{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }}
An angle, which can be expressed in both degrees and radians, measures the distance of turn in a rotation. Because rotations follow a circular path, angles can be used when analyzing circles and vice versa. In fact, there's a special relationship between right triangles and a circle of radius $1.$

Right triangles with certain acute angle measures are considered noteworthy. These include the following angle relationships.

- $30_{∘}$ - $60_{∘}$ - $90_{∘}$
- $45_{∘}$ - $45_{∘}$ - $90_{∘}$
- $60_{∘}$ - $30_{∘}$ - $90_{∘}$

Although the first and third relationships contain the same angles, they are considered **different** because the reference angle in each is different — $30_{∘}$ in the first and $60_{∘}$ in the third. The sine, cosine, and tangents of these triangles can be useful when finding unknown side lengths.

Angle $θ$ | $30_{∘}$ | $45_{∘}$ | $60_{∘}$ |
---|---|---|---|

$sin(θ)$ | $21 $ | $22 $ | $23 $ |

$cos(θ)$ | $23 $ | $22 $ | $21 $ |

$tan(θ)$ | $33 $ | $1$ | $3 $ |

These special measures are justified below.

Consider the $30_{∘}$ - $60_{∘}$ - $90_{∘}$ triangle. Suppose an equilateral triangle has side lengths of $1.$

Bisecting the apex angle yields the following $30_{∘}$ - $60_{∘}$ - $90_{∘}$ triangle.

The value of $h$ — the length of the third side — can be found using the Pythagorean Theorem. Here, $a=21 ,b=h,$ and $c=1.$

$a_{2}+b_{2}=c_{2}$

SubstituteValuesSubstitute values

$(21 )_{2}+h_{2}=1_{2}$

CalcPowCalculate power

$41 +h_{2}=1$

SubEqn$LHS−41 =RHS−41 $

$h_{2}=43 $

SqrtEqn$LHS =RHS $

$h=43 $

DivSqrt$b a =ba $

$h=4 3 $

CalcRootCalculate root

$h=23 $

Since, $h=23 ,$ the $30_{∘}$ - $60_{∘}$ - $90_{∘}$ triangle can be redrawn as follows.

Using $θ=30_{∘},$ the following relationships can be concluded.

$ sin(30_{∘})= 21 /1=21 cos(30_{∘})=23 /1=23 tan(30_{∘})=21 /23 =33 $

Using the same triangle, the values for a $60_{∘}$ - $30_{∘}$ - $90_{∘}$ triangle can be determined.

$ sin(60_{∘})=23 /1=23 cos(60_{∘})= 21 /1=21 tan(60_{∘})=23 /21 =3 $

Suppose an isosceles triangle has a hypotenuse of $1$ and base angles that measure $45_{∘}.$

Because the triangle is isosceles, its legs have equal measure. Because they are unknown, $x$ can be used to represent them. The Pythagorean Theorem can be used to determine the value of $x.$

$a_{2}+b_{2}=c_{2}$

SubstituteExpressionsSubstitute expressions

$x_{2}+x_{2}=1_{2}$

SimpPowTermSimplify power and terms

$2x_{2}=1$

DivEqn$LHS/2=RHS/2$

$x_{2}=21 $

SqrtEqn$LHS =RHS $

$x=±21 $

$x>0$

$x=21 $

SqrtQuot$ba =b a $

$x=2 1 $

ExpandFrac$ba =b⋅2 a⋅2 $

$x=22 $

The $45_{∘}$ - $45_{∘}$ - $45_{∘}$ triangle can be redrawn as follows.

Using $θ=45_{∘},$ the following relationships can be concluded.

$ sin(45_{∘})=22 /1=22 cos(45_{∘})=22 /1=22 tan(45_{∘})=22 /22 =1 $

As was shown above, because the hypotenuse of these special right triangles is $1,$

$sin(θ)=oppandcos(θ)=adj,forθ=30_{∘},45_{∘},60_{∘}.$

Therefore, these values can be used to determine unknown side lengths.By placing right triangles with one of their legs along the $x$-axis, the coordinates of the vertex **not** on the axis can be found using the legs of the triangle.

In the first quadrant, the coordinates are the lengths of the triangle's legs. If the length of the hypotenuse is $1$ the legs of the special right triangles are given directly by the sine and cosine value of the angle at the origin and the coordinates are **also** the sine and cosine values.

Since the hypotenuse of these triangles is $1,$ the marked points all lie on a circle with radius $1$ and a center at the origin.

This is called *the unit circle*. The coordinates of any point on this circle are the cosine and sine values of the angle the segment from the point to the origin creates with the positive $x$-axis.

$x=cos(θ)andy=sin(θ)$

A radian is, like a degree, an angle unit. If the arc length of a circle sector is equal to the radius of the circle, the angle created is $1$ radian (rad), which corresponds to roughly $57.3_{∘}.$

Animate 1 rad

If the arc length is $2$ radii, the angle is $2$ radians, etc. Thus, radians describe the number of radii an angle creates on a circle.

The relation between degrees and radians can be described in different ways. One is $180_{∘}=πrad.$
In calculations, even if the angle is given in radians, "rad" is seldom written. Instead, no unit marker indicates radians. That means that in the expressions
$cos(64_{∘})andcos(5)$ the first angle is given in degrees and the other in radians. At first glance, radians might seem inconvenient, but in certain circumstances, they make calculations simpler. Radians are also the SI unit for angles.

Use special right triangles and the unit circle to find the following trigonometric values. $sin(45_{∘})tan(π)cos(150_{∘})$

Show Solution

Let's find the trigonometric values one at a time.

The value $sin(45_{∘})$ is the $y$-coordinate of the point on the unit circle that, together with the positive $x$-axis, creates the angle $45_{∘}.$

Here, we have a right triangle with the angles $45_{∘}$ and $45_{∘}.$ The hypotenuse is $1,$ so we know that the legs of the triangle are both $22 .$ That means the point is $(22 ,22 ).$

Therefore, $sin(45_{∘})$ is $22 .$

Notice there is no degree symbol. This means the angle is given in radians. $π$ radians is half a revolution, meaning $180_{∘}.$

$tan(θ)$ does not correspond with a point on the unit circle, but it can be found using $sin(θ)$ and $cos(θ).$ $tan(θ)=cos(θ)sin(θ) $ The point corresponding to $π$ rad has the coordinates $(-1,0).$ Thus, $cos(π)=-1$ and $sin(π)=0.$$tan(π)=cos(π)sin(π) $

${\textstyle \ifnumequal{180}{0}{\cos\left(0\right)=1}{}\ifnumequal{180}{30}{\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{45}{\cos\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{60}{\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}}{}\ifnumequal{180}{90}{\cos\left(\dfrac{\pi}{2}\right)=0}{}\ifnumequal{180}{120}{\cos\left(\dfrac{2\pi}{3}\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{180}{135}{\cos\left(\dfrac{3\pi}{4}\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{150}{\cos\left(\dfrac{5\pi}{6}\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{180}{\cos\left(\pi\right)=\text{-} 1}{}\ifnumequal{180}{210}{\cos\left(\dfrac{7\pi}6\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{180}{225}{\cos\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{240}{\cos\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {1}2}{}\ifnumequal{180}{270}{\cos\left(\dfrac{3\pi}{2}\right)=0}{}\ifnumequal{180}{300}{\cos\left(\dfrac{5\pi}3\right)=\dfrac{1}2}{}\ifnumequal{180}{315}{\cos\left(\dfrac{7\pi}4\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{330}{\cos\left(\dfrac{11\pi}6\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{180}{360}{\cos\left(2\pi\right)=1}{}}$, ${\textstyle \ifnumequal{180}{0}{\sin\left(0\right)=0}{}\ifnumequal{180}{30}{\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{180}{45}{\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{60}{\sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{90}{\sin\left(\dfrac{\pi}{2}\right)=1}{}\ifnumequal{180}{120}{\sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{180}{135}{\sin\left(\dfrac{3\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{180}{150}{\sin\left(\dfrac{5\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{180}{180}{\sin\left(\pi\right)=0}{}\ifnumequal{180}{210}{\sin\left(\dfrac{7\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{180}{225}{\sin\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{240}{\sin\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{180}{270}{\sin\left(\dfrac{3\pi}{2}\right)=\text{-} 1}{}\ifnumequal{180}{300}{\sin\left(\dfrac{5\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{180}{315}{\sin\left(\dfrac{7\pi}4\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{180}{330}{\sin\left(\dfrac{11\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{180}{360}{\sin\left(2\pi\right)=0}{}}$

$tan(π)=-10 $

CalcQuotCalculate quotient

$tan(π)=0$

Let's draw the angle $150_{∘}$ from the positive $x$-axis and mark the point it corresponds to on the unit circle.

An angle of $180_{∘}$ would span from the positive $x$-axis to the negative. Hence, a $30_{∘}$ special right triangle can be drawn from the negative $x$-axis to the terminal side of the $150_{∘}$ angle. Recall that this triangle has legs that measure $21 $ and $23 .$

Since $cos(150_{∘})$ is the $x$-coordinate of the point, $cos(150_{∘})=-23 .$

Since both degrees and radians are used to measure angles, it's useful to be able to translate between them.

The arc length for one revolution around a circle is also known as the circumference of the circle, $2πr.$ By dividing this with the length of the arc corresponding to one radian, $r,$ the number of radians for one revolution is obtained. $r2πr =2π$ Thus, one revolution is $2π$ radians. In degrees, this is $360_{∘},$ leading to the relation $360_{∘}=2π$ rad. Dividing both sides by $2$ gives $180_{∘}=πrad.$ Thus, $180_{∘}$ correspond to $π$ radians.

From the relation $180_{∘}=π$ rad it's possible to find two rules, by dividing both sides by either $180$ or $π.$

To get an expression for $1$ rad, divide both sides by $π.$

$180_{∘}=πrad$

DivEqn$LHS/π=RHS/π$

$π180_{∘} =ππ rad$

CalcQuotCalculate quotient

$π180_{∘} =1rad$

RearrangeEqnRearrange equation

$1rad=π180_{∘} $

$1$ radian corresponds to $π180 ≈57.3_{∘}.$

Convert $45_{∘}$ to radians and $2π rad$ to degrees.

Show Solution

To convert an angle from degrees to radians we can use $1_{∘}=180π rad.$ $45_{∘}$ should be multiplied with $180π $ rad.

$45_{∘}$

$1_{∘}=180π rad$

$45⋅180π rad$

MoveLeftFacToNum$a⋅cb =ca⋅b $

$18045π rad$

ReduceFrac$ba =b/45a/45 $

$4π rad$

$2π ⋅π180_{∘} $

MultFracMultiply fractions

$2ππ⋅180_{∘} $

CancelCommonFacCancel out common factors

$2180_{∘} $

CalcQuotCalculate quotient

$90_{∘}$

{{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

{{ exercise.headTitle }}

{{ 'ml-heading-exercise' | message }} {{ focusmode.exercise.exerciseName }}