Find Coordinates (x;y) with only the angle, (0;0), and the length of the hypotenuse

[Koalanet21]

My very basic question is : What is the formula to find the coordinates (x;y) (noted ?;?) using the angle, (0;0), and the length of the hypotenuse?

 

[tkhunny]

Do you have a definition for sine and cosine? This will solve the problem.

Oh, also, 90 - 42 = 48. You MAY need that.

 

[Koalanet21]

The problem is that I need to work with coordinates and I don't fully know how to do that. So, with an angle of 48° and an hypotenuse of 1, starting from (0;0), what are going be x and y at the "top" of the hypotenuse? (?;?) Which formula is needed?

 

[pka]

Hint: what do these have to do with it?
\(\left\{ \begin{gathered}
x = \cos (48^\circ ) \hfill \\
y = \sin (48^\circ ) \hfill \\
\end{gathered} \right.\)

 

[tkhunny]

Asked and answered. Sine and Cosine of the correct angle.

My trouble, here, is that you ALREADY KNOW the length of the hypotenuse. Why are you still asking for it?

 

[Koalanet21]

Thank you! And if the hypotenuse was, let's say, 2.6, what would be the coordinates x and y?

 

[pka]

Thank you! And if the hypotenuse was, let's say, 2.6, what would be the coordinates x and y?

Your class should know that a circle centred at the origin with radius \(r\) is \((r\cos(t),\sin(t))\). Do you have that in the notes?
Now if the radius were \(2.6\) and the angle were \(\dfrac{4\pi}{5}\) then the point is \(\left( {2.6\cos \left( {\frac{{4\pi }}{5}} \right),2.6\sin \left( {\frac{{4\pi }}{5}} \right)} \right)\)

 

[tkhunny]

H = Length of Hypotenuse
x = measure of desired angle

H*sin(x) = length of one side. (You have to decide which by selecting the correct angle.)
H*cos(x) = length of other side.

The original Problem has H = 1.
You have asked, What if H = 2.6?

 

[Steven G]

The point that you want for now is (x,y). So you need to find x and y. Well you were told that x = cos(48) and y = sin(48). hence a quick calculation will give you x and y. Now do you understand why x = cos(48) and y = sin(48)?