WAEC SSCE - General Mathematics - 1988 (Essay)

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When a sector is rolled into a cone, the arc length of the sector becomes the circumference of the cone's base, and the sector radius becomes the slant height.

(a) Arc length \(=\dfrac{270}{360}\times2\pi\times7=\dfrac{3}{4}\times2\times\dfrac{22}{7}\times7=\dfrac{3}{4}\times44=33\text{ cm}.\)

Base circumference \(=2\pi r=33\):

\[2\times\frac{22}{7}\times r=33\Rightarrow\frac{44}{7}r=33\Rightarrow r=\frac{33\times7}{44}=5.25\text{ cm}.\]

Base radius \(=5.25\) cm.

(b) Base area \(=\pi r^{2}=\dfrac{22}{7}\times(5.25)^{2}=\dfrac{22}{7}\times27.5625=86.625\text{ cm}^{2}\).

\(\approx 87\text{ cm}^{2}\) (nearest square centimetre).

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(i) Bar chart of company production

The bar chart below represents the number of items produced in each year. The horizontal axis shows the years and the vertical axis shows the number of items produced.

(ii) Average production

Year	1978	1979	1980	1981	1982
Items produced	4100	2500	1500	1800	9200

\[\text{Average production}=\frac{4100+2500+1500+1800+9200}{5}=\frac{19100}{5}=3820\]

Therefore, the average production for the five-year period is 3,820 items.

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Radius \(R=6400\) km, \(\pi=3.142\), latitude \(46^{\circ}\N\), \(\cos46^{\circ}=0.6947\).

(i) Length of the parallel of latitude \(46^{\circ}\)N

\[L=2\pi R\cos46^{\circ}=2(3.142)(6400)(0.6947)=27{,}940\text{ km}\approx\mathbf{27{,}900\text{ km}}\ (\text{nearest }100).\]

(ii) Distance K to Q (same latitude). \(K\) at \(130^{\circ}\W\), \(Q\) at \(103^{\circ}\W\): difference \(=27^{\circ}\).

\[d_{KQ}=\frac{27}{360}\times27{,}940=2095\text{ km}\approx\mathbf{2100\text{ km}}\ (\text{nearest }100).\]

(iii) Distance Q to P along the parallel. \(Q\) at \(103^{\circ}\W\), \(P\) at \(23^{\circ}\E\): difference \(=103^{\circ}+23^{\circ}=126^{\circ}\).

\[d_{QP}=\frac{126}{360}\times27{,}940=9779\text{ km}\approx\mathbf{9780\text{ km}}\ (\text{nearest }10).\]

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From the diagram, S lies on \(PR\) with \(\angle PQR = 90^\circ\) (right angle at Q) and \(\angle PSQ = 90^\circ\), so \(QS\) is the altitude from Q onto the hypotenuse \(PR\). The given lengths are \(|PS| = 9\text{ cm}\), \(|SR| = 16\text{ cm}\) and \(|SQ| = x\).

(a) Value of x by a trigonometric ratio

Let \(\angle QPS = \alpha\). In right triangle \(PSQ\):

\[\tan\alpha = \frac{QS}{PS} = \frac{x}{9}\]

In right triangle \(QSR\), the angle \(\angle SQR\) equals \(\alpha\) (both equal \(90^\circ - \angle R\)), so

\[\tan\alpha = \frac{SR}{QS} = \frac{16}{x}\]

Equating the two expressions:

\[\frac{x}{9} = \frac{16}{x} \;\Rightarrow\; x^2 = 9\times 16 = 144\]\[x = 12\text{ cm}\]

(b)(i) Size of \(\angle QRS\) to the nearest degree

In right triangle \(QSR\):

\[\tan(\angle QRS) = \frac{QS}{SR} = \frac{12}{16} = 0.75\]\[\angle QRS = \tan^{-1}(0.75) = 36.87^\circ \approx 37^\circ\]

(b)(ii) \(|PQ|\)

In right triangle \(PSQ\), by Pythagoras:

\[|PQ|^2 = |PS|^2 + |QS|^2 = 9^2 + 12^2 = 81 + 144 = 225\]\[|PQ| = \sqrt{225} = 15\text{ cm}\]

Therefore \(x = 12\text{ cm}\), \(\angle QRS \approx 37^\circ\) and \(|PQ| = 15\text{ cm}\).

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(a) \(A=\{m,a,p,e\}\), \(B=\{a,e,i,o,u\}\), \(C=\{l,m,n,o,p,q,r,s,t,u\}\).

• (i) \(A\cup B=\{a,e,i,m,o,p,u\}\).
• (ii) \(A\cup C=\{a,e,l,m,n,o,p,q,r,s,t,u\}\).
• (iii) \(B\cap C=\{o,u\}\), so \(A\cup(B\cap C)=\{a,e,m,o,p,u\}\).

(b) Total \(=400\); Biology \(n(B)=300\); Chemistry \(n(C)=190\); neither \(=70\).

(ii) At least one \(=400-70=\mathbf{330}\) students.

(i) By inclusion-exclusion, both \(=n(B)+n(C)-n(B\cup C)=300+190-330=\mathbf{160}\) students.

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(a) The angle sum of a triangle is two right angles.

Let \(\triangle ABC\) have interior angles \(a,b,c\) at \(A,B,C\). Through \(A\) draw a line \(XY\) parallel to \(BC\).

• \(\angle XAB=\angle ABC=b\) (alternate angles, \(XY\parallel BC\)).
• \(\angle YAC=\angle ACB=c\) (alternate angles, \(XY\parallel BC\)).

Angles on the straight line \(XY\) at \(A\): \(\angle XAB+\angle BAC+\angle YAC=180^{\circ}\), i.e. \(b+a+c=180^{\circ}\). Hence \(a+b+c=180^{\circ}=\) two right angles. \(\blacksquare\)

(b) In \(\triangle LMN\): \(\angle LMN=46^{\circ}\), \(\angle MLN=80^{\circ}\), so \(\angle LNM=180^{\circ}-46^{\circ}-80^{\circ}=54^{\circ}\).

\(NM\) is produced to \(P\), so \(\angle LNP=180^{\circ}-\angle LNM=126^{\circ}\) (angles on a straight line). Its bisector \(NQ\) gives \(\angle LNQ=\tfrac12(126^{\circ})=63^{\circ}\).

Since \(M,L,Q\) are collinear (\(Q\) on \(ML\) produced), \(\angle NLQ=180^{\circ}-\angle NLM=180^{\circ}-80^{\circ}=100^{\circ}\).

In \(\triangle LNQ\): \(\angle LQN=180^{\circ}-\angle LNQ-\angle NLQ=180^{\circ}-63^{\circ}-100^{\circ}=\mathbf{17^{\circ}}.\)

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(a) \(\sin\alpha=0.3907\Rightarrow\alpha=23^{\circ}\).

• (i) \(\tan\alpha=\tan23^{\circ}\approx0.4245.\)
• (ii) \(\cos\alpha=\cos23^{\circ}\approx0.9205\) (also \(\cos\alpha=\sqrt{1-0.3907^{2}}=\sqrt{0.8474}=0.9205\)).

(b) The ladder reaches a height up the wall of \[h=4.5\sin50^{\circ}=4.5\times0.7660=3.447\text{ m}.\] The bottom of the window is \(4\text{ m}\) above the ground, so the gap between the top of the ladder and the window bottom is \[4-3.447=0.553\text{ m}\approx55\text{ cm}.\]

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