Circles, Arcs and Special Triangles

Part 1 · Hook

Why this matters

A 12-inch pizza cut into 8 slices gives each slice a crust about 4.7 inches long. The crust is an arc, and its length is a fixed fraction of the whole circle. Arcs, circles on a grid and two famous right triangles are the last pieces you need before sine and cosine become functions.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In a right triangle, the side opposite angle $A$ is $4$ and the hypotenuse is $8$ . What is $\sin A$ ?

$\frac{1}{2}$
$2$
$\frac{\sqrt{3}}{2}$
$4$

Show the answer

Sine is opposite over hypotenuse: $\frac{4}{8} = \frac{1}{2}$ .

Correct: $\frac{1}{2}$ :
$2$ :
$\frac{\sqrt{3}}{2}$ :
$4$ :

2. Expand $(x + 3)^2$ .

$x^2 + 6x + 9$
$x^2 + 9$
$x^2 + 3x + 9$
$x^2 + 6x + 6$

Show the answer

$(x + 3)(x + 3)$ $= x^2$ $+ 3x$ $+ 3x$ $+ 9$ $= x^2$ $+ 6x$ $+ 9$ .

Correct: $x^2 + 6x + 9$ :
$x^2 + 9$ :
$x^2 + 3x + 9$ :
$x^2 + 6x + 6$ :

Part 4 · See it

See it first

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. — Learn these two triangles by heart: their side ratios give exact values of sine and cosine at $30^\circ$ , $45^\circ$ and $60^\circ$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A right triangle with two $45^\circ$ angles has equal legswith legs $1$ , the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$
Cut an equilateral triangle of side $2$ in halfyou get a 30-60-90 triangle with short leg $1$ , hypotenuse $2$ and long leg $\sqrt{2^2 - 1^2} = \sqrt{3}$
A central angle of $\theta$ degrees takes the fraction $\frac{\theta}{360}$ of the circleits arc has length $s = \frac{\theta}{360} \cdot 2\pi r$
Every point of a circle is the same distance $r$ from the center $(h, k)$ the Pythagorean theorem gives its equation, $(x - h)^2 + (y - k)^2 = r^2$

Part 6 · Key ideas

Key ideas

45-45-90: sides $1 : 1 : \sqrt{2}$ . 30-60-90: sides $1 : \sqrt{3} : 2$ , with the short leg across from $30^\circ$ .
So $\sin 30^\circ = \frac{1}{2}$ , $\cos 30^\circ = \frac{\sqrt{3}}{2}$ , $\sin 45^\circ$ $= \cos 45^\circ$ $= \frac{\sqrt{2}}{2}$ , $\sin 60^\circ = \frac{\sqrt{3}}{2}$ , $\cos 60^\circ = \frac{1}{2}$ .
Arc length is a fraction of the circumference: $s = \frac{\theta}{360} \cdot 2\pi r$ for an angle of $\theta$ degrees.
The circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$ . The right side is $r^2$ , not $r$ .

Part 7 · Misconception

A common mistake

The wrong idea: In a 30-60-90 triangle the side across from $60^\circ$ is twice the side across from $30^\circ$ , because the angle is twice as big.

What actually happens: Sides do not grow in step with angles. The sides are $1$ , $\sqrt{3}$ and $2$ : the hypotenuse, across from $90^\circ$ , is twice the short leg, and the side across from $60^\circ$ is $\sqrt{3} \approx 1.73$ times it.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. A 30-60-90 triangle has hypotenuse $10$ . What is the exact length of the leg across from the $60^\circ$ angle?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Scale $1 : \sqrt{3} : 2$ by 5 so that the hypotenuse is 10: the legs are $5$ and $5\sqrt{3}$ , and $5\sqrt{3}$ is across from $60^\circ$ .

Answer: $5\,\sqrt{3}$ ≈ 8.660

2. What is the exact value of $\sin 60^\circ$ ?

$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$
$\sqrt{3}$
$\frac{\sqrt{2}}{2}$

Show the answer

In the 30-60-90 triangle the side across from $60^\circ$ is $\sqrt{3}$ and the hypotenuse is $2$ , so $\sin 60^\circ = \frac{\sqrt{3}}{2}$ .

Correct: $\frac{\sqrt{3}}{2}$ : Opposite $\sqrt{3}$ over hypotenuse $2$ .
$\frac{1}{2}$ : $\frac{1}{2}$ is $\sin 30^\circ$ (or $\cos 60^\circ$ ): it uses the short leg.
$\sqrt{3}$ : $\sqrt{3}$ is opposite over adjacent, $\tan 60^\circ$ .
$\frac{\sqrt{2}}{2}$ : $\frac{\sqrt{2}}{2}$ belongs to the 45-45-90 triangle.

3. A circle has radius 9 cm. What is the exact length, in centimeters, of the arc cut off by a central angle of $80^\circ$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in cm.

Show the answer

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

Answer: $4\,\pi$ ≈ 12.566 cm

4. Which is an equation of the circle with center $(-3, 4)$ and radius $5$ ?

$(x + 3)^2 + (y - 4)^2 = 25$
$(x - 3)^2 + (y + 4)^2 = 25$
$(x + 3)^2 + (y - 4)^2 = 5$
$(x - 3)^2 + (y - 4)^2 = 25$

Show the answer

$(x - h)^2 + (y - k)^2 = r^2$ with $h = -3$ , $k = 4$ , $r = 5$ : $(x + 3)^2 + (y - 4)^2 = 25$ .

Correct: $(x + 3)^2 + (y - 4)^2 = 25$ : $x - (-3) = x + 3$ , and $r^2 = 25$ .
$(x - 3)^2 + (y + 4)^2 = 25$ : Both signs are flipped: this circle is centered at $(3, -4)$ .
$(x + 3)^2 + (y - 4)^2 = 5$ : The right side is the radius squared, $25$ , not the radius.
$(x - 3)^2 + (y - 4)^2 = 25$ : This circle is centered at $(3, 4)$ ; subtracting $-3$ gives $x + 3$ .

5. On the circle $x^2 + y^2 = 1$ , start at $(1, 0)$ and turn $30^\circ$ counterclockwise about the origin. What are the coordinates of the point you reach?

$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$
$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$
$\left(\frac{1}{2}, \frac{\sqrt{2}}{2}\right)$

Show the answer

Drop a segment from the point to the $x$ -axis: a 30-60-90 triangle with hypotenuse 1. The horizontal leg is $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and the height is $\sin 30^\circ = \frac{1}{2}$ .

Correct: $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ : $(\cos 30^\circ, \sin 30^\circ)$ $= \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ .
$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ : This is the point at $60^\circ$ : the coordinates are swapped.
$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$ : This is the point at $45^\circ$ .
$\left(\frac{1}{2}, \frac{\sqrt{2}}{2}\right)$ : $\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{3}{4}$ , so this point is not even on the circle.

Part 9 · Summary

Summary

The 45-45-90 triangle has sides

1 : 1 : \sqrt{2}

and the 30-60-90 triangle

1 : \sqrt{3} : 2

; they give exact sines and cosines at

30^\circ

,

45^\circ

and

60^\circ

. An arc cut off by a central angle of

\theta

degrees is the fraction

\frac{\theta}{360}

of the circumference

2\pi r

. The circle with center

(h, k)

and radius

r

is

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

.

Part 10 · Up next

What comes next

Part 11 · Connections