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2003 WAEC Mathematics Theory (a) Simplify : \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\) (b) Solve the following equations correct to...

Mathematics
WAEC 2003

(a) Simplify : \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\)

(b) Solve the following equations correct to one decimal place. 

(i) \(\tan (\theta + 25)° = 5.145\)

(ii) \(5\cos \theta - 1 = 0\), where \(0° \leq \theta \leq 90°\).

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Explanation

(a) \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\)

= \((5^{4})^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 5^{2}\)

= \(5^{\frac{3}{2}} \times 5^{\frac{1}{2}} \div 5^{2}\)

= \(5^{\frac{3}{2} + \frac{1}{2} - 2}\)

= \(5^{0} = 1\)

(b)(i) \(\tan (\theta + 25)° = 5.145\)

\(\theta + 25 = \tan^{-1} (5.145)\)

\(\implies \theta + 25 = 79°\)

\(\theta  = 79° - 25° = 54°\) 

(ii) \(5 \cos \theta - 1 = 0\)

\(\implies 5 \cos \theta = 1\)

\(\cos \theta = \frac{1}{5} = 0.2\)

\(\theta = \cos^{-1} (0.2) = 78.463°\)

\(\approxeq 78.5°\) (to one d.p)


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