A) \[{{120}^{o}}\]and \[{{300}^{o}}\]
B) \[{{60}^{o}}\]and \[{{120}^{o}}\]
C) \[{{120}^{o}}\]and \[{{240}^{o}}\]
D) \[{{60}^{o}}\]and \[{{240}^{o}}\]
Correct Answer: C
Solution :
Given, \[\cos \theta =\frac{-1}{2}\] and \[{{0}^{o}}<\theta <{{360}^{o}}\]. We know that \[\cos {{60}^{o}}=\frac{1}{2}\] and \[\cos ({{180}^{o}}-{{60}^{o}})\] \[=-\cos {{60}^{o}}=-\frac{1}{2}\] or \[\cos {{120}^{o}}=-\frac{1}{2}\]. Similarly \[\cos ({{180}^{o}}+{{60}^{o}})\] \[=-\cos {{60}^{o}}=-\frac{1}{2}\] or \[\cos {{240}^{o}}=-\frac{1}{2}.\] Therefore \[\theta ={{120}^{o}}\] and \[{{240}^{o}}\].
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