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?

Guest Aug 27, 2019

#1**+2 **

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**

!

Guest Aug 27, 2019

#2**+3 **

**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?**

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

heureka Aug 27, 2019

#3**+2 **

Even though the answers don't differ much....heureka's is correct

The first answer would be the angular wirth of a chord of 0.6 units if this chord were in a circle with a radius of 31.9 units......however.....this chord is not quite 31.9 units from the center of the circle....so the angular width is just a little larger than the true measure

We can verify this by using the Law of Cosines

arccos [ (.6^2 - 2(31.9^2))/ ( -2*31.9^2) ] = θ ≈ 1.077679450357°

Heureka's answer moves this chord outside the circle, but still tangent to the circle.....and it will be exactly 31.9 units from the circle's center

CPhill
Aug 27, 2019