Why are reference angles necessary in trig

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Reference angles in trigonometry are essential for simplifying calculations by providing the acute angle corresponding to any given angle. They are defined as the difference between the angle and the nearest multiple of 180° (π), ensuring they always fall between 0° and 90°. Trigonometric tables traditionally only include values for angles up to 90°, making reference angles a practical tool for finding sine, cosine, and tangent values for larger angles. This method was particularly useful before calculators, as it allowed for easier lookup and calculation. Overall, while not strictly necessary, reference angles greatly enhance convenience in trigonometric computations.
Luke77
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Why are reference angles necessary in trigonometry. I understand they are the acute version of an obtuse angle.
 
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Hi Luke77! :smile:
Luke77 said:
Why are reference angles necessary in trigonometry.

I've never come across them before. :confused:

Apparently, they're the difference between the angle and the nearest multiple of 180° (π).

(so they're always between 0° and 90°)

In other words, trig tables only give you measurements up to 90°, so you find the "reference angle", look up the trig of that in the trig table, and if necessary multiply by minus-one. :smile:

(see http://www.mathwarehouse.com/trigonometry/reference-angle/finding-reference-angle.php for some examples)
 
Reference angles aren't "necessary". They are convenient, especially back in the years 'BC' (Before Calculators) when you had to look up trig functions in tables. Rather than have tables that extended to, say, 135 degrees, 225 degrees, and 315 degrees, they just went from 0 to 90 degrees so you could look up the "reference angle", 45 degrees, for all those and then apply the correct sign yourself.
 
Thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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