Détails de la réponse

(a) Sides \(7:5:7\). The angle included between the two equal sides (each \(7k\)) is opposite the side \(5k\). By the cosine rule:

\[\cos\theta=\frac{7^{2}+7^{2}-5^{2}}{2(7)(7)}=\frac{49+49-25}{98}=\frac{73}{98}=0.7449.\]

\[\theta=\cos^{-1}(0.7449)=41.9^{\circ}\approx\mathbf{42^{\circ}}\ (\text{nearest degree}).\]

(b) Sum of interior angles \(=(n-2)\times180^{\circ}=1440^{\circ}\):

\[n-2=\frac{1440}{180}=8\Rightarrow \mathbf{n=10\text{ sides}}.\]

(ii) Each exterior angle \(=\dfrac{360^{\circ}}{n}=\dfrac{360^{\circ}}{10}=\mathbf{36^{\circ}}.\)

Détails de la réponse

(a) \(O\) is the centre, so \(|OA|=|OB|=r\). Perimeter of \(\triangle AOB\):

\[2r+|AB|=52.2\Rightarrow 2r+24.2=52.2\Rightarrow r=14\text{ cm}.\]

By the cosine rule in \(\triangle AOB\):

\[\cos(\angle AOB)=\frac{r^{2}+r^{2}-|AB|^{2}}{2r^{2}}=\frac{196+196-585.64}{392}=\frac{-193.64}{392}=-0.4939.\]

\[\angle AOB=\cos^{-1}(-0.4939)=119.6^{\circ}\approx\mathbf{120^{\circ}}\ (\text{nearest degree}).\]

(b) Volume of tank \(=60\times80\times100=480{,}000\text{ cm}^{3}\). Half filled:

\[\tfrac12\times480{,}000=240{,}000\text{ cm}^{3}=\frac{240{,}000}{1000}=\mathbf{240\text{ litres}}\quad(1000\text{ cm}^{3}=1\text{ litre}).\]

Détails de la réponse

(a) Substitute \(x=1,\ y=-2\):

\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]

\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]

From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).

Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).

(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):

\[(4r-3)(3r+4)=(2r-1)(6r+1)\]

\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]

Détails de la réponse

(a) Height of the building.

Let \(d\) be the horizontal distance between the tower and the building. The observer is at the top of the tower, height \(72\text{ m}\).

Bottom of the building (same level as the foot of the tower) has an angle of depression of \(62^{\circ}\), so the vertical drop is the full \(72\text{ m}\):

\[\tan 62^{\circ}=\frac{72}{d}\Rightarrow d=\frac{72}{\tan 62^{\circ}}=\frac{72}{1.8807}=38.28\text{ m}\]

Top of the building has an angle of depression of \(51^{\circ}\), so the drop from the tower top down to the building top is:

\[\text{drop}=d\tan 51^{\circ}=38.28\times 1.2349=47.27\text{ m}\]

The height of the building is the tower height minus this drop:

\[h=72-47.27=24.73\text{ m}\]

Correct to 2 significant figures, the height of the building is \(25\text{ m}\).

(b) Circle, centre O, radius 30 cm, \(\angle POR=120^{\circ}\), \(\pi=3.142\).

(i) Length of chord PR.

Using the cosine rule in triangle \(POR\) with \(|OP|=|OR|=30\text{ cm}\):

\[|PR|^{2}=30^{2}+30^{2}-2(30)(30)\cos 120^{\circ}\]

\[=900+900-1800(-0.5)=1800+900=2700\]

\[|PR|=\sqrt{2700}=51.96\text{ cm}\approx 52.0\text{ cm}\]

(ii) Length of arc PQR.

From the diagram, \(Q\) lies on the major arc, so arc \(PQR\) uses the reflex angle:

\[\text{Reflex }\angle POR=360^{\circ}-120^{\circ}=240^{\circ}\]

\[\text{Arc }PQR=\frac{240}{360}\times 2\pi r=\frac{240}{360}\times 2\times 3.142\times 30\]

\[=\frac{240}{360}\times 188.52=0.6667\times 188.52=125.7\text{ cm}\]

(iii) Perimeter of the shaded portion.

The shaded region is the major segment, bounded by the chord \(PR\) and the major arc \(PQR\):

\[\text{Perimeter}=|PR|+\text{arc }PQR=51.96+125.7=177.7\text{ cm}\]

Correct to 3 significant figures, the perimeter of the shaded portion is \(178\text{ cm}\).

Détails de la réponse

```html

Construction of Triangle PQR and the Required Loci

Given: Construct a triangle PQR such that

• PQ = 8 cm,
• PR = 7 cm,
• ∠QPR = 105°.

(a)(i) Construction of ΔPQR

1. Using a ruler, draw a line segment PQ = 8 cm.
2. At point P, construct an angle of 105°:
   • Construct a 90° angle at P by drawing a perpendicular to PQ.
   • Construct a 120° ray at P using equal arcs with the compasses.
   • Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
3. On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
4. Join R to Q. This gives the required triangle ΔPQR.

(a)(ii) Locus L<sub>1</sub> of points equidistant from P and Q

The locus of points equidistant from P and Q is the perpendicular bisector of PQ.

1. With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
2. Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
3. Join the two arc-intersection points with a ruler.

This straight line is labelled L<sub>1</sub>.

(a)(iii) Locus l<sub>2</sub> of points equidistant from Q and R

The locus of points equidistant from Q and R is the perpendicular bisector of QR.

1. Draw equal arcs centred at Q and R, using a radius greater than half of QR.
2. Join the two points where the arcs intersect.

This straight line is labelled l<sub>2</sub>.

(b)(i) Point of Intersection

Label the point where L<sub>1</sub> and l<sub>2</sub> intersect as T.

Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.

```

Détails de la réponse

(a) Simplify \(\sqrt{1001_{two}}\), leaving the answer in base two.

First convert \(1001_{two}\) to base ten:

\[1001_{two}=1(2^3)+0(2^2)+0(2^1)+1(2^0)=8+0+0+1=9_{ten}.\]

Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:

\[3_{ten}=1(2^1)+1(2^0)=11_{two}.\]

Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).

(b) Find \(z\) in terms of \(x\) and \(y\).

In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):

\[|KP|=\tfrac{1}{2}|PQ|=\tfrac{z}{2}.\]

Applying Pythagoras' theorem to right-angled triangle \(OKP\):

\[|OP|^2=|OK|^2+|KP|^2\]\[x^2=y^2+\left(\frac{z}{2}\right)^2\]\[\left(\frac{z}{2}\right)^2=x^2-y^2\]\[\frac{z}{2}=\sqrt{x^2-y^2}\]

Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).

(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).

(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:

\[\angle PSQ=\tfrac{1}{2}\times 160^\circ=80^\circ.\]

In triangle \(PQS\), the three angles sum to \(180^\circ\):

\[\angle QPS=180^\circ-\angle PQS-\angle PSQ=180^\circ-40^\circ-80^\circ=\mathbf{60^\circ}.\]

(ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.

• Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
• \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
• \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).

The four arcs around the circle sum to \(360^\circ\):

\[\text{arc }PQ+\text{arc }QR+\text{arc }RS+\text{arc }SP=360^\circ\]\[160^\circ+90^\circ+\text{arc }RS+80^\circ=360^\circ\]\[\text{arc }RS=30^\circ.\]

\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):

\[\angle RQS=\tfrac{1}{2}\times30^\circ=\mathbf{15^\circ}.\]

Détails de la réponse

(a) Write both sides in base 3: \(81=3^{4},\ 27=3^{3}\).

\[\frac{1}{81^{x-2}}=3^{-4(x-2)},\qquad 27^{1-x}=3^{3(1-x)}.\]

Equating indices:

\[-4(x-2)=3(1-x)\Rightarrow -4x+8=3-3x\Rightarrow -x=-5\Rightarrow x=5.\]

(b) Rationalise each term:

\[\frac{5}{\sqrt7-\sqrt3}=\frac{5(\sqrt7+\sqrt3)}{7-3}=\frac{5(\sqrt7+\sqrt3)}{4},\qquad \frac{1}{\sqrt7+\sqrt3}=\frac{\sqrt7-\sqrt3}{4}.\]

\[\text{Sum}=\frac{5\sqrt7+5\sqrt3+\sqrt7-\sqrt3}{4}=\frac{6\sqrt7+4\sqrt3}{4}=\frac{3\sqrt7+2\sqrt3}{2}.\]

Détails de la réponse

\(R=6400\) km, \(\pi=\tfrac{22}{7}\). \(K(60^{\circ}\N,50^{\circ}\W)\).

(a)(i) Longitude of L (due east of \(K\), along latitude \(60^{\circ}\N\)). Radius of that parallel uses \(\cos60^{\circ}=\tfrac12\):

\[KL=\frac{\theta}{360}\times2\pi R\cos60^{\circ}\Rightarrow 3520=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\frac12.\]

Full parallel \(=\dfrac{44}{7}\times3200=20114.3\) km, so \(\theta=\dfrac{3520\times360}{20114.3}=63^{\circ}\) (eastward).

Starting at \(50^{\circ}\W\) and moving \(63^{\circ}\) east: \(-50^{\circ}+63^{\circ}=+13^{\circ}\). Longitude of \(L=13^{\circ}\E\).

(a)(ii) Latitude of N (due north of \(K\), along a meridian). Full meridian circle \(=2\pi R=\dfrac{44}{7}\times6400=40228.6\) km.

\[KN=\frac{\phi}{360}\times40228.6=10951\Rightarrow\phi=\frac{10951\times360}{40228.6}=98^{\circ}.\]

From \(60^{\circ}\N\), going \(98^{\circ}\) north first reaches the North Pole after \(30^{\circ}\), then continues \(68^{\circ}\) down the opposite meridian. Latitude of \(N=90^{\circ}-68^{\circ}=22^{\circ}\N\) (on the \(130^{\circ}\E\) meridian).

(b) Let total income be \(I\). Tax-free \(=20\%\), so taxable \(=0.8I\); tax \(=25\) kobo per naira \(=25\%\) of taxable:

\[0.25\times0.8I=1200\Rightarrow 0.2I=1200\Rightarrow I=\mathbf{N6000}.\]

Détails de la réponse

(a) Factor the first bracket:

\[\frac{x^{2}}{2}-x+\frac12=\frac12\left(x^{2}-2x+1\right)=\frac12(x-1)^{2}.\]

\[\therefore\ \frac12(x-1)^{2}\cdot\frac{1}{x-1}=\frac{x-1}{2}.\]

(b) \(P\) is 40 km from \(Q\) on bearing \(061^{\circ}\) (measured from north).

(i) North of Q: \(40\cos61^{\circ}=40(0.4848)=19.4\text{ km}\ (1\text{ d.p.}).\)

(ii) East of Q: \(40\sin61^{\circ}=40(0.8746)=35.0\text{ km}\ (1\text{ d.p.}).\)

(c) Let the son's share be \(s\). Each daughter gets \(\tfrac34 s\); three daughters get \(3\times\tfrac34 s=\tfrac94 s\).

\[s+\frac94 s=5720\Rightarrow\frac{13}{4}s=5720\Rightarrow s=5720\times\frac{4}{13}=\mathbf{N1760}.\]

Détails de la réponse

Détails de la réponse

(a) Area of the minor segment

The chord \(XY\) subtends \(\theta = 96^\circ\) at the centre and the radius is \(r = 5\text{ cm}\). The minor segment is the minor sector minus triangle \(OXY\).

Area of minor sector (with \(\pi = \tfrac{22}{7}\)):

\[\frac{\theta}{360}\pi r^2 = \frac{96}{360}\times\frac{22}{7}\times 5^2 = 20.95\text{ cm}^2\]

Area of triangle OXY:

\[\frac{1}{2}r^2\sin\theta = \frac{1}{2}\times 5^2\times\sin 96^\circ = 12.43\text{ cm}^2\]

Area of minor segment:

\[20.95 - 12.43 = 8.52\text{ cm}^2\]

Correct to three significant figures, the area of the minor segment is \(\mathbf{8.52\text{ cm}^2}\).

(b) Total area of the shaded portions

The square has side 14 cm, so the inscribed circle has radius \(r = 7\text{ cm}\).

Shaded sector of the circle (angle \(80^\circ\)):

\[\frac{80}{360}\times\frac{22}{7}\times 7^2 = 34.22\text{ cm}^2\]

Shaded portion of the square (the four corners between the square and the circle):

\[\text{Area of square} = 14^2 = 196\text{ cm}^2\]

\[\text{Area of circle} = \frac{22}{7}\times 7^2 = 154\text{ cm}^2\]

\[196 - 154 = 42\text{ cm}^2\]

Total shaded area:

\[42 + 34.22 = 76.22 \approx 76.2\text{ cm}^2\]

Détails de la réponse

Détails de la réponse