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How do you solve for the second forces magnitude? The 17' and 30' threw me off. enter image description here

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Look for example at $29^\circ 17'$. That means $29$ degrees and $17$ minutes. But a minute is one-sixtieth of a degree, so $17$ minutes is $\frac{17}{60}$ degrees, approximately $0.2833333$ degrees. Therefore $29^\circ 17'$ is approximately $29.2833333$ degrees.

Similarly, $76^\circ 30'$ just means $76+\frac{30}{60}$ degrees, that is, $76.5$ degrees.

So in the "normal" formulas that you would use to find the magnitude of the resultant force, just use the appropriate decimal approximation. Some calculators allow direct input in degrees and minutes, and do the conversion to decimal automatically.

In old-fashioned astronomy, there is a finer subdivision still, the second. There are $60$ seconds in a minute. So $1$ second is $\frac{1}{3600}$ degrees. If you are told that an angle is $29^\circ 17'\, 34''$, that means $29+\frac{17}{60}+\frac{34}{3600}$ degrees. You could use a calculator to get a good decimal approximation to this, and then proceed as usual.

Remark: The use of minutes and seconds to measure angles is slowly (too slowly!) fading.

However, you may need to acquire some ability to transform a decimal degree answer, such as $42.372$ degrees, to degrees, minutes, and perhaps even seconds. (It may be that in your answers, angles are expected to be given in degrees, minutes, and even seconds.) Sometimes a hybrid notation is used, dropping seconds, as in $46^\circ 27.4'$.

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  • $\begingroup$ There is some historical value to 'earth-based' units. One second of latitude corresponds to a nautical mile, which makes for convenient estimates. The separation of scales reduces magnitude errors as well. However, as a child, computing in multiple bases was no fun (money had multiple bases too). $\endgroup$
    – copper.hat
    Commented May 25, 2012 at 7:03

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