Help!

Help!

Suppose the width of your index finger is 0.6 inches and the length of your arm is 31.9 inches. Based on these measurements, what will be the angular width (in degrees) of your index finder held at arm’s length?

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\( IF=\frac{L\cdot \pi}{180°}\cdot \alpha \\ \alpha =\frac{IF\cdot180°}{L\cdot\pi}=\frac{0.6''\cdot180°}{31.9''\cdot\pi}\)

\(\color{black} \alpha =1.07766356451°=1°\ 4'\ 39.5888''\)   incorrect

\(sin(\frac{\alpha}{2})=\frac{IF}{2\cdot L}=\frac{0.6''}{2\cdot 31.9''}\\ \frac{\alpha}{2}=arc sin(\frac{0.6}{2\cdot 31.9})\\ \alpha=2\cdot arc sin(\frac{0.6}{2\cdot 31.9})\\ \)

\(\alpha=1.07767945°=1°\ 4'\ 39.646'' \)      correct

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Suppose the width of your index finger is 0.6 inches and the length of your arm is 31.9 inches.

Based on these measurements, what will be the angular width (in degrees) of your index finder held at arm’s length?

\(\begin{array}{|rcll|} \hline \mathbf{\tan{\dfrac{\alpha}{2}} } &=& \mathbf{\dfrac{\dfrac{0.6}{2}}{31.9} } \\ \\ \tan{\dfrac{\alpha}{2}} &=& \dfrac{0.3}{31.9} \\ \\ \tan{\dfrac{\alpha}{2}} &=& 0.00940438871 \\ \\ \dfrac{\alpha}{2} &=& \arctan{ 0.00940438871 } \\\\ \dfrac{\alpha}{2} &=& 0.53881589788 \\ \\ \alpha &=& 2\cdot 0.53881589788 \\ \mathbf{\alpha} &=& \mathbf{1.07763179577^\circ} \\ \hline \end{array}\)