# Measurement of Angles

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

GPS coordinates are given in decimal degrees:

latitude: 36.52569°, longitude: -5.04784°

Write the coordinates in degrees, minutes and seconds.

### Example 2

New York City is located at

latitude: 40°43'50.196", longitude: -73°56'6.871"

Write the coordinates in decimal degrees.

### Example 3

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

### Example 4

Find the radian measure of an angle of 10000°.

### 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:

$$latitude: 36.52569^\circ\;,$$ $$longitude: - 5.04784^\circ.$$

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

$$latitude: {40^\circ} 43^\prime 50.196^{\prime\prime},\;$$ $$longitude: -{73^\circ} 56^\prime 6.871^{\prime\prime}$$

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 $$\alpha = \frac{\ell }{R},$$ 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 $$2\pi \times \frac{{20}}{{60}} = \frac{{2\pi }}{3}\,rad.$$ 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 $$A = \frac{\alpha {R^2}}{2}.$$ 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 $$R = \frac{{2A}}{\ell }.$$