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Question

(B) The interior angle $	heta$ of a regular pentagon is given by $	heta = \frac{(n-2) 	imes 180^{\circ}}{n}$,where $n=5$. So,$	heta = \frac{(5-2)

Vedclass · Accepted Answer

$|\operatorname{cosec} 18^{\circ}|$

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(B) The interior angle $	heta$ of a regular pentagon is given by $	heta = \frac{(n-2) 	imes 180^{\circ}}{n}$,where $n=5$. So,$	heta = \frac{(5-2) 	imes 180^{\circ}}{5} = \frac{3 	imes 180^{\circ}}{5} = 108^{\circ}$. The magnitude of the vector $\vec{v} = (\sin 	heta) \hat{i} + (\cos 	heta) \hat{j} + (	an 	heta) \hat{k}$ is $|\vec{v}| = \sqrt{\sin^2 	heta + \cos^2 	heta + 	an^2 	heta}$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $|\vec{v}| = \sqrt{1 + 	an^2 	heta} = \sqrt{\sec^2 	heta} = |\sec 	heta|$. Substituting $	heta = 108^{\circ}$,we get $|\sec 108^{\circ}|$. Since $\sec 108^{\circ} = \sec(180^{\circ} - 72^{\circ}) = -\sec 72^{\circ} = -\operatorname{cosec} 18^{\circ}$,the magnitude is $|-\operatorname{cosec} 18^{\circ}| = |\operatorname{cosec} 18^{\circ}|$. Thus,the correct option is $B$.

If $\theta$ is the interior angle of a regular pentagon,then $|(\sin \theta) \hat{i}+(\cos \theta) \hat{j}+(\tan \theta) \hat{k}|=$

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