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

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Aicharya · Accepted Answer

Let $\mathrm{f}(x)=\sin x$ $	herefore \quad \mathrm{f}^{\prime}(x)=\cos x$ Here, $\mathrm{a}=60^{\circ}$ and $\begin{aligned} & \mathrm{h}=10^{\prime \prime}=\left(\frac{1}{360}ight)^0=\frac{1}{360} 	imes 0.0175^{\mathrm{c}}=0.000049^{\mathrm{c}} \ & f(a)=\sin \left(60^{\circ}ight)=\frac{\sqrt{3}}{2}=\frac{1.732}{2}=0.866 \ & \mathrm{f}^{\prime}(\mathrm{a})=\cos (60)=\frac{1}{2}=0.5 \ & 	herefore \quad \mathrm{f}(\mathrm{a}+\mathrm{h}) \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \ & 	herefore \quad \sin \left(60^{\circ} 0^{\prime} 10^{\prime \prime} \approx 0.866+0.000049 	imes 0.5ight. \ & \approx 0.866024 \ & \end{aligned}$

The approximate value of $\sin \left(60^{\circ} 0^{\prime} 10^{\prime \prime}\right)$ is (given that…

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