Question 25

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waecmaths question: 

Given that $\sin {{60}^{\circ }}=\tfrac{\sqrt{3}}{2}\text{ and }\cos {{60}^{\circ }}=\tfrac{1}{2}$ evaluate $\frac{1-\sin {{60}^{\circ }}}{1+\cos {{60}^{\circ }}}$

Option A: 

$\frac{2+\sqrt{3}}{3}$

Option B: 

$\frac{1-\sqrt{3}}{3}$

Option C: 

$\frac{1+\sqrt{3}}{3}$

Option D: 

$\frac{2-\sqrt{3}}{3}$

waecmaths solution: 

$\frac{1-\sin {{60}^{\circ }}}{1+\cos {{60}^{\circ }}}=\frac{1-\tfrac{\sqrt{3}}{2}}{1+\tfrac{1}{2}}=\frac{\tfrac{2-\sqrt{3}}{2}}{\tfrac{2+1}{2}}=\frac{2-\sqrt{3}}{3}$

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