(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 \(0° \leq \theta \leq 90°\).
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)