9. a telecommunication mast and a pillar stand on the same horizontal level. from the mid - point, o, between the foot of the pillar, s, and the foot of the mast, t, the angle of elevation of the top of the mast g, is 70°. from the point o, the angle of elevation of the top of the pillar, h, is 28°. if the pillar is 10 m high:
(a) represent the information in a diagram;
(b) calculate, correct to the nearest whole number:
(i) the height of the mast;
(ii) |hg|.
10. (a) the points m, n and l lie on a circle with centre o. an angle subtended at the centre by the minor arc mn is 88°.
(i) illustrate the information in a diagram.
(ii) if l is a point on the minor arc, determine the value of the angle subtended by the chord mn at l.
(b) a tiler, coffie, is able to lay 400 tiles each day in 8 hours. another tiler, boakye, works at 3/2 times coffies rate when laying same kind of tiles.
(i) how many tiles can boakye lay in 8 hours?
(ii) if both coffie and boakye decide to work together, determine the time required to lay 2000 tiles.
11. a cardboard is in the form of an isosceles triangle with vertices abc. |ab| = |ac|=(2p + q)m, |bc|=(3p + 5)m and the perimeter of the cardboard is (7q + 3)m.
(a) illustrate the information in a diagram.
(b) given that p:q = 2:3, find, correct to one decimal place, the:
(i) value of p and q;
(ii) area of the cardboard.
12. the following are the marks obtained by 50 students in an english test.
27 31 35 52 81 50 48 15 59 6
17 41 30 94 23 38 40 70 76 18
52 61 43 68 56 57 62 82 54 25
18 35 9 74 59 46 71 4 74 33
61 16 50 45 41 45 28 22 26 45
(a) using the class interval, 1 - 10, 11 - 20, 21 - 30, ..., construct a cumulative frequency table for the distribution.
(b) draw a cumulative frequency curve for the distribution.
(c) use the curve to find the upper quartile.
(d) if only 10% of the students passed the test, use the curve to determine the pass mark.

Question

Accepted Answer

### 9.
# Explanation:
## Step1: Set up right - triangle relationships for height of mast
Let the distance from $O$ to the foot of the pillar and mast be $x$. For the pillar of height $h_{1}=10$ m with angle of elevation $\theta_{1} = 28^{\circ}$, we have $\tan\theta_{1}=\frac{h_{1}}{x}$, so $\tan28^{\circ}=\frac{10}{x}$, and $x = \frac{10}{\tan28^{\circ}}$. For the mast of height $h_{2}$ with angle of elevation $\theta_{2}=70^{\circ}$, $\tan\theta_{2}=\frac{h_{2}}{x}$.
## Step2: Calculate the height of the mast
Since $x=\frac{10}{\tan28^{\circ}}$ and $\tan70^{\circ}=\frac{h_{2}}{x}$, then $h_{2}=x\tan70^{\circ}=\frac{10\tan70^{\circ}}{\tan28^{\circ}}$. We know that $\tan70^{\circ}\approx2.7475$ and $\tan28^{\circ}\approx0.5317$. So $h_{2}=\frac{10\times2.7475}{0.5317}\approx52$ m.
## Step3: Calculate $|HG|$
We use the three - dimensional distance formula in the vertical and horizontal plane. First, the vertical distance between the top of the mast and pillar is $h = h_{2}-h_{1}$, and the horizontal distance between the mast and pillar is $2x$. By the Pythagorean theorem, $|HG|=\sqrt{(h_{2} - h_{1})^{2}+(2x)^{2}}$. Since $h_{2}-h_{1}=52 - 10 = 42$ m and $x=\frac{10}{\tan28^{\circ}}\approx18.8$ m, $2x\approx37.6$ m. Then $|HG|=\sqrt{42^{2}+37.6^{2}}=\sqrt{1764 + 1413.76}=\sqrt{3177.76}\approx56$ m.
# Answer:
(i) 52 m
(ii) 56 m

### 10.
## (a)
# Explanation:
## Step1: Draw the circle diagram for part (i)
Draw a circle with center $O$. Mark points $M$, $N$ and $L$ on the circle. Draw the central angle $\angle MON = 88^{\circ}$.
## Step2: Calculate the angle subtended by the minor arc at point $L$ for part (ii)
The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at the circumference. So if the central angle $\angle MON=88^{\circ}$, the angle subtended by the minor arc $MN$ at point $L$ is $\frac{1}{2}\angle MON = 44^{\circ}$.
# Answer:
(i) Draw a circle with center $O$, points $M$, $N$ and $L$ on it and $\angle MON = 88^{\circ}$
(ii) $44^{\circ}$
## (b)
# Explanation:
## Step1: Calculate Boakye's rate for part (i)
Coffie can lay 400 tiles in 8 hours. Boakye works at $\frac{3}{2}$ times Coffie's rate. So Boakye can lay $400\times\frac{3}{2}=600$ tiles in 8 hours.
## Step2: Calculate the combined rate and time for part (ii)
The combined rate of Coffie and Boakye is $400 + 600=1000$ tiles per 8 hours. Let $t$ be the time required to lay 2000 tiles. The rate $r = 1000$ tiles per 8 hours, and we know that $r=\frac{\text{number of tiles}}{\text{time}}$. So $\frac{1000}{8}=\frac{2000}{t}$, and $t = 16$ hours.
# Answer:
(i) 600 tiles
(ii) 16 hours

### 11.
## (a)
# Explanation:
## Step1: Draw the isosceles - triangle diagram
Draw an isosceles triangle $ABC$ with $|AB| = |AC|=(2p + q)$ m and $|BC|=(3p + 5)$ m.
## (b)
# Explanation:
## Step1: Use the ratio $p:q = 2:3$ to express $q$ in terms of $p$
Since $p:q=2:3$, then $q=\frac{3}{2}p$. The perimeter of the isosceles triangle $P=2(2p + q)+(3p + 5)=(7p + 3)$ m. Substitute $q=\frac{3}{2}p$ into the perimeter formula: $2(2p+\frac{3}{2}p)+(3p + 5)=7p + 3$. Expand the left - hand side: $4p+3p+3p + 5=7p + 3$, $10p+5=7p + 3$, $10p-7p=3 - 5$, $3p=-2$, $p =-\frac{2}{3}$. This is wrong. Let's correct it. The perimeter $P = 2(2p + q)+(3p + 5)=7q+3$. Substitute $q=\frac{3}{2}p$ into $P$: $2(2p+\frac{3}{2}p)+(3p + 5)=7\times\frac{3}{2}p+3$. Expand: $4p + 3p+3p+5=\frac{21}{2}p+3$, $10p + 5=\frac{21}{2}p+3$, $10p-\frac{21}{2}p=3 - 5$, $\frac{20p-21p}{2}=-2$, $-\frac{p}{2}=-2$, $p = 4$. Then $q=\frac{3}{2}\times4 = 6$.
## Step2: Calculate the area of the isosceles triangle
First, find the semi - perimeter $s=\frac{2(2p + q)+(3p + 5)}{2}$. Substitute $p = 4$ and $q = 6$: $|AB|=|AC|=2\times4 + 6=14$ m, $|BC|=3\times4+5 = 17$ m. $s=\frac{2\times14 + 17}{2}=\frac{28+17}{2}=\frac{45}{2}$ m. By Heron's formula $A=\sqrt{s(s - a)(s - b)(s - c)}$, $A=\sqrt{\frac{45}{2}(\frac{45}{2}-14)(\frac{45}{2}-14)(\frac{45}{2}-17)}=\sqrt{\frac{45}{2}\times\frac{17}{2}\times\frac{17}{2}\times\frac{11}{2}}=\sqrt{\frac{45\times17\times17\times11}{16}}\approx107.7$ $m^{2}$
# Answer:
(i) $p = 4$, $q = 6$
(ii) $107.7$ $m^{2}$

### 12.
# Explanation:
## Step1: Construct the cumulative - frequency table for part (a)
| Class Interval | Frequency | Cumulative Frequency |
|---|---|---|
| $1 - 10$ | 7 | 7 |
| $11 - 20$ | 11 | 18 |
| $21 - 30$ | 10 | 28 |
| $31 - 40$ | 8 | 36 |
| $41 - 50$ | 7 | 43 |
| $51 - 60$ | 4 | 47 |
| $61 - 70$ | 3 | 50 |
## Step2: Draw the cumulative - frequency curve for part (b)
Plot the upper - class limits on the x - axis and the cumulative frequencies on the y - axis and draw a smooth curve through the points $(10,7)$, $(20,18)$, $(30,28)$, $(40,36)$, $(50,43)$, $(60,47)$, $(70,50)$.
## Step3: Find the upper quartile for part (c)
The upper quartile $Q_{3}$ corresponds to the 75th percentile. Since $n = 50$, $0.75n=37.5$. Reading from the cumulative - frequency curve, the upper quartile is approximately 43.
## Step4: Find the pass mark for part (d)
If 10% of the students passed, then 90% failed. $0.9n = 45$. Reading from the cumulative - frequency curve, the pass mark is approximately 62.
# Answer:
(a) See the cumulative - frequency table above
(b) Draw a curve through the points $(10,7)$, $(20,18)$, $(30,28)$, $(40,36)$, $(50,43)$, $(60,47)$, $(70,50)$
(c) 43
(d) 62

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QUESTION IMAGE

Question

Explanation:

9.

Step1: Set up right - triangle relationships for height of mast

Step2: Calculate the height of the mast

Step3: Calculate \(|HG|\)

Step1: Draw the circle diagram for part (i)

Step2: Calculate the angle subtended by the minor arc at point \(L\) for part (ii)

Step1: Calculate Boakye's rate for part (i)

Step2: Calculate the combined rate and time for part (ii)

Answer:

10.

(a)