Q.What must be the radius of a circular running path, around which an athlete must run $5$ times in order to describe $1$ km?

An angle is generated by rotating a ray (the initial side) about its endpoint (the vertex) until it reaches a final position (the terminal side). The direction of rotation gives the angle its sign: anticlockwise rotation is positive, clockwise rotation is negative. One full sweep of the ray back to its own starting position is one complete rotation.

Three systems for measuring an angle:

• Sexagesimal (degree) system — a right angle split into $90$ degrees ($^\circ$); $1^\circ=60$ minutes ($'$); $1'=60$ seconds ($''$).
• Centesimal system — a right angle split into $100$ grades ($^g$); each grade into $100$ minutes; each of those into $100$ seconds. Rarely used in practice.
• Circular (radian) system — measured via an arc-length-to-radius ratio, developed fully below. The system used throughout higher mathematics.

Degree measure in detail. One complete rotation is $360^\circ$, so $1^\circ=\dfrac{1}{360}$ of a full turn; $1'=\dfrac{1}{60}$ of a degree; $1''=\dfrac{1}{60}$ of a minute $=\dfrac{1}{3600}$ of a degree. Two angles are congruent if equal in measure, complementary if they sum to $90^\circ$, supplementary if they sum to $180^\circ$, and conjugate if both lie in $[0^\circ,360^\circ)$ and sum to $360^\circ$.

Standard position and quadrants. An angle is in standard position when its vertex is at the origin and its initial side lies along the positive $x$-axis; it is then a first/second/third/fourth-quadrant angle according to where its terminal side falls. An angle whose terminal side falls exactly on an axis is a quadrantal angle — always a multiple of $90^\circ$ ($0^\circ,90^\circ,180^\circ,270^\circ,360^\circ,\ldots$).

Coterminal angles. Two angles in standard position with the same terminal side (however many extra full rotations were used to get there) are coterminal. If $\alpha,\beta$ are coterminal, $\beta=\alpha+k(360^\circ)$ for some integer $k$. To find the coterminal angle of any given angle inside the standard range $[0^\circ,360^\circ)$, repeatedly add or subtract $360^\circ$ until the result lands in that range.

Why degrees aren't enough. Right-triangle trigonometry, and degree measure along with it, only ever produces acute angles. But calculus, physics and chemistry need trigonometric functions defined on the whole real line. The fix is to measure an angle using a length ratio instead of an arbitrary human convention like "$360$ parts to a turn."

Definition — the radian. For a circle of radius $r$, let an angle $\theta$ at the centre subtend an arc of length $s$. Then the radian measure of $\theta$ is

Because all circles are similar, this ratio depends only on the angle, never on which particular circle is used to measure it. When $s=r$ exactly, $\theta=1$ radian — the angle subtended by an arc exactly as long as the radius. Since $s$ and $r$ are both lengths, $\theta$ is a pure (unitless) number, which is why a plain number like $\dfrac{\pi}{3}$ or $2$, with no symbol attached, is automatically read as being in radians.

Degree $\leftrightarrow$ radian conversion. A full rotation is both $360^\circ$ (by definition of the degree) and $2\pi$ radians (the circumference of the unit circle, since $s=r\theta$ with $r=1,\theta=2\pi$ gives $s=2\pi$). Equating the two:

More generally, $x^\circ = \dfrac{\pi x}{180}$ radians and $x$ radians $=\left(\dfrac{180x}{\pi}\right)^{\!\circ}$.

Why radians are preferred. Formulas involving angles are simply cleaner in radian measure — compare the sector-area formula $\dfrac{\pi r^2\theta}{360^\circ}$ (degrees) with $\dfrac{1}{2}r^2\theta$ (radians): no leftover $\pi$ or $360$ in the radian version. This is exactly why calculus formulas (derivatives/integrals of $\sin,\cos$, etc.) are only clean when the angle is in radians.

Worked conversion example (own numbers). Convert $150^\circ$ to radians and $\dfrac{5\pi}{4}$ radians to degrees.

• $150^\circ = \dfrac{\pi}{180}\times150 = \dfrac{5\pi}{6}$ radians.
• $\dfrac{5\pi}{4}$ radians $= \dfrac{180}{\pi}\times\dfrac{5\pi}{4} = \dfrac{180\times5}{4}=225^\circ$.