Finding time difference between two arriving wave fronts

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member 731016
Homework Statement
Please see below
Relevant Equations
S = Rθ
For part(b),
Diagram of a sector of Earth with points A and B on the surface at 60.0 degrees apart, showing a P wave path through the body and a Rayleigh wave path along the surface, labeled Figure P16.2.

The solution is,
Worked solution, part (b): P-wave path length 2R sin 30.0° = 2(6.37 × 10^6 m) sin 30.0° = 6.37 × 10^6 m, circled in red; then times at speeds 7 800 m/s and 4 500 m/s, and the time difference 666 s = 11.1 min.

However, where did they get the formula shown in red from?

Many thanks!
 
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  • Worked solution calculating P-wave travel time of about 817 s over 6.37 × 10^6 m at 7800 m/s, and Rayleigh-wave travel time of about 1482 s over 6.67 × 10^6 m at 4500 m/s, giving a time difference of 666 s = 11.1 min.
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##AB=2R\sin30^o.## Do you see why? Hint: Draw a perpendicular from the center to the base AB. It splits AB into two equal parts. What is the length of each part?
 
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kuruman said:
##AB=2R\sin30^o.## Do you see why? Hint: Draw a perpendicular from the center to the base AB. It splits AB into two equal parts. What is the length of each part?
Thanks I see it now @kuruman ! Is the reason why they used radians instead of degrees in the arc length formula is because radians is a ratio of the length while degrees is not.

Many thanks!
 
What you call the arc length formula is actually the definition of the angle as the ratio of the arc length to the radius. As such it has no dimensions.
 
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