(a) If logo a = 1.3010 and log\(_{10}\)b - 1.4771.... | WAEC SSCE - General Mathematics - 2019 (Essay)

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(a) If logo a = 1.3010 and log\(_{10}\)b - 1.4771. find the value of ab

(b)

In the diagram. O is the centre of the circle,< ACB = 39\(^o\) and < CBE = 62\(^o\). Find: (i) the interior angle AOC;

(ii) angle BAC.

Answer

Answer Details

(a) Finding \(ab\)

Given \(\log_{10} a = 1.3010\) and \(\log_{10} b = 1.4771\). Using the law \(\log(ab) = \log a + \log b\):

\[\log_{10}(ab) = 1.3010 + 1.4771 = 2.7781\]

Taking the antilogarithm,

\[ab = 10^{2.7781} = 600\]

(As a check, \(a = 10^{1.3010} = 20\) and \(b = 10^{1.4771} = 30\), so \(ab = 600\).) Hence \(ab = 600\).

(b) Circle with centre \(O\)

From the diagram, \(A\), \(B\), \(C\) lie on the circle, \(O\) is the centre, \(ABE\) is a straight line, \(\angle ACB = 39^\circ\) and \(\angle CBE = 62^\circ\).

(i) Interior angle \(AOC\)

Since \(ABE\) is a straight line, \(\angle ABC\) is the supplement of \(\angle CBE\):

\[\angle ABC = 180^\circ - 62^\circ = 118^\circ\]

The inscribed angle \(\angle ABC\) stands on chord \(AC\). The angle subtended at the centre by the same chord (the reflex angle at \(O\), on the opposite side of \(AC\) from \(B\)) is twice the inscribed angle:

\[\text{reflex } \angle AOC = 2 \times 118^\circ = 236^\circ\]

Therefore the required interior (non-reflex) angle \(AOC\) is

\[\angle AOC = 360^\circ - 236^\circ = \boxed{124^\circ}\]

(ii) Angle \(BAC\)

In triangle \(ABC\), the three angles sum to \(180^\circ\), with \(\angle ACB = 39^\circ\) and \(\angle ABC = 118^\circ\):

\[\angle BAC = 180^\circ - 118^\circ - 39^\circ = \boxed{23^\circ}\]

Hence \(\angle AOC = 124^\circ\) and \(\angle BAC = 23^\circ\).

Read lesson note on Circles (WAEC)

Read lesson note on Logarithms (WAEC)

Circles

Logarithms

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