WAEC SSCE - General Mathematics - 1990 (Essay)

Answer Details

(a) The angles of a triangle sum to two right angles.

Let triangle ABC have interior angles at A, B and C. Through C draw a line XY parallel to AB, with X and Y on opposite sides of C.

• \(\angle XCA = \angle CAB\) (alternate angles, XY \(\parallel\) AB, transversal CA).
• \(\angle YCB = \angle CBA\) (alternate angles, XY \(\parallel\) AB, transversal CB).

The angles on the straight line XY at C add to two right angles:

\[ \angle XCA + \angle ACB + \angle YCB = 180^\circ. \]

Replacing the two alternate angles gives

\[ \angle CAB + \angle ACB + \angle CBA = 180^\circ, \]

so the three interior angles of the triangle sum to two right angles. \(\blacksquare\)

(b) Let \(\angle ACE = \angle ECB = c\) (CE bisects \(\angle ACB\)), and write \(\angle CAB = A\) and \(\angle CBA = B\).

\(\angle CAE = A\). The exterior angle at B, \(\angle CBD\), equals the sum of the two remote interior angles:

\[ \angle CBD = A + \angle ACB = A + 2c. \]

Therefore

\[ \angle CAE + \angle CBD = A + (A + 2c) = 2A + 2c. \tag{1} \]

Now \(\angle CEB\) is the exterior angle of triangle ACE at E, so it equals the sum of the two remote interior angles \(\angle CAE\) and \(\angle ACE\):

\[ \angle CEB = A + c \implies 2\,\angle CEB = 2A + 2c. \tag{2} \]

From (1) and (2),

\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]

Answer Details

(a) Let \(H\), \(I\) and \(Y\) represent the sets of students who speak Hausa, Igbo and Yoruba respectively. Since the given pairwise intersections include those who speak all three languages, the Venn diagram is:

(b) Since every student speaks at least one of the languages,

\[n(H\cup I\cup Y)=40.\]

Using the inclusion-exclusion principle,

\[40=25+16+21-8-11-6+x\]

\[40=37+x\]

\[x=3.\]

Therefore, 3 students speak all three languages.

Hence the numerical entries in the Venn diagram are:

\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]

\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]

Answer Details

Rewrite the inequalities in terms of y:

Draw the three boundary lines as solid lines, since equality is included. Their pairwise intersections are:

The required solution set is the common region below \(y=2x+4\) and above both \(y=2-\frac{x}{3}\) and \(y=7x-9\). It is the shaded triangular region shown.

Hence, the solution region is the closed triangle with vertices

Answer Details

None

Answer Details

Take \(R = 6400\) km and \(\pi = 3.142\).

(a) Distance P to Q along the parallel of latitude \(40^\circ\)N. The longitude difference is \(38^\circ\text{E} + 22^\circ\text{W} = 60^\circ\). The radius of the parallel is \(R\cos 40^\circ\), so

\[ PQ = \frac{60}{360} \times 2\pi R \cos 40^\circ = \frac{1}{6} \times 2 \times 3.142 \times 6400 \times 0.7660 = 5134\ \text{km}. \]

(b) Distance Q to T along the meridian \(22^\circ\)W. The latitude difference is \(40^\circ - 28^\circ = 12^\circ\), and along a meridian the radius is \(R\):

\[ QT = \frac{12}{360} \times 2\pi R = \frac{1}{30} \times 2 \times 3.142 \times 6400 = 1340.6\ \text{km}. \]

(c) Average speed from P to T via Q over 12 hours. Total distance:

\[ PQ + QT = 5134 + 1340.6 = 6474.6\ \text{km}. \] \[ \text{Average speed} = \frac{6474.6}{12} = 539.6 \approx 540\ \text{km/h (to 3 s.f.).} \]

Answer Details

(a) Volume of a right circular cone \(= \dfrac{1}{3}\pi r^2 h\), with \(r = 4\) cm, \(h = 6\) cm, \(\pi = 3.142\):

\[ V = \frac{1}{3} \times 3.142 \times 4^2 \times 6 = \frac{1}{3} \times 3.142 \times 16 \times 6 = 3.142 \times 32 = 100.544\ \text{cm}^3. \]

\(V \approx 100.5\ \text{cm}^3\).

(b) The tank is a hemisphere of diameter 10 m, so radius \(R = 5\) m.

\[ \text{Volume of tank} = \frac{2}{3}\pi R^3 = \frac{2}{3} \times 3.142 \times 5^3 = \frac{2}{3} \times 3.142 \times 125 = 261.83\ \text{m}^3. \]

The pipe has radius \(20\ \text{cm} = 0.2\ \text{m}\) and water flows at 2 m/s. Volume delivered per second:

\[ \text{Flow rate} = \pi (0.2)^2 \times 2 = 3.142 \times 0.04 \times 2 = 0.25136\ \text{m}^3/\text{s}. \]

Time to fill:

\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]

Answer Details

Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.

Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with

\[ NT = 7 - 3 = 4\ \text{m}. \]

The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:

(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]

(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]

(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where

\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]