Question

Determine the percent uncertainty in$\theta$,and in$\sin \theta$, when (a)$\theta =15°\pm 0.5°$, (b)$\theta =75°\pm 0.5°$.

Short answer

The values of percent uncertainty in $\theta$and $\sin \theta ,$respectively, are:

(a) 3% and 3%

(b) 0.7% and 0.2%

Step-by-step solution

Step 1. (a) Determine the uncertainty in θ

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:

$P=\left(\frac{\Delta \theta }{\theta }\right)\times 100$

Here, $\Delta \theta$is the change in angle.

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

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

Step 2. (a) Calculate the percent uncertainty in sinθ

The change in sine angle is calculated as:

$\begin{array}{l}\Delta \sin \theta =\sin 15.5°-\sin 15°\\ \Delta \sin \theta =0.267-0.259\\ \Delta \sin \theta =0.0084\end{array}$

The relation to calculate the uncertainty is given by:

$P_0=\left(\frac{\Delta \sin \theta }{\sin \theta }\right)\times 100$

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

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

Thus, $P=3.33\%$and $P_0=3.08\%$are the required percent uncertainties for $\theta =15°\pm 0.5°$.

Step 3. (b) Determine the percent uncertainty in θ

The relation to calculate the uncertainty is given by:

$P=\left(\frac{\Delta \theta }{\theta }\right)\times 100$

Here, $\Delta \theta$is the change in angle.

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

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

Step 4. (b) Calculate the percent uncertainty in sinθ

The change in sine angle is calculated as:

$\begin{array}{l}\Delta \sin \theta =\sin 75.5°-\sin 75°\\ \Delta \sin \theta =0.968-0.966\\ \Delta \sin \theta =2\times {10}^{-3}\end{array}$

The relation to calculate the uncertainty is given by:

$P_{\mathrm{sin\theta }}=\left(\frac{\Delta \sin \theta }{\sin \theta }\right)\times 100$

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

$\begin{array}{l}{P}_{P_{\mathrm{sin\theta }}}=\left(\frac{2\times {10}^{-3}}{0.966}\right)\times 100\\ P_{P_{\mathrm{sin\theta }}}=0.207\%\\ P_{P_{\mathrm{sin\theta }}}\approx 0.2\%\end{array}$