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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?

 Aug 27, 2019
 #1
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+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?

..

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

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

 

laugh  !

 Aug 27, 2019
edited by asinus  Aug 27, 2019
edited by asinus  Aug 27, 2019
 #2
avatar+23041 
+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?

 

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

 

laugh

 Aug 27, 2019
 #3
avatar+102913 
+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

 

 

cool cool cool

CPhill  Aug 27, 2019

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