The sine, cosine and tangent of 210o are respectively

Mathematics

JAMB 1987

A. \(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
B. \(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{3}\)
C. \(\frac{1}{2}\), \(\frac{\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}\)
D. \(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)

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Correct Answer: Option D

Explanation

210o = 180o - 210o = - 30o

From ratio of sides, sin -30o = -\(\frac{1}{2}\)

Cos 210o = 180o - 210o = -30o

= cos -30o = \(\frac{-3}{2}\)

But tan 30o = \(\frac{1}{\sqrt{3}}\), rationalizing this

= \(\frac{1}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{3}}{3}\)

∴ = \(\frac{-1}{2}\), \(\frac{\sqrt{-3}}{2}\), \(\frac{\sqrt{3}}{3}\)

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