(a) Simplify : \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)
(b) Given that \(\sin x = \frac{2}{3}\), evaluate, leaving your answer in surd form and without using tables or calculator, \(\tan x - \cos x\).
(b) Given that \(\sin x = \frac{2}{3}\), evaluate, leaving your answer in surd form and without using tables or calculator, \(\tan x - \cos x\).
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Correct Answer: Option n
Explanation:
(a) \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)
\(1\frac{1}{4} + \frac{7}{9} = \frac{5}{4} + \frac{7}{9}\)
= \(\frac{45 + 28}{36}\)
= \(\frac{73}{36}\)
\(1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64} = \frac{13}{9} - (\frac{8}{3} \times \frac{9}{64})\)
= \(\frac{13}{9} - \frac{3}{8}\)
= \(\frac{104 - 27}{72}\)
= \(\frac{77}{72}\)
\(\therefore \frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}} = \frac{73}{36} \div \frac{77}{72}\)
= \(\frac{73}{36} \times \frac{72}{77}\)
= \(\frac{146}{77}\)
= \(1\frac{69}{77}\)
(b) \(\sin x = \frac{2}{3}\)
In \(\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\)
\(\therefore |BC|^{2} = 9 - 4 = 5\)
\(|BC| = \sqrt{5}\)
\(\tan x = \frac{2}{\sqrt{5}} ; \cos x = \frac{\sqrt{5}}{3}\)
\(\therefore \tan x - \cos x = \frac{2}{\sqrt{5}} - \frac{\sqrt{5}}{3}\)
= \(\frac{6 - 5}{3\sqrt{5}}\)
= \(\frac{1}{3\sqrt{5}}\)
Rationalizing, we have
\(\frac{1}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{15}\)
(a) \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)
\(1\frac{1}{4} + \frac{7}{9} = \frac{5}{4} + \frac{7}{9}\)
= \(\frac{45 + 28}{36}\)
= \(\frac{73}{36}\)
\(1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64} = \frac{13}{9} - (\frac{8}{3} \times \frac{9}{64})\)
= \(\frac{13}{9} - \frac{3}{8}\)
= \(\frac{104 - 27}{72}\)
= \(\frac{77}{72}\)
\(\therefore \frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}} = \frac{73}{36} \div \frac{77}{72}\)
= \(\frac{73}{36} \times \frac{72}{77}\)
= \(\frac{146}{77}\)
= \(1\frac{69}{77}\)
(b) \(\sin x = \frac{2}{3}\)
In \(\Delta ABC, 3^{2} = 2^{2} + |BC|^{2}\)
\(\therefore |BC|^{2} = 9 - 4 = 5\)
\(|BC| = \sqrt{5}\)
\(\tan x = \frac{2}{\sqrt{5}} ; \cos x = \frac{\sqrt{5}}{3}\)
\(\therefore \tan x - \cos x = \frac{2}{\sqrt{5}} - \frac{\sqrt{5}}{3}\)
= \(\frac{6 - 5}{3\sqrt{5}}\)
= \(\frac{1}{3\sqrt{5}}\)
Rationalizing, we have
\(\frac{1}{3\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{15}\)