WAEC SSCE - General Mathematics - 2016 (Essay)
Tambaya 1 Rahoto
(a) Without using Mathematical tables or calculators, evaluate \(\frac{0.09 \times 1.21}{3.3 \times 0.00025}\), leaving the answer in standard form (Scientific Notation).
(b) A principal of GH¢5,600 was deposited for 3 years at compound interest. If the interest earned was GH¢1,200, find, correct to 3 significant figures, the interest rate per annum.
(a) Evaluate \(\dfrac{0.09\times1.21}{3.3\times0.00025}\).
Numerator: \(0.09\times1.21=0.1089\).
Denominator: \(3.3\times0.00025=0.000825\).
\[\frac{0.1089}{0.000825}=132\]In standard form: \(132=1.32\times10^{2}\).
\(=1.32\times10^{2}\)
(b) Principal \(P=\text{GH}\!\cent5600\); interest earned in 3 years \(=1200\), so the amount is \(A=5600+1200=6800\).
Compound interest: \(A=P(1+r)^{3}\).
\[6800=5600(1+r)^{3}\Rightarrow (1+r)^{3}=\frac{6800}{5600}=1.2143\]\[1+r=\sqrt[3]{1.2143}=1.0669\Rightarrow r=0.0669\]Rate \(\approx 6.69\%\) per annum (3 s.f.).
Bayanin Amsa
(a) Evaluate \(\dfrac{0.09\times1.21}{3.3\times0.00025}\).
Numerator: \(0.09\times1.21=0.1089\).
Denominator: \(3.3\times0.00025=0.000825\).
\[\frac{0.1089}{0.000825}=132\]In standard form: \(132=1.32\times10^{2}\).
\(=1.32\times10^{2}\)
(b) Principal \(P=\text{GH}\!\cent5600\); interest earned in 3 years \(=1200\), so the amount is \(A=5600+1200=6800\).
Compound interest: \(A=P(1+r)^{3}\).
\[6800=5600(1+r)^{3}\Rightarrow (1+r)^{3}=\frac{6800}{5600}=1.2143\]\[1+r=\sqrt[3]{1.2143}=1.0669\Rightarrow r=0.0669\]Rate \(\approx 6.69\%\) per annum (3 s.f.).
Tambaya 2 Rahoto
(a) Find the sum of the Arithmetic Progression (AP) 1, 3, 5,..., 101.
(b) Out of the 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all 3 means of transport. The number, x, of travellerswho travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train :
(i) represent this information in a Venn diagram;
(ii) calculate the
(1) value of x ; (2) number who travelled by at least two means.
(a) Sum of the AP
For \(1,3,5,\ldots,101\):
\(a=1\), \(d=2\), and \(l=101\).
\[ 101=1+(n-1)2 \] \[ 100=2(n-1)\Rightarrow n-1=50\Rightarrow n=51 \] \[ S_n=\frac{n}{2}(a+l) =\frac{51}{2}(1+101) =\frac{51}{2}(102) =2601 \]The sum is \(2601\).
(b)(i) Venn diagram
Let \(B\), \(T\), and \(C\) represent travellers who used bus, train, and car respectively. Since bus only and bus-and-car only are both \(x\), use the bus total:
\[ x+7+x+8=47 \] \[ 2x+15=47\Rightarrow 2x=32\Rightarrow x=16 \]The train-only region is:
\[ 30-(7+3+8)=12 \]As all \(95\) travellers are represented in the diagram, the car-only region is:
\[ 95-(16+12+16+7+3+8)=33 \](ii)(1) The value of \(x\) is \(16\).
(ii)(2) Travellers who used at least two means of transport
“At least two” includes each region for exactly two means, as well as the region for all three means:
\[ 7+16+3+8=34 \]\(34\) travellers travelled by at least two means of transport.
Examination reminder: When finding “at least two” in a Venn diagram, include the central intersection because those people used all three means, which is also at least two.
Bayanin Amsa
(a) Sum of the AP
For \(1,3,5,\ldots,101\):
\(a=1\), \(d=2\), and \(l=101\).
\[ 101=1+(n-1)2 \] \[ 100=2(n-1)\Rightarrow n-1=50\Rightarrow n=51 \] \[ S_n=\frac{n}{2}(a+l) =\frac{51}{2}(1+101) =\frac{51}{2}(102) =2601 \]The sum is \(2601\).
(b)(i) Venn diagram
Let \(B\), \(T\), and \(C\) represent travellers who used bus, train, and car respectively. Since bus only and bus-and-car only are both \(x\), use the bus total:
\[ x+7+x+8=47 \] \[ 2x+15=47\Rightarrow 2x=32\Rightarrow x=16 \]The train-only region is:
\[ 30-(7+3+8)=12 \]As all \(95\) travellers are represented in the diagram, the car-only region is:
\[ 95-(16+12+16+7+3+8)=33 \](ii)(1) The value of \(x\) is \(16\).
(ii)(2) Travellers who used at least two means of transport
“At least two” includes each region for exactly two means, as well as the region for all three means:
\[ 7+16+3+8=34 \]\(34\) travellers travelled by at least two means of transport.
Examination reminder: When finding “at least two” in a Venn diagram, include the central intersection because those people used all three means, which is also at least two.
Tambaya 3 Rahoto
| Scores | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 25 | 30 | x | 28 | 40 | 32 |
The table shows the outcome when a die is thrown a number of times. If the probability of obtaining a 3 is 0.225;
(a) How many times was the die thrown?
(b) Calculate the probability that a trial chosen at random gives a score of an even number or a prime number.
Given distribution.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 25 | 30 | x | 28 | 40 | 32 |
(a) Number of throws. Let \(N\) be the total number of throws. Then
\[N = 25+30+x+28+40+32 = 155 + x.\]
The probability of obtaining a 3 is \(\dfrac{x}{N} = 0.225\), so
\[x = 0.225(155 + x) \Rightarrow x = 34.875 + 0.225x \Rightarrow 0.775x = 34.875 \Rightarrow x = 45.\]
Therefore \(N = 155 + 45 = \mathbf{200}\). The die was thrown 200 times.
The completed frequencies are: 25, 30, 45, 28, 40, 32.
(b) P(even number or prime number). Even scores: \(\{2, 4, 6\}\); prime scores: \(\{2, 3, 5\}\). Their union is \(\{2, 3, 4, 5, 6\}\) (only the score 1 is excluded). Adding their frequencies:
\[30 + 45 + 28 + 40 + 32 = 175.\]
\[P(\text{even or prime}) = \frac{175}{200} = \frac{7}{8} = 0.875.\]
Equivalently, \(1 - P(1) = 1 - \dfrac{25}{200} = 0.875\).
Bayanin Amsa
Given distribution.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 25 | 30 | x | 28 | 40 | 32 |
(a) Number of throws. Let \(N\) be the total number of throws. Then
\[N = 25+30+x+28+40+32 = 155 + x.\]
The probability of obtaining a 3 is \(\dfrac{x}{N} = 0.225\), so
\[x = 0.225(155 + x) \Rightarrow x = 34.875 + 0.225x \Rightarrow 0.775x = 34.875 \Rightarrow x = 45.\]
Therefore \(N = 155 + 45 = \mathbf{200}\). The die was thrown 200 times.
The completed frequencies are: 25, 30, 45, 28, 40, 32.
(b) P(even number or prime number). Even scores: \(\{2, 4, 6\}\); prime scores: \(\{2, 3, 5\}\). Their union is \(\{2, 3, 4, 5, 6\}\) (only the score 1 is excluded). Adding their frequencies:
\[30 + 45 + 28 + 40 + 32 = 175.\]
\[P(\text{even or prime}) = \frac{175}{200} = \frac{7}{8} = 0.875.\]
Equivalently, \(1 - P(1) = 1 - \dfrac{25}{200} = 0.875\).
Tambaya 4 Rahoto
The weight (in kg) of 50 contestants at a competition is as follows:
65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64.
(a) Construct a frequenct table for the discrete data.
(b) Calculate, correct to 2 decimal places, the;
(i) mean ; (ii) standard deviation of the data.
(a) Frequency table
Tallying the 50 weights gives:
| Weight, \(x\) (kg) | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 5 | 12 | 14 | 10 | 6 | 2 |
Total frequency \(=1+5+12+14+10+6+2=50\).
(b)(i) Mean
| \(x\) | \(f\) | \(fx\) | \(fx^{2}\) |
|---|---|---|---|
| 62 | 1 | 62 | 3844 |
| 63 | 5 | 315 | 19845 |
| 64 | 12 | 768 | 49152 |
| 65 | 14 | 910 | 59150 |
| 66 | 10 | 660 | 43560 |
| 67 | 6 | 402 | 26934 |
| 68 | 2 | 136 | 9248 |
| Total | 50 | 3253 | 211733 |
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3253}{50}=65.06\text{ kg}\]
(b)(ii) Standard deviation
\[\text{S.D.}=\sqrt{\frac{\sum fx^{2}}{\sum f}-\left(\frac{\sum fx}{\sum f}\right)^{2}}=\sqrt{\frac{211733}{50}-(65.06)^{2}}\]
\[=\sqrt{4234.66-4232.8036}=\sqrt{1.8564}=1.36\text{ kg (2 d.p.)}\]
Bayanin Amsa
(a) Frequency table
Tallying the 50 weights gives:
| Weight, \(x\) (kg) | 62 | 63 | 64 | 65 | 66 | 67 | 68 |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 5 | 12 | 14 | 10 | 6 | 2 |
Total frequency \(=1+5+12+14+10+6+2=50\).
(b)(i) Mean
| \(x\) | \(f\) | \(fx\) | \(fx^{2}\) |
|---|---|---|---|
| 62 | 1 | 62 | 3844 |
| 63 | 5 | 315 | 19845 |
| 64 | 12 | 768 | 49152 |
| 65 | 14 | 910 | 59150 |
| 66 | 10 | 660 | 43560 |
| 67 | 6 | 402 | 26934 |
| 68 | 2 | 136 | 9248 |
| Total | 50 | 3253 | 211733 |
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3253}{50}=65.06\text{ kg}\]
(b)(ii) Standard deviation
\[\text{S.D.}=\sqrt{\frac{\sum fx^{2}}{\sum f}-\left(\frac{\sum fx}{\sum f}\right)^{2}}=\sqrt{\frac{211733}{50}-(65.06)^{2}}\]
\[=\sqrt{4234.66-4232.8036}=\sqrt{1.8564}=1.36\text{ kg (2 d.p.)}\]
Tambaya 5 Rahoto
(a) 
In the diagram, < KLM = x, < LMK = y, < KJH = r and < KGF = 110°. If 2x = r = y, find the value of x.
(b) Ten boys and twelve girls collected donations for a project. The total amount collected by the boys was N600.00 gretaer than that collected by the girls. If the average collection of the boys was N100.00 greater than the average collection of the girls, how much was collected by the two groups?
(a) Finding x.
From the diagram, J-H-G-F is a straight line along the bottom. The line J-K-L is straight (K lies on JL), and the line K-M-G is straight (M lies on KG). The marked angles are \(\angle KLM = x\) at L, \(\angle LMK = y\) at M, \(\angle KJH = r\) at J, and \(\angle KGF = 110^\circ\) at G, with the condition \(2x = r = y\).
Step 1: angle the line KG makes with the base at G. Since J-G-F is straight, \(\angle KGF\) and \(\angle KGJ\) are on a straight line:
\[\angle KGJ = 180^\circ - 110^\circ = 70^\circ\]Step 2: angle at K in triangle KLM. The angles of \(\triangle KLM\) sum to \(180^\circ\):
\[\angle LKM = 180^\circ - x - y\]Because J-K-L is a straight line, the angle on the other side of K (angle JKG, which is the angle of triangle JKG at K) is supplementary:
\[\angle JKG = 180^\circ - \angle LKM = 180^\circ - (180^\circ - x - y) = x + y\]Step 3: use triangle JKG. Its vertices are J and G on the base and K above. Its angles are \(\angle KJG = r\), \(\angle JKG = x+y\) and \(\angle KGJ = 70^\circ\), summing to \(180^\circ\):
\[r + (x + y) + 70^\circ = 180^\circ \quad\Rightarrow\quad r + x + y = 110^\circ\]Step 4: apply \(r = 2x\) and \(y = 2x\):
\[2x + x + 2x = 110^\circ\]\[5x = 110^\circ \quad\Rightarrow\quad x = 22^\circ\]x = 22\(^\circ\). (Then \(r = 44^\circ\), \(y = 44^\circ\).)
(b) Total amount collected by the two groups.
Let the total collected by the girls be \(G\). Then the boys collected \(G + 600\) (N600 more).
- Average of the 12 girls: \(\dfrac{G}{12}\)
- Average of the 10 boys: \(\dfrac{G + 600}{10}\)
The boys' average is N100 more than the girls' average:
\[\frac{G + 600}{10} = \frac{G}{12} + 100\]Multiply through by 60 (the LCM of 10 and 12):
\[6(G + 600) = 5G + 6000\]\[6G + 3600 = 5G + 6000\]\[G = 2400\]So the girls collected \(\text{N}2400\) and the boys collected \(2400 + 600 = \text{N}3000\).
\[\text{Total} = 2400 + 3000 = \text{N}5400.00\]Total collected = N5,400.00. (Check: girls' average \(=200\), boys' average \(=300\), difference \(=100\).)
Bayanin Amsa
(a) Finding x.
From the diagram, J-H-G-F is a straight line along the bottom. The line J-K-L is straight (K lies on JL), and the line K-M-G is straight (M lies on KG). The marked angles are \(\angle KLM = x\) at L, \(\angle LMK = y\) at M, \(\angle KJH = r\) at J, and \(\angle KGF = 110^\circ\) at G, with the condition \(2x = r = y\).
Step 1: angle the line KG makes with the base at G. Since J-G-F is straight, \(\angle KGF\) and \(\angle KGJ\) are on a straight line:
\[\angle KGJ = 180^\circ - 110^\circ = 70^\circ\]Step 2: angle at K in triangle KLM. The angles of \(\triangle KLM\) sum to \(180^\circ\):
\[\angle LKM = 180^\circ - x - y\]Because J-K-L is a straight line, the angle on the other side of K (angle JKG, which is the angle of triangle JKG at K) is supplementary:
\[\angle JKG = 180^\circ - \angle LKM = 180^\circ - (180^\circ - x - y) = x + y\]Step 3: use triangle JKG. Its vertices are J and G on the base and K above. Its angles are \(\angle KJG = r\), \(\angle JKG = x+y\) and \(\angle KGJ = 70^\circ\), summing to \(180^\circ\):
\[r + (x + y) + 70^\circ = 180^\circ \quad\Rightarrow\quad r + x + y = 110^\circ\]Step 4: apply \(r = 2x\) and \(y = 2x\):
\[2x + x + 2x = 110^\circ\]\[5x = 110^\circ \quad\Rightarrow\quad x = 22^\circ\]x = 22\(^\circ\). (Then \(r = 44^\circ\), \(y = 44^\circ\).)
(b) Total amount collected by the two groups.
Let the total collected by the girls be \(G\). Then the boys collected \(G + 600\) (N600 more).
- Average of the 12 girls: \(\dfrac{G}{12}\)
- Average of the 10 boys: \(\dfrac{G + 600}{10}\)
The boys' average is N100 more than the girls' average:
\[\frac{G + 600}{10} = \frac{G}{12} + 100\]Multiply through by 60 (the LCM of 10 and 12):
\[6(G + 600) = 5G + 6000\]\[6G + 3600 = 5G + 6000\]\[G = 2400\]So the girls collected \(\text{N}2400\) and the boys collected \(2400 + 600 = \text{N}3000\).
\[\text{Total} = 2400 + 3000 = \text{N}5400.00\]Total collected = N5,400.00. (Check: girls' average \(=200\), boys' average \(=300\), difference \(=100\).)
Tambaya 6 Rahoto
(a) Find the equation of a straight line which passes through the point (2, -3) and is parallel to the line \(2x + y = 6\).
(b) The operation \(\Delta\) is defined on the set T = {2, 3, 5, 7} by \(x \Delta y = (x + y + xy) mod 8\).
(i) Construct modulo 8 table for the operation \(\Delta\) on the set T.
(ii) Use the the table to find: (a) \(2 \Delta (5 \Delta 7)\) ; (b) \(2 \Delta n = 5 \Delta 7\).
(a) Equation of the line
\(2x+y=6\Rightarrow y=-2x+6\), so its gradient is \(-2\). A parallel line has the same gradient \(m=-2\) and passes through \((2,-3)\):
\[y-(-3)=-2(x-2)\Rightarrow y+3=-2x+4\Rightarrow y=-2x+1\quad\text{i.e. }2x+y=1.\]
(b)(i) With \(x\,\Delta\,y=(x+y+xy)\bmod 8\) on \(T=\{2,3,5,7\}\):
| \(\Delta\) | 2 | 3 | 5 | 7 |
|---|---|---|---|---|
| 2 | 0 | 3 | 1 | 7 |
| 3 | 3 | 7 | 7 | 7 |
| 5 | 1 | 7 | 3 | 7 |
| 7 | 7 | 7 | 7 | 7 |
(e.g. \(2\Delta5=(2+5+10)\bmod 8=17\bmod 8=1\).)
(ii)(a) \(5\Delta7=7\), so \(2\Delta(5\Delta7)=2\Delta7=7\).
(ii)(b) \(5\Delta7=7\), so we need \(2\Delta n=7\). From the \(2\)-row, \(2\Delta7=7\), hence \(n=7\).
Bayanin Amsa
(a) Equation of the line
\(2x+y=6\Rightarrow y=-2x+6\), so its gradient is \(-2\). A parallel line has the same gradient \(m=-2\) and passes through \((2,-3)\):
\[y-(-3)=-2(x-2)\Rightarrow y+3=-2x+4\Rightarrow y=-2x+1\quad\text{i.e. }2x+y=1.\]
(b)(i) With \(x\,\Delta\,y=(x+y+xy)\bmod 8\) on \(T=\{2,3,5,7\}\):
| \(\Delta\) | 2 | 3 | 5 | 7 |
|---|---|---|---|---|
| 2 | 0 | 3 | 1 | 7 |
| 3 | 3 | 7 | 7 | 7 |
| 5 | 1 | 7 | 3 | 7 |
| 7 | 7 | 7 | 7 | 7 |
(e.g. \(2\Delta5=(2+5+10)\bmod 8=17\bmod 8=1\).)
(ii)(a) \(5\Delta7=7\), so \(2\Delta(5\Delta7)=2\Delta7=7\).
(ii)(b) \(5\Delta7=7\), so we need \(2\Delta n=7\). From the \(2\)-row, \(2\Delta7=7\), hence \(n=7\).
Tambaya 7 Rahoto
(a) Using the method of completing the square, solve, correct to 2 decimal places, \(\frac{x - 2}{4} = \frac{x + 2}{2x}\).
(b) 
In the diagram, PQRST is a circle with centre O. If PS is a diameter, RS//QT, and < QTS = 52°, find :
(i) < SQT ; (ii) < PQT.
(a) Completing the square.
Clear the fractions in \(\dfrac{x-2}{4}=\dfrac{x+2}{2x}\) by cross-multiplying:
\[2x(x-2)=4(x+2)\]
\[2x^{2}-4x=4x+8\]
\[2x^{2}-8x-8=0\]
Divide through by 2 to make the coefficient of \(x^{2}\) equal to 1:
\[x^{2}-4x-4=0 \quad\Rightarrow\quad x^{2}-4x=4\]
Add the square of half the coefficient of \(x\), i.e. \(\left(\tfrac{-4}{2}\right)^{2}=4\), to both sides:
\[x^{2}-4x+4=4+4\]
\[(x-2)^{2}=8\]
\[x-2=\pm\sqrt{8}=\pm 2.8284\]
\[x=2+2.8284=4.83 \quad\text{or}\quad x=2-2.8284=-0.83\]
\[\boxed{x=4.83 \text{ or } x=-0.83}\ \text{(to 2 d.p.)}\]
(b) Circle PQRST, centre O.
From the diagram: PS is a diameter, the tick marks show \(QR=RS\), the arrows show \(RS\parallel QT\), and \(\angle QTS=52^\circ\).
(i) \(\angle SQT\).
\(\angle QTS=52^\circ\) is an angle at the circumference standing on the arc \(QRS\), so
\[\text{arc } QR+\text{arc } RS=2\times 52^\circ=104^\circ.\]
Since \(QR=RS\), the equal chords cut off equal arcs, so \(\text{arc } QR=\text{arc } RS=52^\circ\).
Because \(RS\parallel QT\), the arcs between the parallel chords are equal, giving \(\text{arc } ST=\text{arc } QR=52^\circ\).
\(\angle SQT\) stands at the circumference on arc \(ST\):
\[\angle SQT=\tfrac{1}{2}\,\text{arc } ST=\tfrac{1}{2}\times 52^\circ=\boxed{26^\circ}.\]
(ii) \(\angle PQT\).
PS is a diameter, so \(\angle PQS=90^\circ\) (angle in a semicircle).
The ray \(QT\) lies between \(QP\) and \(QS\), so
\[\angle PQT=\angle PQS-\angle SQT=90^\circ-26^\circ=\boxed{64^\circ}.\]
Bayanin Amsa
(a) Completing the square.
Clear the fractions in \(\dfrac{x-2}{4}=\dfrac{x+2}{2x}\) by cross-multiplying:
\[2x(x-2)=4(x+2)\]
\[2x^{2}-4x=4x+8\]
\[2x^{2}-8x-8=0\]
Divide through by 2 to make the coefficient of \(x^{2}\) equal to 1:
\[x^{2}-4x-4=0 \quad\Rightarrow\quad x^{2}-4x=4\]
Add the square of half the coefficient of \(x\), i.e. \(\left(\tfrac{-4}{2}\right)^{2}=4\), to both sides:
\[x^{2}-4x+4=4+4\]
\[(x-2)^{2}=8\]
\[x-2=\pm\sqrt{8}=\pm 2.8284\]
\[x=2+2.8284=4.83 \quad\text{or}\quad x=2-2.8284=-0.83\]
\[\boxed{x=4.83 \text{ or } x=-0.83}\ \text{(to 2 d.p.)}\]
(b) Circle PQRST, centre O.
From the diagram: PS is a diameter, the tick marks show \(QR=RS\), the arrows show \(RS\parallel QT\), and \(\angle QTS=52^\circ\).
(i) \(\angle SQT\).
\(\angle QTS=52^\circ\) is an angle at the circumference standing on the arc \(QRS\), so
\[\text{arc } QR+\text{arc } RS=2\times 52^\circ=104^\circ.\]
Since \(QR=RS\), the equal chords cut off equal arcs, so \(\text{arc } QR=\text{arc } RS=52^\circ\).
Because \(RS\parallel QT\), the arcs between the parallel chords are equal, giving \(\text{arc } ST=\text{arc } QR=52^\circ\).
\(\angle SQT\) stands at the circumference on arc \(ST\):
\[\angle SQT=\tfrac{1}{2}\,\text{arc } ST=\tfrac{1}{2}\times 52^\circ=\boxed{26^\circ}.\]
(ii) \(\angle PQT\).
PS is a diameter, so \(\angle PQS=90^\circ\) (angle in a semicircle).
The ray \(QT\) lies between \(QP\) and \(QS\), so
\[\angle PQT=\angle PQS-\angle SQT=90^\circ-26^\circ=\boxed{64^\circ}.\]
Tambaya 8 Rahoto
(a) If \(\frac{3p + 4q}{3p - 4q} = 2\), find \(p : q\).
(b) 
The diagram shows the cross section of a bridge with a semi-circular hollow in the middle. If the perimeter of the cross section is 34 cm, calculate the :
(i) length PQ; (ii) area of the cross section.
[Take \(\pi = \frac{22}{7}\)].
(a) Find \(p:q\).
\[\frac{3p+4q}{3p-4q}=2\]Cross multiply:
\[3p+4q=2(3p-4q)=6p-8q\]\[3p+4q-6p+8q=0\]\[-3p+12q=0\Rightarrow 3p=12q\Rightarrow p=4q\]\[p:q=\mathbf{4:1}\](b) The bridge cross-section.
From the diagram, \(PQRU\) is a rectangle with the two vertical sides \(PU=QR=4\text{ m}\), and a semicircular hollow is cut from the base. Along the base the two flat pieces are \(UT=SR=2\text{ m}\), and \(TS\) is the diameter of the semicircle. Let the radius be \(r\), so \(TS=2r\).
(i) Length \(PQ\).
The outline (perimeter) is the top \(PQ\), the two sides, the two flat base pieces, and the semicircular arc:
\[\text{Perimeter}=PQ+PU+QR+UT+SR+\text{arc}(TS)\]Since \(PQRU\) is a rectangle, \(PQ=UR=UT+TS+SR=2+2r+2=2r+4\). With arc \(=\pi r\):
\[34=(2r+4)+4+4+2+2+\pi r\]\[34=2r+16+\pi r\]\[18=r\left(2+\tfrac{22}{7}\right)=r\cdot\tfrac{36}{7}\]\[r=\frac{18\times7}{36}=3.5\text{ cm}\]Therefore
\[PQ=2r+4=2(3.5)+4=\mathbf{11\text{ cm}}\](ii) Area of the cross-section.
Area \(=\) rectangle \(-\) semicircle:
\[=PQ\times4-\tfrac{1}{2}\pi r^{2}=11\times4-\tfrac{1}{2}\cdot\tfrac{22}{7}\cdot(3.5)^{2}\]\[=44-\tfrac{1}{2}\cdot\tfrac{22}{7}\cdot12.25=44-19.25=\mathbf{24.75\text{ cm}^{2}}\](The heights and base pieces read \(4\) and \(2\) from the figure; with the stated perimeter \(34\) this gives a consistent \(r=3.5\).)
Bayanin Amsa
(a) Find \(p:q\).
\[\frac{3p+4q}{3p-4q}=2\]Cross multiply:
\[3p+4q=2(3p-4q)=6p-8q\]\[3p+4q-6p+8q=0\]\[-3p+12q=0\Rightarrow 3p=12q\Rightarrow p=4q\]\[p:q=\mathbf{4:1}\](b) The bridge cross-section.
From the diagram, \(PQRU\) is a rectangle with the two vertical sides \(PU=QR=4\text{ m}\), and a semicircular hollow is cut from the base. Along the base the two flat pieces are \(UT=SR=2\text{ m}\), and \(TS\) is the diameter of the semicircle. Let the radius be \(r\), so \(TS=2r\).
(i) Length \(PQ\).
The outline (perimeter) is the top \(PQ\), the two sides, the two flat base pieces, and the semicircular arc:
\[\text{Perimeter}=PQ+PU+QR+UT+SR+\text{arc}(TS)\]Since \(PQRU\) is a rectangle, \(PQ=UR=UT+TS+SR=2+2r+2=2r+4\). With arc \(=\pi r\):
\[34=(2r+4)+4+4+2+2+\pi r\]\[34=2r+16+\pi r\]\[18=r\left(2+\tfrac{22}{7}\right)=r\cdot\tfrac{36}{7}\]\[r=\frac{18\times7}{36}=3.5\text{ cm}\]Therefore
\[PQ=2r+4=2(3.5)+4=\mathbf{11\text{ cm}}\](ii) Area of the cross-section.
Area \(=\) rectangle \(-\) semicircle:
\[=PQ\times4-\tfrac{1}{2}\pi r^{2}=11\times4-\tfrac{1}{2}\cdot\tfrac{22}{7}\cdot(3.5)^{2}\]\[=44-\tfrac{1}{2}\cdot\tfrac{22}{7}\cdot12.25=44-19.25=\mathbf{24.75\text{ cm}^{2}}\](The heights and base pieces read \(4\) and \(2\) from the figure; with the stated perimeter \(34\) this gives a consistent \(r=3.5\).)
Tambaya 9 Rahoto
(a) Solve : \(7(x + 4) - \frac{2}{3}(x - 6) \leq 2[x - 3(x + 5)]\)
(b) A transport company has a total of 20 vehicles made up of tricycle and taxicabs. Each tricycle carries 2 passengers while each taxicab carries four passengers. If the 20 vehicles carry a total of 66 passengers at a time, how many tricycles does the company have?
(a) Solve \(7(x+4)-\tfrac{2}{3}(x-6)\le 2[x-3(x+5)]\).
Right side: \(x-3(x+5)=x-3x-15=-2x-15\), so \(2(-2x-15)=-4x-30\).
Left side: \(7x+28-\tfrac{2}{3}(x-6)=7x+28-\tfrac{2}{3}x+4=\tfrac{19}{3}x+32\).
Inequality: \(\tfrac{19}{3}x+32\le -4x-30\). Multiply through by 3:
\[19x+96\le -12x-90\Rightarrow 31x\le -186\Rightarrow x\le -6\]\(x\le -6\)
(b) Let \(t\) = tricycles and \(c\) = taxicabs.
Vehicles: \(t+c=20\).
Passengers: \(2t+4c=66\).
From the first, \(c=20-t\). Substitute:
\[2t+4(20-t)=66\Rightarrow 2t+80-4t=66\Rightarrow -2t=-14\Rightarrow t=7\]The company has 7 tricycles (and 13 taxicabs).
Bayanin Amsa
(a) Solve \(7(x+4)-\tfrac{2}{3}(x-6)\le 2[x-3(x+5)]\).
Right side: \(x-3(x+5)=x-3x-15=-2x-15\), so \(2(-2x-15)=-4x-30\).
Left side: \(7x+28-\tfrac{2}{3}(x-6)=7x+28-\tfrac{2}{3}x+4=\tfrac{19}{3}x+32\).
Inequality: \(\tfrac{19}{3}x+32\le -4x-30\). Multiply through by 3:
\[19x+96\le -12x-90\Rightarrow 31x\le -186\Rightarrow x\le -6\]\(x\le -6\)
(b) Let \(t\) = tricycles and \(c\) = taxicabs.
Vehicles: \(t+c=20\).
Passengers: \(2t+4c=66\).
From the first, \(c=20-t\). Substitute:
\[2t+4(20-t)=66\Rightarrow 2t+80-4t=66\Rightarrow -2t=-14\Rightarrow t=7\]The company has 7 tricycles (and 13 taxicabs).
Tambaya 10 Rahoto
(a) Copy and complete the table of values, correct to one decimal place, for the relation \(y = 3\sin x + 2\cos x\) for \(0° \leq x \leq 360°\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° | 330° | 360° |
| y | 3.0 | 1.6 | -2.0 | -3.6 | -3.0 | 2.0 |
(b) Using scales of 2cm to 30°mon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation \(y = 3\sin x + 2\cos x\) for \(0°\leq x \leq 360°\).
(c) Use the graph to solve :
(i) \(3\sin x + 2\cos x = 0\)
(ii) \(2 + 2\cos x + 3\sin x = 0\).
(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\), correct to one decimal place.
Each missing value is obtained by direct substitution:
\(x = 0^\circ:\ 3\sin 0 + 2\cos 0 = 3(0) + 2(1) = 2.0\)
\(x = 30^\circ:\ 3(0.500) + 2(0.866) = 1.5 + 1.7 = 3.2\)
\(x = 60^\circ:\ 3(0.866) + 2(0.500) = 2.6 + 1.0 = 3.6\)
\(x = 150^\circ:\ 3(0.500) + 2(-0.866) = 1.5 - 1.7 = -0.2\)
\(x = 210^\circ:\ 3(-0.500) + 2(-0.866) = -1.5 - 1.7 = -3.2\)
\(x = 300^\circ:\ 3(-0.866) + 2(0.500) = -2.6 + 1.0 = -1.6\)
\(x = 330^\circ:\ 3(-0.500) + 2(0.866) = -1.5 + 1.7 = 0.2\)
The completed table is:
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° | 330° | 360° |
| y | 2.0 | 3.2 | 3.6 | 3.0 | 1.6 | -0.2 | -2.0 | -3.2 | -3.6 | -3.0 | -1.6 | 0.2 | 2.0 |
(b) Draw the graph of \(y = 3\sin x + 2\cos x\) for \(0^\circ \leq x \leq 360^\circ\).
Using scales of 2 cm to \(30^\circ\) on the x-axis and 2 cm to 1 unit on the y-axis, the thirteen points from the table are plotted and joined with a smooth curve. The curve is a single wave that rises to a maximum of about \(3.6\) near \(x = 60^\circ\) and falls to a minimum of about \(-3.6\) near \(x = 240^\circ\). The dashed horizontal line \(y = -2\) is added for part (c)(ii).
(c) Use the graph to solve:
(i) \(3\sin x + 2\cos x = 0\)
This means \(y = 0\); the solutions are the points where the curve cuts the x-axis. Reading these from the graph gives
\[x \approx 147^\circ \quad \text{and} \quad x \approx 325.5^\circ.\]
(ii) \(2 + 2\cos x + 3\sin x = 0\)
Rearrange so that the graphed expression stands alone:
\[3\sin x + 2\cos x = -2, \quad \text{i.e.} \quad y = -2.\]
Draw the horizontal line \(y = -2\) and read where it meets the curve. The line cuts the curve at
\[x \approx 180^\circ \quad \text{and} \quad x \approx 292.5^\circ.\]
Bayanin Amsa
(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\), correct to one decimal place.
Each missing value is obtained by direct substitution:
\(x = 0^\circ:\ 3\sin 0 + 2\cos 0 = 3(0) + 2(1) = 2.0\)
\(x = 30^\circ:\ 3(0.500) + 2(0.866) = 1.5 + 1.7 = 3.2\)
\(x = 60^\circ:\ 3(0.866) + 2(0.500) = 2.6 + 1.0 = 3.6\)
\(x = 150^\circ:\ 3(0.500) + 2(-0.866) = 1.5 - 1.7 = -0.2\)
\(x = 210^\circ:\ 3(-0.500) + 2(-0.866) = -1.5 - 1.7 = -3.2\)
\(x = 300^\circ:\ 3(-0.866) + 2(0.500) = -2.6 + 1.0 = -1.6\)
\(x = 330^\circ:\ 3(-0.500) + 2(0.866) = -1.5 + 1.7 = 0.2\)
The completed table is:
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° | 330° | 360° |
| y | 2.0 | 3.2 | 3.6 | 3.0 | 1.6 | -0.2 | -2.0 | -3.2 | -3.6 | -3.0 | -1.6 | 0.2 | 2.0 |
(b) Draw the graph of \(y = 3\sin x + 2\cos x\) for \(0^\circ \leq x \leq 360^\circ\).
Using scales of 2 cm to \(30^\circ\) on the x-axis and 2 cm to 1 unit on the y-axis, the thirteen points from the table are plotted and joined with a smooth curve. The curve is a single wave that rises to a maximum of about \(3.6\) near \(x = 60^\circ\) and falls to a minimum of about \(-3.6\) near \(x = 240^\circ\). The dashed horizontal line \(y = -2\) is added for part (c)(ii).
(c) Use the graph to solve:
(i) \(3\sin x + 2\cos x = 0\)
This means \(y = 0\); the solutions are the points where the curve cuts the x-axis. Reading these from the graph gives
\[x \approx 147^\circ \quad \text{and} \quad x \approx 325.5^\circ.\]
(ii) \(2 + 2\cos x + 3\sin x = 0\)
Rearrange so that the graphed expression stands alone:
\[3\sin x + 2\cos x = -2, \quad \text{i.e.} \quad y = -2.\]
Draw the horizontal line \(y = -2\) and read where it meets the curve. The line cuts the curve at
\[x \approx 180^\circ \quad \text{and} \quad x \approx 292.5^\circ.\]
Tambaya 11 Rahoto
Using ruler and a pair of compasses only,
(a) construct :
(i) \(\Delta\)XYZ such that |XY| = 10cm, < XYZ = 30° and < YXZ = 45°.
(ii) locus, \(l_{1}\), of points equidistant from Y and Z.
(iii) locus, \(l_{2}\), of points parallel to XY through Z.
(b) Locate M, the point of intersection of \(l_{1}\) and \(l_{2}\).
(c) Measure < ZMY.
Construction using ruler and a pair of compasses only
(a)(i) Triangle XYZ. Draw the base \(|XY| = 10\text{ cm}\). At \(X\) construct \(\angle YXZ = 45^{\circ}\): first raise a perpendicular at \(X\) to obtain \(90^{\circ}\), then bisect that right angle to get \(45^{\circ}\). At \(Y\) construct \(\angle XYZ = 30^{\circ}\): first construct a \(60^{\circ}\) angle (an arc of radius equal to \(|XY_1|\) stepped off along the base), then bisect it to get \(30^{\circ}\). The ray from \(X\) and the ray from \(Y\) meet at \(Z\). This fixes \(\triangle XYZ\), and the third angle is \[\angle XZY = 180^{\circ} - 45^{\circ} - 30^{\circ} = 105^{\circ}.\]
(a)(ii) Locus \(l_1\). The set of points equidistant from \(Y\) and \(Z\) is the perpendicular bisector of \(YZ\). With centre \(Y\) and then centre \(Z\), using the same radius (a little more than half of \(|YZ|\)), draw arcs that cross above and below \(YZ\); the straight line through the two crossings is \(l_1\).
(a)(iii) Locus \(l_2\). The line through \(Z\) parallel to \(XY\). Since \(XY\) is the base, drop a perpendicular from \(Z\) to \(XY\) and at \(Z\) construct a right angle to that perpendicular; the resulting line through \(Z\) is parallel to \(XY\) and is \(l_2\).
(b) Mark \(M\), the point where \(l_1\) and \(l_2\) intersect.
The complete construction is shown below (measured values: \(|XZ| \approx 5.2\text{ cm}\), \(|YZ| \approx 7.3\text{ cm}\)).
(c) Measurement of \(\angle ZMY\). Because \(M\) lies on \(l_1\), it is equidistant from \(Y\) and \(Z\), so \(|MY| = |MZ|\) and \(\triangle MYZ\) is isosceles. On measuring the angle at \(M\) with a protractor: \[\boxed{\angle ZMY = 120^{\circ}.}\]
Bayanin Amsa
Construction using ruler and a pair of compasses only
(a)(i) Triangle XYZ. Draw the base \(|XY| = 10\text{ cm}\). At \(X\) construct \(\angle YXZ = 45^{\circ}\): first raise a perpendicular at \(X\) to obtain \(90^{\circ}\), then bisect that right angle to get \(45^{\circ}\). At \(Y\) construct \(\angle XYZ = 30^{\circ}\): first construct a \(60^{\circ}\) angle (an arc of radius equal to \(|XY_1|\) stepped off along the base), then bisect it to get \(30^{\circ}\). The ray from \(X\) and the ray from \(Y\) meet at \(Z\). This fixes \(\triangle XYZ\), and the third angle is \[\angle XZY = 180^{\circ} - 45^{\circ} - 30^{\circ} = 105^{\circ}.\]
(a)(ii) Locus \(l_1\). The set of points equidistant from \(Y\) and \(Z\) is the perpendicular bisector of \(YZ\). With centre \(Y\) and then centre \(Z\), using the same radius (a little more than half of \(|YZ|\)), draw arcs that cross above and below \(YZ\); the straight line through the two crossings is \(l_1\).
(a)(iii) Locus \(l_2\). The line through \(Z\) parallel to \(XY\). Since \(XY\) is the base, drop a perpendicular from \(Z\) to \(XY\) and at \(Z\) construct a right angle to that perpendicular; the resulting line through \(Z\) is parallel to \(XY\) and is \(l_2\).
(b) Mark \(M\), the point where \(l_1\) and \(l_2\) intersect.
The complete construction is shown below (measured values: \(|XZ| \approx 5.2\text{ cm}\), \(|YZ| \approx 7.3\text{ cm}\)).
(c) Measurement of \(\angle ZMY\). Because \(M\) lies on \(l_1\), it is equidistant from \(Y\) and \(Z\), so \(|MY| = |MZ|\) and \(\triangle MYZ\) is isosceles. On measuring the angle at \(M\) with a protractor: \[\boxed{\angle ZMY = 120^{\circ}.}\]
Tambaya 12 Rahoto
(a) 
In the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS.
(b) A cylinder tin, 7cm high, is closed at one end. If its total surface area is 462\(cm^{2}\), calculate its radius. [Take \(\pi = \frac{22}{7}\)].
(a) Finding \(\angle VUS\).
The points V, R, S, T, U lie on a circle, and line MRQ is a tangent at R. Given \(\angle RTS = 28^\circ\) and \(\angle VRM = 46^\circ\).
Step 1 - use the tangent-chord (alternate segment) property on chord RV. The angle between the tangent RM and the chord RV equals the inscribed angle in the alternate segment standing on RV. Hence the arc VR (the intercepted arc) satisfies
\[\angle VRM = \tfrac{1}{2}(\text{arc } VR) \;\Rightarrow\; \text{arc } VR = 2 \times 46^\circ = 92^\circ.\]
Step 2 - use the inscribed angle on chord RS. \(\angle RTS = 28^\circ\) is the angle at the circumference standing on chord RS, so
\[\text{arc } RS = 2 \times 28^\circ = 56^\circ.\]
Step 3 - find \(\angle VUS\). \(\angle VUS\) is the inscribed angle at U standing on chord VS. It equals half the arc VS that does not contain U; that arc runs from V through R to S:
\[\text{arc } VRS = \text{arc } VR + \text{arc } RS = 92^\circ + 56^\circ = 148^\circ.\]
\[\angle VUS = \tfrac{1}{2}(148^\circ) = 74^\circ.\]
\[\boxed{\angle VUS = 74^\circ.}\]
(b) Radius of the cylindrical tin.
The tin is closed at one end only, so its total surface area is the curved surface plus one circular end:
\[A = 2\pi r h + \pi r^2 = 462,\qquad h = 7,\ \pi = \tfrac{22}{7}.\]
Substitute:
\[2 \times \tfrac{22}{7} \times r \times 7 + \tfrac{22}{7} r^2 = 462.\]
\[44r + \tfrac{22}{7}r^2 = 462.\]
Multiply every term by 7:
\[308r + 22r^2 = 3234.\]
Divide through by 22:
\[14r + r^2 = 147 \;\Rightarrow\; r^2 + 14r - 147 = 0.\]
Solve:
\[r = \frac{-14 \pm \sqrt{14^2 + 4(147)}}{2} = \frac{-14 \pm \sqrt{196 + 588}}{2} = \frac{-14 \pm \sqrt{784}}{2} = \frac{-14 \pm 28}{2}.\]
Taking the positive root: \(r = \dfrac{14}{2} = 7\text{ cm}.\)
Check: \(2(\tfrac{22}{7})(7)(7) + (\tfrac{22}{7})(7^2) = 308 + 154 = 462\text{ cm}^2.\) \(\;\checkmark\)
\[\boxed{r = 7\text{ cm}.}\]
Bayanin Amsa
(a) Finding \(\angle VUS\).
The points V, R, S, T, U lie on a circle, and line MRQ is a tangent at R. Given \(\angle RTS = 28^\circ\) and \(\angle VRM = 46^\circ\).
Step 1 - use the tangent-chord (alternate segment) property on chord RV. The angle between the tangent RM and the chord RV equals the inscribed angle in the alternate segment standing on RV. Hence the arc VR (the intercepted arc) satisfies
\[\angle VRM = \tfrac{1}{2}(\text{arc } VR) \;\Rightarrow\; \text{arc } VR = 2 \times 46^\circ = 92^\circ.\]
Step 2 - use the inscribed angle on chord RS. \(\angle RTS = 28^\circ\) is the angle at the circumference standing on chord RS, so
\[\text{arc } RS = 2 \times 28^\circ = 56^\circ.\]
Step 3 - find \(\angle VUS\). \(\angle VUS\) is the inscribed angle at U standing on chord VS. It equals half the arc VS that does not contain U; that arc runs from V through R to S:
\[\text{arc } VRS = \text{arc } VR + \text{arc } RS = 92^\circ + 56^\circ = 148^\circ.\]
\[\angle VUS = \tfrac{1}{2}(148^\circ) = 74^\circ.\]
\[\boxed{\angle VUS = 74^\circ.}\]
(b) Radius of the cylindrical tin.
The tin is closed at one end only, so its total surface area is the curved surface plus one circular end:
\[A = 2\pi r h + \pi r^2 = 462,\qquad h = 7,\ \pi = \tfrac{22}{7}.\]
Substitute:
\[2 \times \tfrac{22}{7} \times r \times 7 + \tfrac{22}{7} r^2 = 462.\]
\[44r + \tfrac{22}{7}r^2 = 462.\]
Multiply every term by 7:
\[308r + 22r^2 = 3234.\]
Divide through by 22:
\[14r + r^2 = 147 \;\Rightarrow\; r^2 + 14r - 147 = 0.\]
Solve:
\[r = \frac{-14 \pm \sqrt{14^2 + 4(147)}}{2} = \frac{-14 \pm \sqrt{196 + 588}}{2} = \frac{-14 \pm \sqrt{784}}{2} = \frac{-14 \pm 28}{2}.\]
Taking the positive root: \(r = \dfrac{14}{2} = 7\text{ cm}.\)
Check: \(2(\tfrac{22}{7})(7)(7) + (\tfrac{22}{7})(7^2) = 308 + 154 = 462\text{ cm}^2.\) \(\;\checkmark\)
\[\boxed{r = 7\text{ cm}.}\]
Tambaya 13 Rahoto
(a) 
In the diagram, PQST is a parallelogram, PR is a straight line, |TS| = 8cm, |SM| = 6cm and area of triangle PSR = \(36 cm^{2}\). Find the value of |QR|.
(b) A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angle of 45° and 60° respectively. If the tree is 10.65m high, calculate, correct to 3 significant figures, the height of the flagpole.
(a) Find |QR|.
From the diagram, \(PQST\) is a parallelogram, and \(P,\,Q,\,R\) lie on one straight line. The opposite sides of the parallelogram are equal, so:
\[|PQ|=|TS|=8\text{ cm}\]
Triangle \(PSR\) has its base along the line \(PR\), and \(SM=6\text{ cm}\) is the perpendicular height from \(S\) to that base.
Using the area of a triangle:
\[\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}\]
\[36=\frac{1}{2}\times |PR|\times 6=3\,|PR|\]
\[|PR|=\frac{36}{3}=12\text{ cm}\]
Since \(|PR|=|PQ|+|QR|\):
\[12=8+|QR|\Rightarrow |QR|=4\text{ cm}\]
(b) Height of the flagpole.
Let the horizontal distance from the tree to the flagpole be \(d\). The bird is at the top of the tree, at height \(10.65\text{ m}\).
Bottom of the flagpole (ground level) is seen at an angle of depression of \(60^{\circ}\). The vertical drop is the full \(10.65\text{ m}\):
\[\tan 60^{\circ}=\frac{10.65}{d}\Rightarrow d=\frac{10.65}{\tan 60^{\circ}}=\frac{10.65}{1.7321}=6.149\text{ m}\]
Top of the flagpole is seen at an angle of depression of \(45^{\circ}\). The vertical drop from the bird down to the top of the flagpole is:
\[\text{drop}=d\tan 45^{\circ}=6.149\times 1=6.149\text{ m}\]
The height of the flagpole is the tree height minus this drop:
\[h=10.65-6.149=4.501\text{ m}\]
Correct to 3 significant figures, the height of the flagpole is \(4.50\text{ m}\).
Bayanin Amsa
(a) Find |QR|.
From the diagram, \(PQST\) is a parallelogram, and \(P,\,Q,\,R\) lie on one straight line. The opposite sides of the parallelogram are equal, so:
\[|PQ|=|TS|=8\text{ cm}\]
Triangle \(PSR\) has its base along the line \(PR\), and \(SM=6\text{ cm}\) is the perpendicular height from \(S\) to that base.
Using the area of a triangle:
\[\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}\]
\[36=\frac{1}{2}\times |PR|\times 6=3\,|PR|\]
\[|PR|=\frac{36}{3}=12\text{ cm}\]
Since \(|PR|=|PQ|+|QR|\):
\[12=8+|QR|\Rightarrow |QR|=4\text{ cm}\]
(b) Height of the flagpole.
Let the horizontal distance from the tree to the flagpole be \(d\). The bird is at the top of the tree, at height \(10.65\text{ m}\).
Bottom of the flagpole (ground level) is seen at an angle of depression of \(60^{\circ}\). The vertical drop is the full \(10.65\text{ m}\):
\[\tan 60^{\circ}=\frac{10.65}{d}\Rightarrow d=\frac{10.65}{\tan 60^{\circ}}=\frac{10.65}{1.7321}=6.149\text{ m}\]
Top of the flagpole is seen at an angle of depression of \(45^{\circ}\). The vertical drop from the bird down to the top of the flagpole is:
\[\text{drop}=d\tan 45^{\circ}=6.149\times 1=6.149\text{ m}\]
The height of the flagpole is the tree height minus this drop:
\[h=10.65-6.149=4.501\text{ m}\]
Correct to 3 significant figures, the height of the flagpole is \(4.50\text{ m}\).
