Maths / Introduction to Trigonometry / Trigonometric ratios for 60 degrees

QUESTION

Find the value of :  $\dpi{120} \fn_jvn \large \frac{cos30^0+sin60^0}{1+sin30^0+cos60^0}$

EXPLANATION
Explain TypeExplanation Content
Text

We know that $cos30^0=\frac{\sqrt{3}}{2},sin60^0=\frac{\sqrt{3}}{2},=\frac{1}{2}$ and $cos60^0=\frac{1}{2}$

$\therefore$       $\frac{cos30^0+sin60^0}{1+sin30^0+cos60^0}=\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}$

$\frac{2\left ( \frac{\sqrt{3}}{2} \right )}{2}=\frac{\sqrt{3}}{2}$

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