(a) Using mathematical tables, find ; (i) \(2 \sin 63.35°\) ; (ii) \(\log \cos 44.74°\); (b) Find the value of K given that \(\log K - \log (K - 2) = \log 5...

Assessment: WAEC SSCE - General Mathematics - 1995 (Essay) Subject: General Mathematics

Question 1 Report

(a) Using mathematical tables, find ; (i) \(2 \sin 63.35°\) ; (ii) \(\log \cos 44.74°\);

(b) Find the value of K given that \(\log K - \log (K - 2) = \log 5\);

(c) Use logarithm tables to evaluate \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\)

Answer Details

(a)(i) From tables, \(\sin 63.35° = 0.8937\).

\(2\sin 63.35° = 2 \times 0.8937 = \mathbf{1.787}\)

(ii) \(\cos 44.74° = 0.7103\).

\(\log \cos 44.74° = \log 0.7103 = \bar{1}.8515 \;(= -0.1485)\)

(b) \(\log K - \log(K-2) = \log 5\)

\(\log\!\left(\dfrac{K}{K-2}\right) = \log 5 \ \Rightarrow\ \dfrac{K}{K-2} = 5\)

\(K = 5(K-2) = 5K - 10 \ \Rightarrow\ 4K = 10 \ \Rightarrow\ \mathbf{K = 2.5}\)

(c) Evaluate \(\dfrac{(3.68)^2 \times 6.705}{\sqrt{0.3581}}\) using logarithms.

ExpressionLog
\((3.68)^2\)\(2 \times 0.5658 = 1.1316\)
\(6.705\)\(0.8264\)
Numerator total\(1.9580\)
\(\sqrt{0.3581}\)\(\tfrac{1}{2}(\bar{1}.5540) = \bar{1}.7770\)

\(\text{Log of answer} = 1.9580 - \bar{1}.7770 = 1.9580 + 0.2230 = 2.1810\)

Antilog \(2.1810 = \mathbf{151.7}\) (4 s.f.).

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