The approximate value of sin(60^{circ} 0^{pri | vedclass

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(C) Let $f(x) = \sin x$. Then $f^{\prime}(x) = \cos x$. Here,$a = 60^{\circ}$ and $h = 10^{\prime \prime}$. Since $1^{\circ} = 60^{\prime}$ and $1^{\p

Vedclass · Accepted Answer

$0.8660243$

Vedclass · Answer

(C) Let $f(x) = \sin x$. Then $f^{\prime}(x) = \cos x$. Here,$a = 60^{\circ}$ and $h = 10^{\prime \prime}$. Since $1^{\circ} = 60^{\prime}$ and $1^{\prime} = 60^{\prime \prime}$,we have $1^{\circ} = 3600^{\prime \prime}$. Thus,$h = \frac{10}{3600}^{\circ} = \frac{1}{360}^{\circ}$. Converting $h$ to radians: $h = \frac{1}{360} 	imes 0.0175^{c} \approx 0.0000486^{c}$. Now,$f(a) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2} = \frac{1.732}{2} = 0.866$. And $f^{\prime}(a) = \cos(60^{\circ}) = 0.5$. Using the linear approximation formula $f(a+h) \approx f(a) + h f^{\prime}(a)$: $\sin(60^{\circ} 0^{\prime} 10^{\prime \prime}) \approx 0.866 + (0.0000486 	imes 0.5)$. $\approx 0.866 + 0.0000243 = 0.8660243$.

The approximate value of $\sin(60^{\circ} 0^{\prime} 10^{\prime \prime})$ is (given that $\sqrt{3}=1.732, 1^{\circ}=0.0175^{c}$):

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