91Ó°ÊÓ

The percentage of uncertainty is the calculation of a quantity is given by a change in the value of the quantity to the original value of the quantity.

The percentage of uncertainty in theta will be the change in the value of theta to its original value.

The relation to calculate the uncertainty is given by:

$P=\left(\frac{\mathrm{Δ}\mathrm{θ}}{\mathrm{θ}}\right)×100$

Here, $\mathrm{Δ}\mathrm{θ}$is the change in angle.

On plugging the values in the above relation, you get:

$\begin{array}{l}P=\left(\frac{0.5°}{15°}\right)×100\\ P=3.33\%\\ P≈3\%\end{array}$

The change in sine angle is calculated as:

$\begin{array}{l}\mathrm{Δ}\sin \mathrm{θ}=\sin 15.5°-\sin 15°\\ \mathrm{Δ}\sin \mathrm{θ}=0.267-0.259\\ \mathrm{Δ}\sin \mathrm{θ}=0.0084\end{array}$

The relation to calculate the uncertainty is given by:

$P_0=\left(\frac{\mathrm{Δ}\sin \mathrm{θ}}{\sin \mathrm{θ}}\right)×100$

On plugging the values in the above relation, you get:

$\begin{array}{l}{P}_0=\left(\frac{0.008}{0.259}\right)×100\\ P_0=3.08\%\\ P_0≈3\%\end{array}$

Thus, $P=3.33\%$and $P_0=3.08\%$are the required percent uncertainties for $\mathrm{θ}=15°Â\pm 0.5°$.

The relation to calculate the uncertainty is given by:

$P=\left(\frac{\mathrm{Δ}\mathrm{θ}}{\mathrm{θ}}\right)×100$

Here, $\mathrm{Δ}\mathrm{θ}$is the change in angle.

On plugging the values in the above relation, you get:

$\begin{array}{l}{P}_{\mathrm{θ}}=\left(\frac{0.5°}{75°}\right)×100\\ P_{\mathrm{θ}}=0.66\%\\ P_{\mathrm{θ}}≈0.7\%\end{array}$

The change in sine angle is calculated as:

$\begin{array}{l}\mathrm{Δ}\sin \mathrm{θ}=\sin 75.5°-\sin 75°\\ \mathrm{Δ}\sin \mathrm{θ}=0.968-0.966\\ \mathrm{Δ}\sin \mathrm{θ}=2×{10}^{-3}\end{array}$

The relation to calculate the uncertainty is given by:

$P_{\mathrm{²õ¾\pm ²Ôθ}}=\left(\frac{\mathrm{Δ}\sin \mathrm{θ}}{\sin \mathrm{θ}}\right)×100$

On plugging the values in the above relation, you get:

$\begin{array}{l}{P}_{P_{\mathrm{²õ¾\pm ²Ôθ}}}=\left(\frac{2×{10}^{-3}}{0.966}\right)×100\\ P_{P_{\mathrm{²õ¾\pm ²Ôθ}}}=0.207\%\\ P_{P_{\mathrm{²õ¾\pm ²Ôθ}}}≈0.2\%\end{array}$