MHB Significant figures: Law of cosines

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In the discussion about calculating the length |AB| in a right triangle with sides of 735 m and 420 m, and an angle of 50°, the calculated length is approximately 565.48 m. The main concern revolves around determining the appropriate number of significant figures for the answer, given the ambiguity in the significant figures of the measurements. It is suggested that the angle 50° has the least significant digits, leading to the conclusion that the answer should be reported as 565 m to reflect this uncertainty. The conversation also touches on the idea that for simple trigonometry problems, the precision of the answer may not require extensive error analysis. Ultimately, the consensus leans towards using two significant digits, with the final answer being 565 m.
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We have a right triangle with sides of length 735 and 420 m where the intermediate angle of aforementioned sides is 50°. The task is simple: Calculate |AB|. Here is a picture:

View attachment 1024

Upon calculating |AB|, we arrive at approximately 565.48 m. My problem is how many significant figures one ought to have in the answer. General practice is to have as many digits as the given data with least amount of significant digits.

But in this case, how many digits can we view the data 420 and 50 respectively to have? Integral values with zeros preceding the decimal point can at times be quite ambiguous. It would not make any sense to say "9 000 000" has seven significant digits since it is given without context.

So, 420 either has two significant digits or three; 50 either has one significant digit or two. 565.48 with three significant digits is 565, two significant digits 570 and one significant digit 600. My intuition tells me to answer 565, but I am really not sure which cases of significant digits to confidently rule out.
 

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sweatingbear said:
We have a right triangle with sides of length 735 and 420 m where the intermediate angle of aforementioned sides is 50°. The task is simple: Calculate |AB|. Here is a picture:

View attachment 1024

Upon calculating |AB|, we arrive at approximately 565.48 m. My problem is how many significant figures one ought to have in the answer. General practice is to have as many digits as the given data with least amount of significant digits.

But in this case, how many digits can we view the data 420 and 50 respectively to have? Integral values with zeros preceding the decimal point can at times be quite ambiguous. It would not make any sense to say "9 000 000" has seven significant digits since it is given without context.

So, 420 either has two significant digits or three; 50 either has one significant digit or two. 565.48 with three significant digits is 565, two significant digits 570 and one significant digit 600. My intuition tells me to answer 565, but I am really not sure which cases of significant digits to confidently rule out.

I suggest to err on the side of caution.
The least significant is the $50^\circ$, which without any extra information we should interpret as $50.0 \pm 0.5^\circ$.

As a result the answer would be approximately $565 \pm 5\text{ m}$.
To err on the side of caution, I would indeed write this down as $565\text{ m}$, which leaves the actual precision still somewhat ambiguous, but at least not as less than the original measurements.

Note that in actual lab work (where precision is important), the standard error of each measurement is recorded.
The way the errors propagate is analyzed and the final results are reported with a specification of the expected error.
If we assume errors of $\pm 0.5\text{ m}$ respectively $\pm 0.5^\circ$, analysis shows that the final error would be $\pm 3.7\text{ m}$.
 
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Thanks for the reply.

Frankly the problem is merely a simple trigonometry problem and no such where the considering of measurement uncertainty, specifying error bounds or similar courses of actions are necessary to heed (although debatable, but I digress).

So you would take it that 50 is the data with least significant digits? All right I am with you on that, but the significant figures are ambiguous. One significant digit is, in my eyes, an exaggerated (and rather erroneous) approximation of the length. It seems two significant digits is the way to go, but my gut still would have wanted me to answer 565 as opposed to 570.
 
sweatingbear said:
Thanks for the reply.

Frankly the problem is merely a simple trigonometry problem and no such where the considering of measurement uncertainty, specifying error bounds or similar courses of actions are necessary to heed (although debatable, but I digress).

So you would take it that 50 is the data with least significant digits? All right I am with you on that, but the significant figures are ambiguous. One significant digit is, in my eyes, an exaggerated (and rather erroneous) approximation of the length. It seems two significant digits is the way to go, but my gut still would have wanted me to answer 565 as opposed to 570.

For merely a simple trigonometry problem, it's not really important.
So go with your gut.

Either way, it is ambiguous what the precision of either 565 or 570 is.
To properly specify a precision of 2 digits without going into specifying error bounds, you're supposed to write $5.7 \cdot 10^2\text{ m}$.
But yeah, for a simple trigonometry problem that is over the top.
 
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