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Question: \[\frac{\sin 70{^\circ} + \cos 40{^\circ}}{\cos 70{^\circ} + \sin 40{^\circ}} =\]...

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

A

1

B

13\frac{1}{\sqrt{3}}

C

3\sqrt{3}

D

12\frac{1}{2}

Answer

3\sqrt{3}

Explanation

Solution

sin⁡70o+cos⁡40ocos⁡70o+sin⁡40o\frac{\sin 70^{o} + \cos 40^{o}}{\cos 70^{o} + \sin 40^{o}}

=sin⁡70∘+sin⁡50∘sin⁡20∘+sin⁡40∘=2sin⁡60∘cos⁡10∘2sin⁡30∘cos⁡(−10∘)= \frac{\sin 70{^\circ} + \sin 50{^\circ}}{\sin 20{^\circ} + \sin 40{^\circ}} = \frac{2\sin 60{^\circ}\cos 10{^\circ}}{2\sin 30{^\circ}\cos( - 10{^\circ})}

=sin⁡60osin⁡30o=32.21=3= \frac{\sin 60^{o}}{\sin 30^{o}} = \frac{\sqrt{3}}{2}.\frac{2}{1} = \sqrt{3}.