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

Let \(\mathrm{f}(x)=\sin x\)
\(\therefore \mathrm{f}^{\prime}(x)=\cos x\)
Here, \(\mathrm{a}=60^{\circ}\) and
\(\mathrm{h}=10^{\prime \prime}=\left(\frac{1}{360}\right)^0=\frac{1}{360} \times 0.0175^{\mathrm{c}}=0.000049^{\mathrm{c}} \)
\( f(a)=\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}=\frac{1.732}{2}=0.866 \)
\(\mathrm{f}^{\prime}(\mathrm{a})=\cos (60)=\frac{1}{2}=0.5 \)
\( \therefore \mathrm{f}(\mathrm{a}+\mathrm{h}) \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \)
\(\therefore \sin \left(60^{\circ} 0^{\prime} 10^{\prime \prime} \approx 0.866+0.000049 \times 0.5\right. \)
\(\approx 0.866024\)