Question

If sinθ = $\frac{1}{2}$, then cotθ = ?
(a) $\frac{1}{\sqrt{3}}$
(b) $\sqrt{3}$
(c) $\frac{\sqrt{3}}{2}$
(d) 1

Answer 1

(b) $\sqrt{3}$

Given: sinθ = $\frac{1}{2}$, but sinθ = $\frac{\mathrm{BC}}{\mathrm{AC}}$
So, $\frac{\mathrm{BC}}{\mathrm{AC}}$ = $\frac{1}{2}$
Thus, BC = k and AC = 2k


Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
AB² = AC² $-$ BC²
AB² = (2k)² $-$ (k)²
AB² = 3k²
AB = $\sqrt{3}$k
So, tanθ = $\frac{\mathrm{BC}}{\mathrm{AB}}$ = $\frac{k}{\sqrt{3}k}$ = $\frac{1}{\sqrt{3}}$
∴ cotθ = $\frac{1}{\tan \theta }$ = $\sqrt{3}$