of the control tower.
Let the pilot be at a height \(H\) above the ground, with the horizontal distance to the tower equal to 8.5 km. The angle of depression to the base is \(33^\circ\) and to the top is \(30^\circ\) (the top is higher, so its depression is smaller).
Height of pilot above the ground, using the base:
\[\tan 33^\circ = \frac{H}{8.5} \Rightarrow H = 8.5\tan 33^\circ = 8.5(0.6494) = 5.520\text{ km}.\]
(a) Shortest distance from the pilot to the base. This is the straight line (line of sight):
\[d = \frac{8.5}{\cos 33^\circ} = \frac{8.5}{0.8387} = 10.135\text{ km} \approx \mathbf{10.1\text{ km}}.\]
(b) Height of the control tower. The height of the pilot above the top of the tower is
\[8.5\tan 30^\circ = 8.5(0.5774) = 4.908\text{ km}.\]
Therefore the tower height is
\[H - 8.5\tan 30^\circ = 5.520 - 4.908 = 0.612\text{ km}.\]
The height of the control tower \(\approx \mathbf{0.612\text{ km}}\) (about 612 m), correct to three significant figures.
Bayanin Amsa
Let the pilot be at a height \(H\) above the ground, with the horizontal distance to the tower equal to 8.5 km. The angle of depression to the base is \(33^\circ\) and to the top is \(30^\circ\) (the top is higher, so its depression is smaller).
Height of pilot above the ground, using the base:
\[\tan 33^\circ = \frac{H}{8.5} \Rightarrow H = 8.5\tan 33^\circ = 8.5(0.6494) = 5.520\text{ km}.\]
(a) Shortest distance from the pilot to the base. This is the straight line (line of sight):
\[d = \frac{8.5}{\cos 33^\circ} = \frac{8.5}{0.8387} = 10.135\text{ km} \approx \mathbf{10.1\text{ km}}.\]
(b) Height of the control tower. The height of the pilot above the top of the tower is
\[8.5\tan 30^\circ = 8.5(0.5774) = 4.908\text{ km}.\]
Therefore the tower height is
\[H - 8.5\tan 30^\circ = 5.520 - 4.908 = 0.612\text{ km}.\]
The height of the control tower \(\approx \mathbf{0.612\text{ km}}\) (about 612 m), correct to three significant figures.
Tambaya 11 Rahoto
(a) Solve the equation, correct to two decimal places \(2x^{2} + 7x - 11 = 0\)
(b) Using the substitution \(P = \frac{1}{x}; Q = \frac{1}{y}\), solve the simultaneous equations : \(\frac{2}{x} + \frac{1}{y} = 3 ; \frac{1}{x} - \frac{5}{y} = 7\)
(a) Solve \(2x^2 + 7x - 11 = 0\) using the quadratic formula with \(a = 2, b = 7, c = -11\).
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 + 88}}{4} = \frac{-7 \pm \sqrt{137}}{4}.\]
Now \(\sqrt{137} = 11.7047\), so
\[x = \frac{-7 + 11.7047}{4} = \frac{4.7047}{4} = 1.18,\qquad x = \frac{-7 - 11.7047}{4} = \frac{-18.7047}{4} = -4.68.\]
Hence \(x = 1.18\) or \(x = -4.68\) (2 d.p.).
(b) Let \(P = \dfrac{1}{x}\) and \(Q = \dfrac{1}{y}\). The equations become:
\[2P + Q = 3 \quad(1), \qquad P - 5Q = 7 \quad(2).\]
From (1): \(Q = 3 - 2P\). Substitute into (2):
\[P - 5(3 - 2P) = 7 \Rightarrow P - 15 + 10P = 7 \Rightarrow 11P = 22 \Rightarrow P = 2.\]
Then \(Q = 3 - 2(2) = -1\).
Returning to \(x\) and \(y\): \(x = \dfrac{1}{P} = \dfrac{1}{2}\) and \(y = \dfrac{1}{Q} = -1\).
Therefore \(x = \dfrac{1}{2},\; y = -1\).
Bayanin Amsa
(a) Solve \(2x^2 + 7x - 11 = 0\) using the quadratic formula with \(a = 2, b = 7, c = -11\).
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 + 88}}{4} = \frac{-7 \pm \sqrt{137}}{4}.\]
Now \(\sqrt{137} = 11.7047\), so
\[x = \frac{-7 + 11.7047}{4} = \frac{4.7047}{4} = 1.18,\qquad x = \frac{-7 - 11.7047}{4} = \frac{-18.7047}{4} = -4.68.\]
Hence \(x = 1.18\) or \(x = -4.68\) (2 d.p.).
(b) Let \(P = \dfrac{1}{x}\) and \(Q = \dfrac{1}{y}\). The equations become:
\[2P + Q = 3 \quad(1), \qquad P - 5Q = 7 \quad(2).\]
From (1): \(Q = 3 - 2P\). Substitute into (2):
\[P - 5(3 - 2P) = 7 \Rightarrow P - 15 + 10P = 7 \Rightarrow 11P = 22 \Rightarrow P = 2.\]
Then \(Q = 3 - 2(2) = -1\).
Returning to \(x\) and \(y\): \(x = \dfrac{1}{P} = \dfrac{1}{2}\) and \(y = \dfrac{1}{Q} = -1\).
Therefore \(x = \dfrac{1}{2},\; y = -1\).
Tambaya 12 Rahoto
(a) Using a ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals |PR| = 9cm and |QS| = 6cm, intersecting at K and < QKR = 60°.
(b) Construct a rectangle PABS which is equal in area to PQRS in (a) above and on the same side of PS as PQRS. Measure |PA|.
(a) Constructing \(PQRS\)
The key fact is that the diagonals of a parallelogram bisect each other. Therefore, at their intersection \(K\),
\[ PK=KR=\frac{9}{2}=4.5\text{ cm},\qquad QK=KS=\frac{6}{2}=3\text{ cm}. \]
- Draw \(PR=9\text{ cm}\) and construct its midpoint \(K\).
- At \(K\), construct an angle of \(60^\circ\) with \(KR\) as one arm.
- Mark \(Q\) on this ray so that \(KQ=3\text{ cm}\).
- Extend \(QK\) through \(K\) and mark \(S\) so that \(KS=3\text{ cm}\).
- Join \(P\) to \(Q\), \(Q\) to \(R\), \(R\) to \(S\), and \(S\) to \(P\).
(b) Constructing the equal-area rectangle \(PABS\)
Use \(PS\) as the base of the rectangle. The area of parallelogram \(PQRS\) is
\[ \text{area} = PS \times \text{perpendicular height}. \]
Drop a perpendicular from \(Q\) to \(PS\), meeting \(PS\) at \(H\). The length \(QH\) is the perpendicular height of the parallelogram. Construct a perpendicular to \(PS\) at \(P\), on the same side of \(PS\) as \(Q\), and transfer the length \(QH\) onto it to locate \(A\). Thus, \(PA=QH\). Draw a line through \(A\) parallel to \(PS\), and a line through \(S\) parallel to \(PA\); they meet at \(B\).
This makes
\[ \text{area of }PABS=PS\times PA=PS\times QH=\text{area of }PQRS. \]
For the measurement, the parallelogram area is
\[ \frac{1}{2}(9)(6)\sin 60^\circ =27\sin60^\circ \approx 23.38\text{ cm}^2. \]
Also, in triangle \(PKS\), the angle \(PKS\) is \(60^\circ\), not \(120^\circ\). Therefore,
\[ PS^2=4.5^2+3^2-2(4.5)(3)\cos60^\circ =20.25+9-13.5 =15.75, \]
\[ PS\approx 3.97\text{ cm}. \]
Hence
\[ PA=\frac{23.38}{3.97}\approx 5.89\text{ cm}. \]
\(\boxed{PA\approx 5.9\text{ cm}}\)
The earlier value of \(3.6\text{ cm}\) results from using \(120^\circ\) for \(\angle PKS\). Since \(KP\) and \(KS\) are separated by \(60^\circ\), the correct angle is \(60^\circ\).
Bayanin Amsa
(a) Constructing \(PQRS\)
The key fact is that the diagonals of a parallelogram bisect each other. Therefore, at their intersection \(K\),
\[ PK=KR=\frac{9}{2}=4.5\text{ cm},\qquad QK=KS=\frac{6}{2}=3\text{ cm}. \]
- Draw \(PR=9\text{ cm}\) and construct its midpoint \(K\).
- At \(K\), construct an angle of \(60^\circ\) with \(KR\) as one arm.
- Mark \(Q\) on this ray so that \(KQ=3\text{ cm}\).
- Extend \(QK\) through \(K\) and mark \(S\) so that \(KS=3\text{ cm}\).
- Join \(P\) to \(Q\), \(Q\) to \(R\), \(R\) to \(S\), and \(S\) to \(P\).
(b) Constructing the equal-area rectangle \(PABS\)
Use \(PS\) as the base of the rectangle. The area of parallelogram \(PQRS\) is
\[ \text{area} = PS \times \text{perpendicular height}. \]
Drop a perpendicular from \(Q\) to \(PS\), meeting \(PS\) at \(H\). The length \(QH\) is the perpendicular height of the parallelogram. Construct a perpendicular to \(PS\) at \(P\), on the same side of \(PS\) as \(Q\), and transfer the length \(QH\) onto it to locate \(A\). Thus, \(PA=QH\). Draw a line through \(A\) parallel to \(PS\), and a line through \(S\) parallel to \(PA\); they meet at \(B\).
This makes
\[ \text{area of }PABS=PS\times PA=PS\times QH=\text{area of }PQRS. \]
For the measurement, the parallelogram area is
\[ \frac{1}{2}(9)(6)\sin 60^\circ =27\sin60^\circ \approx 23.38\text{ cm}^2. \]
Also, in triangle \(PKS\), the angle \(PKS\) is \(60^\circ\), not \(120^\circ\). Therefore,
\[ PS^2=4.5^2+3^2-2(4.5)(3)\cos60^\circ =20.25+9-13.5 =15.75, \]
\[ PS\approx 3.97\text{ cm}. \]
Hence
\[ PA=\frac{23.38}{3.97}\approx 5.89\text{ cm}. \]
\(\boxed{PA\approx 5.9\text{ cm}}\)
The earlier value of \(3.6\text{ cm}\) results from using \(120^\circ\) for \(\angle PKS\). Since \(KP\) and \(KS\) are separated by \(60^\circ\), the correct angle is \(60^\circ\).
