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UPSC NDA Math Model Paper 3

Question : 31 of 120

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cos 20° + cos 100° + cos 140° = ?

Solution:

\Rightarrow \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}

= (\cos 20^{\circ} + \cos 140^{\circ}) + \cos 100^{\circ}

As we know that,

\cos B + \cos C = 2\cos

\frac{B+C}{2} \cos \frac{B-C}{2}

\Rightarrow \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}

= 2 \cos 80^{\circ} \cos 60^{\circ} + \cos 100^{\circ}

\Rightarrow \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}

= \cos 80^{\circ} + \cos 100^{\circ}

As we know that,

\cos B+\cos C=2 \cos \frac{B+C}{2} \cos \frac{B-C}{2}

\Rightarrow \cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}

=2 \cos 90^{\circ} \cos 10^{\circ}

\Rightarrow \cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}

=2 \times 0 \times \cos 10^{\circ}=0

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