Circles, Arcs and Special Triangles — AP® Precalculus Notes

The next topic puts angles on a circle and reads sine and cosine as coordinates. Three tools make that easy: two right triangles whose sides you know exactly, the length of an arc, and the equation of a circle.

The two special right triangles

Most right triangles have messy side ratios. Two have ratios you can write exactly, and they appear again and again.

45-45-90. Two equal angles mean two equal legs. With legs $1$ , the Pythagorean theorem gives a hypotenuse of $\sqrt{1 + 1} = \sqrt{2}$ . Any 45-45-90 triangle is a scaled copy: legs $a$ , hypotenuse $a\sqrt{2}$ .

30-60-90. Cut an equilateral triangle with sides $2$ down the middle. Each half has a hypotenuse $2$ , a short leg $1$ (half the base) and a long leg $\sqrt{2^2 - 1^2} = \sqrt{3}$ . The short leg is across from $30^\circ$ , the long leg across from $60^\circ$ .

Left: a 45-45-90 triangle with two legs of length 1 and a hypotenuse of length square root of 2; both acute angles are 45 degrees. Right: a 30-60-90 triangle with the short leg 1 across from the 30 degree angle, the long leg square root of 3 across from the 60 degree angle, and the hypotenuse 2, twice the short leg. — **Figure 1.** The 45-45-90 and 30-60-90 triangles. LevlPrep original diagram.

Reading SOH CAH TOA off these triangles gives exact values:

Angle	$\sin$	$\cos$	$\tan$
$30^\circ$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
$45^\circ$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$
$60^\circ$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$

( $\frac{1}{\sqrt{2}}$ and $\frac{\sqrt{2}}{2}$ are the same number: multiply top and bottom by $\sqrt{2}$ .)

Worked example 1. A 30-60-90 triangle has hypotenuse $10$ . Find both legs.

Scale $1 : \sqrt{3} : 2$ so the hypotenuse is $10$ : multiply by $5$ . The short leg is $5$ and the long leg is $5\sqrt{3}$ .

Arc length

A central angle has its vertex at the center of a circle. An angle of $\theta$ degrees is the fraction $\frac{\theta}{360}$ of a full turn, so it cuts off that fraction of the circumference $2\pi r$ :

$s = \frac{\theta}{360} \cdot 2\pi r$ .

Left: a circle with center O and radius r. A central angle of 60 degrees cuts off an arc, drawn in green, of length s. Because 60 degrees is one sixth of 360 degrees, the arc is one sixth of the circumference 2 pi r. Right: on a coordinate grid, the circle with center (2, −1) and radius 3, whose equation is (x − 2) squared plus (y + 1) squared equals 9. The points (5, −1) and (2, 2) are marked on it, each 3 units from the center. — **Figure 2.** Left: a $60^\circ$ angle cuts off one sixth of the circle. Right: a circle on a grid and its equation. LevlPrep original diagram.

Worked example 2. A circle has radius 9 cm. How long is the arc cut off by a central angle of $80^\circ$ ?

$s$ $= \frac{80}{360} \cdot 2\pi(9)$ $= \frac{2}{9} \cdot 18\pi$ $= 4\pi$ cm, about 12.6 cm.

Notice what happens when the arc is exactly as long as the radius: $\frac{\theta}{360} \cdot 2\pi r = r$ gives $\theta$ $= \frac{360}{2\pi}$ $= \frac{180}{\pi}$ $\approx 57.3^\circ$ , whatever the size of the circle. That angle is about to get its own name.

The equation of a circle

A circle is every point at distance $r$ from its center $(h, k)$ . The horizontal and vertical distances from the center, $x - h$ and $y - k$ , are the legs of a right triangle with hypotenuse $r$ . The Pythagorean theorem gives

$(x - h)^2 + (y - k)^2 = r^2$ .

Watch the signs: a center at $(2, -1)$ gives $(x - 2)^2 + (y + 1)^2 = r^2$ . And the right side is the radius squared: radius $3$ means $= 9$ .

The simplest case is the circle of radius $1$ centered at the origin, $x^2 + y^2 = 1$ . The point on it reached by turning $30^\circ$ up from the positive $x$ -axis is the top corner of a 30-60-90 triangle with hypotenuse $1$ : it is $(\cos 30^\circ, \sin 30^\circ)$ $= \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ . Check: $\frac{3}{4} + \frac{1}{4} = 1$ . That picture is exactly where the next topic starts.