Significant figures: Law of cosines

[SweatingBear]

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 like Serena]

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 $\pm 3.7\text{ m}$.

 

[SweatingBear]

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.

 

[I like Serena]

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.

 

[CellCoree]

As a scientist, it is important to use the correct number of significant figures in calculations to accurately reflect the precision of the data. In this case, we can determine the number of significant figures by looking at the given data and determining the least precise measurement.

Since the side lengths of the triangle are given to the nearest meter, we can assume that the data has two significant figures. Therefore, the answer should also have two significant figures. In this case, the answer should be rounded to 570 m.

However, it is always important to consider the context of the data and use your scientific judgment to determine the appropriate number of significant figures. In this case, if the data was given with more precision (e.g. 735.2 and 420.3), then the answer should also reflect that level of precision and have three significant figures.

In conclusion, the answer to this problem should be 570 m, rounded to two significant figures based on the precision of the given data.