# Sin and cosine calculations are only calculate on acute angles of triangles?

So sine and cosine calculations are only calculate on the acute non-right angles of triangles. Is that correct?

This from math2.org:

Definition 1 Given any angle q (0 £ q £ 90°), we can find the sine or cosine of that angle by constructing a right triangle with one vertex of angle q. The sine is equal to the length of the side opposite to q, divided by the length of the triangle's hypotenuse. The cosine is equal to the length of the side adjacent to q, divided by the length of the triangle's hypotenuse. In this way, we can find the sine or cosine of any q in the range 0 £ q £ 90°.

Definition 2 Draw a unit circle, in that a circle of radius 1, centered at the origin of a 2-dimensional coordinate system. Given an angle q, locate the point on the circle that is located at an angle q from the origin. (According to standard convention, angles are measured counter-clockwise from the positive horizontal axis.) The sin(q) can be defined as the y-coordinate of this point. The cos(q) can be defined as the x-coordinate of this point. In this way, we can find the sine or cosine of any real value of q (q Î Â).

So my questions:

1. What are the sin and cosine exactly? Do you only calculate the sine and cosine of right triangles? What do you if the triangle is not right? Are there still relationships between the sides given an angle?
2. When you calculate sin(85 degrees), what exactly are we calculating? Are we calculating the opposite side/hypotenuse of a right triangle?

SINE RULE: If a triangle has sides of length $a,b,c$ and angles of size $A,B,C$, so that $A$ is the angle opposite side $a$, and so on, then $$\frac{\sin A}a=\frac{\sin B}b=\frac{\sin C}c$$ COSINE RULE: $$c^2=a^2+b^2-2ab\cos C$$ For example, the cosine of a right angle is zero, so the cosine rule for a right-angled triangle is $c^2=a^2+b^2$, which you may know as Pythagoras' Theorem.
• If $A=65^{\circ}$, then you could draw a right-angled triangle, one angle $65^{\circ}$, measure the ratio. The ratio will be $0.9065....$. If another angle is $B=75^{\circ}$, then $\sin B=0.9659....$ The third angle will then be $C=40^{\circ}$ because $A+B+C=180^{\circ}$, and $\sin40=0.6428...$. The sine rule tells you the sides of the triangle are in the ratio $a=0.9065R,b=0.9659R,c=0.6428R$ – Empy2 Jan 22 '16 at 18:08
• Take the right-angled triangle, and then make the adjacent side shorter and shorter. You find the other angle gets closer to $90^{\circ}$. You almost get two right angles. When the cosine is very close to zero, the other angle is very close to $90^{\circ}$. – Empy2 Jan 22 '16 at 18:13