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Given \(\sin \theta =  \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°\), find \(\tan 2\theta\) in surd...

Further Mathematics
WAEC 2009

Given \(\sin \theta =  \frac{\sqrt{3}}{2}, 0° \leq \theta \leq 90°\), find \(\tan 2\theta\) in surd form.

  • A. \(- \sqrt{3}\)
  • B. \(-\frac{\sqrt{3}}{2}\)
  • C. \(\frac{\sqrt{3}}{2}\)
  • D. \(\sqrt{3}\)
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Correct Answer: Option A
Explanation

\(\sin \theta = \frac{\sqrt{3}}{2} \implies opp = \sqrt{3}; hyp = 2\)

\(adj^{2} = 2^{2} - (\sqrt{3})^{2} = 1 \implies adj = 1\)

\(\cos \theta = \frac{1}{2}\)

\(\sin 2\theta = \sin (180 - \theta) = \sin \theta = \frac{\sqrt{3}}{2}\)

\(\cos 2\theta = \cos (180 - \theta) = -\cos \theta = -\frac{1}{2}\)

\(\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\)

= \(- \sqrt{3}\)


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Join your school's WhatsApp group
NECO June/July 2024 - Get offline past questions & answers - Download objective & theory, all in one app 48789
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
NECO June/July 2024 - Get offline past questions & answers - Download objective & theory, all in one app 48789
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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts