Question

$\frac{\sin 70^{\circ }+\cos 40^{\circ }}{\cos 70^{\circ }+\sin 40^{\circ }}$=

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (B)

$\frac{\sin 70^{\circ }+\cos 40^{\circ }}{\cos 70^{\circ }+\sin 40^{\circ }}$
$=\frac{sin\left(90^{\circ }-20^{\circ }\right)+sin\left(90^{\circ }-40^{\circ }\right)}{cos\left(90^{\circ }-20^{\circ }\right)+sin\left(90^{\circ }-50^{\circ }\right)}$
$=\frac{cos20^{\circ }+cos40^{\circ }}{cos70^{\circ }+cos50^{\circ }}$
$=\frac{2cos30^{\circ }cos10^{\circ }}{2cos60^{\circ }cos10^{\circ }}=\frac{2cos30^{\circ }}{2cos60^{\circ }}$
$=\frac{2\left(\sqrt{3}/2\right)}{2\left(\frac{1}{2}\right)}=\sqrt{3}$