# Precalculus

## Trigonometry # Measurement of Angles

### Definitions

In geometry, an angle is defined as a figure formed by two rays sharing a common endpoint. The rays are called the sides of the angle, and the common endpoint is called its vertex.

In trigonometry, an angle is often defined in terms of rotation. Consider a ray that is rotated around its endpoint to create an angle. The ray in the initial position is called the initial side of the angle. Respectively, the ray in the terminal position, after the rotation, is called the terminal side of the angle. The amount of rotation from the initial side to the terminal side determines the measure of the angle.

A positive angle is an angle that is formed by counterclockwise rotation. If an angle is created by clockwise rotation, it has a negative measure.

There are two commonly used ways to measure angles – in degrees and radians.

### Degree Measure of an Angle

The most common measure of an angle is in degrees. One full rotation is equal to $$360^\circ.$$ An angle of $$180^\circ$$ is called a straight angle. An angle whose measure is $$90^\circ$$ is called a right angle.

Hence,

${{1^\circ } = \frac{1}{{360}} \;\text{of a full rotation}, \text{ or }\;}\kern0pt{{1^ \circ } = \frac{1}{{90}} \;\text{of a right angle.}}$

Sometimes we need to measure small angles less than $$1^\circ.$$ For example, the Moon’s angular diameter is about $${\large{\frac{1}{2}}\normalsize}$$ degrees.

For finer measurements, one degree is divided into $$60$$ minutes (of arc), and one minute into $$60$$ seconds (of arc):

${{1^ \circ} = 60^{\prime},\;\;}\kern0pt{1^{\prime} = 60^{\prime\prime},\;\;}\kern0pt{{1^\circ} = 3600^{\prime\prime}.}$

Another example from astronomy: the angular size of Venus ranges from $$9.7^{\prime\prime}$$ to $$66^{\prime\prime}.$$ Venus has the largest angular size at its closest approach to Earth.

#### Example $$1$$

Write the angle $$23^\circ 54^\prime$$ in decimal degrees.

Solution.

Since $$1^\prime = {\left( {\large{\frac{1}{{60}}}\normalsize} \right)^\circ},$$ then

${{23^\circ}54^\prime = {23^\circ} + 54^\prime }={ {23^\circ} + 54 \times {\left( {\frac{1}{{60}}} \right)^\circ} }={ {23^\circ} + {\left( {\frac{{54}}{{60}}} \right)^\circ} }={ {23^\circ} + {0.9^\circ} }={ {23.9^\circ}}$

#### Example $$2$$

Write the angle $$123.456^\circ$$ in degrees, minutes, and seconds.

Solution.

${{123.456^\circ} = {123^\circ} + {0.456^\circ} }={ {123^\circ} + 0.456 \times 60^\prime }={ {123^\circ} + 27.36^\prime }={ {123^\circ} + 27^\prime + 0.36^\prime }={ {123^\circ} + 27^\prime + 0.36 \times 60^{\prime\prime} }={ {123^\circ} + 27^\prime + 21.6^{\prime\prime} }={ {123^\circ}27^\prime 21.6^{\prime\prime}}$

### Radian Measure of an Angle

The radian measure of an angle is defined as the ratio of the arc it cuts off to the radius of a circle centered at the vertex of the angle.

If $$\ell$$ is the arc length and $$R$$ is the radius of the circle, then the central angle $$\alpha$$ that subtends the arc is defined in radians as

Recall that the length of the circumference of a circle of radius $$R$$ is equal to $$2\pi R.$$ So, the radian measure of an angle of $$360^\circ$$ is given by

$\require{cancel}{{360^\circ} = \frac{\ell }{R} }={ \frac{{2\pi \cancel{R}}}{\cancel{R}} }={ 2\pi \left( {rad}\right) .}$

Hence, to convert radian to degrees we can use the following formula:

${1\,{radian} = \frac{{{{360}^\circ}}}{{2\pi }} }={ \frac{{{{180}^\circ}}}{\pi } }\approx{ {57.3^\circ}}$

Consider a circular sector cut from a circle of radius $$R.$$ If the sector has a central angle of $$\alpha,$$ its area is given by the formula

#### Example $$3$$

Convert $$60^\circ$$ into radians.

Solution.

We know that $$1\,radian = \large{\frac{{{{180}^\circ}}}{\pi }}\normalsize.$$ Then $${1^\circ} = \large{\frac{\pi }{{180}}}\normalsize\,radians.$$ This yields:

${{60^\circ} = 60 \cdot \frac{\pi }{{180}} }={ \frac{\pi }{3}\,radians.}$

#### Example $$4$$

Given a circle of radius $$3.$$ Find the length of an arc cut off by a central angle of $$2$$ radians.

Solution.

The arc length is determined by the formula $$\ell = \alpha R,$$ where $$\alpha$$ is the central angle expressed in radians and $$R$$ is the radius of the circle. Substituting the known values, we get

$\ell = \alpha R = 2 \cdot 3 = 6.$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

GPS coordinates are given in decimal degrees: ${latitude: 36.52569^\circ,\;\;}\kern0pt{longitude: – 5.04784^\circ}$ Write the coordinates in degrees, minutes and seconds.

### Example 2

New York City is located at ${latitude: {40^\circ} 43^\prime 50.196^{\prime\prime},\;\;}\kern0pt{longitude: -{73^\circ} 56^\prime 6.871^{\prime\prime}}$ Write the coordinates in decimal degrees.

### Example 3

A wheel has a radius of $$0.3\,m.$$ How many full rotations does the wheel make if it travels $$10\,m?$$

### Example 4

Find the radian measure of an angle of $$10000^\circ.$$

### Example 5

The diameter of human hair is about $$75\,\mu m.$$ How far from the observer is the hair if it is seen at an angle of $$1$$ second?

### Example 6

The minute hand of Big Ban, the Elizabeth Tower clock in London, is $$4.3\,m$$ long. What distance does the tip of the minute hand travel in $$20$$ minutes?

### Example 7

Find the area of a sector with central angle of $$3$$ radian in a circle of radius $$1\,m.$$

### Example 8

A sector in a circle has the area $$A$$ and arc length $$\ell.$$ Find the radius of the circle.

### Example 1.

GPS coordinates are given in decimal degrees: ${latitude: 36.52569^\circ,\;\;}\kern0pt{longitude: – 5.04784^\circ}$ Write the coordinates in degrees, minutes and seconds.

Solution.

Convert the latitude:

${{36.52569^\circ} = {36^\circ} + {0.52569^\circ} }={ {36^\circ} + 0.52569 \times 60^\prime }={ {36^\circ} + 31.5414^\prime }={ {36^\circ} + 31^\prime + 0.5414^\prime }={ {36^\circ} + 31^\prime + 0.5414 \times 60^{\prime\prime} }={ {36^\circ} + 31^\prime + 32.384^{\prime\prime} }={ {36^\circ}31^\prime 32.384^{\prime\prime} }\approx{ {36^\circ}31^\prime 32.4^{\prime\prime}}$

Similarly we find the longitude:

${- {5.04784^\circ} = – {5^\circ} – {0.04784^\circ} }={ – {5^\circ} – 0.04784 \times 60^\prime }={ – {5^\circ} – 2.8704^\prime }={ – {5^\circ} – 2^\prime – 0.8704^\prime }={ – {5^\circ} – 2^\prime – 0.8704 \times 60^{\prime\prime} }={ – {5^\circ} – 2^\prime – 52.224^{\prime\prime} }={ – {5^\circ} 2^\prime 52.224^{\prime\prime} }\approx{ – {5^\circ}2’52.2^{\prime\prime}}$

### Example 2.

New York City is located at ${latitude: {40^\circ} 43^\prime 50.196^{\prime\prime},\;\;}\kern0pt{longitude: -{73^\circ} 56^\prime 6.871^{\prime\prime}}$ Write the coordinates in decimal degrees.

Solution.

Calculate the latitude:

${{40^\circ}43^\prime 50.196” = {40^\circ} + 43^\prime + 50.196^{\prime\prime} }={ {40^\circ} + 43 \times {\left( {\frac{1}{{60}}} \right)^\circ} + 50.196 \times {\left( {\frac{1}{{3600}}} \right)^\circ} }={ {40^\circ} + {0.71667^\circ} + {0.01394^\circ} }={ {40.53061^\circ}}$

The longitude is given by

${- {73^\circ}56^\prime 6.871^{\prime\prime} = – {73^\circ} – 56^\prime – 6.871^{\prime\prime} }={ – {73^\circ} – 56 \times {\left( {\frac{1}{{60}}} \right)^\circ} – 6.871 \times {\left( {\frac{1}{{3600}}} \right)^\circ} }={ – {73^\circ} – {0.93333^\circ} – {0.00191^\circ} }={ – {73.93524^\circ}}$

### Example 3.

A wheel has a radius of $$0.3\,m.$$ How many full rotations does the wheel make if it travels $$10\,m?$$

Solution.

When a wheel of radius $$R$$ makes one full rotation, it travels the distance $$2\pi R.$$ So the number of rotations is determined by the formula

$N = \frac{L}{{2\pi R}},$

where $$L$$ is the distance traveled.

By substituting the values of $$L$$ and $$R,$$ we have

${N = \frac{L}{{2\pi R}} }={ \frac{{10}}{{2\pi \times 0.3}} }\approx{ 5.3}$

Hence, the wheel makes $$5$$ full rotations.

### Example 4.

Find the radian measure of an angle of $$10000^\circ.$$

Solution.

First we determine the number of full rotations $$N.$$ Since one full rotation is $$360^\circ,$$ then

$N = \frac{{{{10000}^\circ}}}{{{{360}^\circ}}} = 27.778$

Each rotation is $$2\pi$$ radians. Therefore, $$10000^\circ$$ is equal to

${{10000^\circ} = 2\pi N }={ 2\pi \times 27.778 }={ 174.53\,rad}$

### Example 5.

The diameter of human hair is about $$75\,\mu m.$$ How far from the observer is the hair if it is seen at an angle of $$1$$ second?

Solution.

Using the formula $$\alpha = \large{\frac{\ell }{R}}\normalsize,$$ we express the distance $$R$$ from the observer to the hair in terms of $$\ell$$ and $$\alpha:$$

$R = \frac{\ell }{\alpha }.$

We write the diameter of hair $$\ell$$ in meters:

${\ell = 75\,\mu m }={ 75 \times {10^{ – 6}}\,m }={ 7.5 \times {10^5}\,m.}$

The angle $$\alpha$$ in radians is given by

${\alpha = 1^{\prime\prime} }={ \frac{\pi }{{180}} \times \frac{1}{{3600}} }={ \frac{\pi }{{648000}} }={ 4.848 \times {10^{ – 6}}\,rad}$

Then

${R = \frac{\ell }{\alpha } }={ \frac{{7.5 \times {{10}^{ – 5}}}}{{4.848 \times {{10}^{ – 6}}}} }={ 1.547 \times 10 }\approx{ 15.5\,m}$

### Example 6.

The minute hand of Big Ban, the Elizabeth Tower clock in London, is $$4.3\,m$$ long. What distance does the tip of the minute hand travel in $$20$$ minutes?

Solution.

The minute hand makes one full rotation $$\left( {2\pi }\,rad \right)$$ in $$60$$ minutes. So in $$20$$ minutes, the hand turns $$2\pi \times \large{\frac{{20}}{{60}}}\normalsize = \large{\frac{{2\pi }}{3}}\normalsize\,rad.$$ The distance moved by the tip of the hand is

${\ell = \alpha R }={ \frac{{2\pi }}{3} \times 4.3 }={ 9\,m.}$

### Example 7.

Find the area of a sector with central angle of $$3$$ radian in a circle of radius $$1\,m.$$

Solution.

We use the formula

$A = \frac{{\alpha {R^2}}}{2}.$

Substitute $$\alpha = 3\,rad,$$ and $$R = 1\,m:$$

${A = \frac{{3 \times {1^2}}}{2} }={ \frac{3}{2} }={ 1.5\,{m^2}.}$

### Example 8.

A sector in a circle has the area $$A$$ and arc length $$\ell.$$ Find the radius of the circle.

Solution.

The arc length $$\ell$$ of a central angle $$\alpha$$ is expressed by the formula $$\ell = \alpha R,$$ where $$R$$ is the radius of the circle. The area of a circular sector is given by $$A = \large{\frac{{\alpha {R^2}}}{2}\normalsize}.$$ If $$A$$ and $$\ell$$ are known, we can find $$\alpha$$ and $$R$$ from these two equations.

${\left\{ \begin{array}{l} \ell = \alpha R\\ A = \frac{{\alpha {R^2}}}{2} \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} \alpha = \frac{\ell }{R}\\ A = \frac{\ell }{R} \cdot \frac{{{R^2}}}{2} \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} \alpha = \frac{\ell }{R}\\ A = \frac{{\ell R}}{2} \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} \alpha = \frac{\ell }{R}\\ R = \frac{{2A}}{\ell } \end{array} \right.,}\;\; \Rightarrow {\left\{ \begin{array}{l} \alpha = \frac{\ell }{{\frac{{2A}}{\ell }}} = \frac{{{\ell ^2}}}{{2A}}\\ R = \frac{{2A}}{\ell } \end{array} \right..}$

Thus, the radius of the circle is given by $$R = \large{\frac{{2A}}{\ell }}\normalsize.$$