The exercises with resolution show a method; the following ask to apply it. Try each one before opening the answer.

Arch measurements

1

Express $225^\circ$ in radians.

See the resolution

Simple rule of three:

$$\begin{array}{rcl} 180^\circ & \longleftrightarrow & \pi \text{ rad} \\ 225^\circ & \longleftrightarrow & x \end{array} \qquad\Longrightarrow\qquad x = \frac{225 \cdot \pi}{180} = \frac{5\pi}{4} \text{ rad}.$$
2

Express in radians: a) $210^\circ$; b) $240^\circ$; c) $270^\circ$; d) $300^\circ$; e) $315^\circ$; f) $330^\circ$.

Ver resposta

a) $\dfrac{7\pi}{6}$ rad b) $\dfrac{4\pi}{3}$ rad c) $\dfrac{3\pi}{2}$ rad d) $\dfrac{5\pi}{3}$ rad e) $\dfrac{7\pi}{4}$ rad f) $\dfrac{11\pi}{6}$ rad

3

Express $\dfrac{11\pi}{6}$ rad in degrees.

See the resolution
$$\begin{array}{rcl} \pi \text{ rad} & \longleftrightarrow & 180^\circ \\ \dfrac{11\pi}{6} \text{ rad} & \longleftrightarrow & x \end{array} \qquad\Longrightarrow\qquad x = \frac{\frac{11\pi}{6} \cdot 180}{\pi} = 330^\circ.$$
4

Express in degrees: a) $\dfrac{\pi}{6}$ rad; b) $\dfrac{\pi}{4}$ rad; c) $\dfrac{\pi}{3}$ rad; d) $\dfrac{2\pi}{3}$ rad; e) $\dfrac{3\pi}{4}$ rad; f) $\dfrac{5\pi}{6}$ rad.

Ver resposta

a) $30^\circ$ · b) $45^\circ$ · c) $60^\circ$ · d) $120^\circ$ · e) $135^\circ$ · f) $150^\circ$

5

An arc of a circle measures 30 cm and the radius of the circle measures 10 cm. Calculate the measure of the arc in radians.

See the resolution

The measurement in radians is the ratio between the length of the arc and the length of the radius:

$$\text{medida de } \overset{\frown}{AB} \text{ em rad} = \frac{\text{comprimento do arco } \overset{\frown}{AB}}{\text{comprimento do raio}} = \frac{30 \text{ cm}}{10 \text{ cm}} = 3 \text{ rad}.$$
6

On a circle of radius 10 cm, an arc $\overset{\frown}{AB}$ is marked such that the chord $\overline{AB}$ measures 10 cm. Calculate the measure of the arc in radians.

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See the resolution

The segment $\overline{AB}$ has the radius: it is the side of the regular hexagon inscribed in the circle. Therefore, the smallest arc $\overset{\frown}{AB}$ is $\dfrac{1}{6}$ of the circumference and measures

$$\frac{1}{6} \times 2\pi \text{ rad} = \frac{\pi}{3} \text{ rad}.$$
7

A degree is divided into $60'$ (60 minutes) and a minute is divided into $60''$ (60 seconds). For example, an arc of $30'$ is an arc of $0{,}5^\circ$. Convert the arcs into radians: a) $22^\circ 30'$; b) $31^\circ 15' 45''$.

See the resolution

a) $22^\circ 30' = 22 \times 60' + 30' = 1350'$ and $180^\circ = 180 \times 60' = 10\,800'$. Then

$$\begin{array}{rcl} 10\,800' & \longleftrightarrow & \pi \text{ rad} \\ 1350' & \longleftrightarrow & x \end{array} \qquad\Longrightarrow\qquad x = \frac{1350 \cdot \pi}{10\,800} = \frac{\pi}{8} \text{ rad}.$$

b) $31^\circ 15' 45'' = 31 \times 3600'' + 15 \times 60'' + 45'' = 112\,545''$ and $180^\circ = 180 \times 3600'' = 648\,000''$. Then

$$\begin{array}{rcl} 648\,000'' & \longleftrightarrow & \pi \text{ rad} \\ 112\,545'' & \longleftrightarrow & x \end{array} \qquad\Longrightarrow\qquad x = \frac{112\,545 \cdot \pi}{648\,000} \approx \frac{112\,545 \cdot 3{,}1416}{648\,000} \approx 0{,}54563 \text{ rad}.$$
8

Convert the arc of 1 rad to degrees.

See the resolution
$$\begin{array}{rcl} 3{,}1416 \text{ rad} & \longleftrightarrow & 180^\circ \\ 1 \text{ rad} & \longleftrightarrow & x \end{array} \qquad\Longrightarrow\qquad x = \frac{180^\circ}{3{,}1416}.$$

Doing the division, $\dfrac{180}{3{,}1416} \approx 57{,}2956$. The integer part gives $57^\circ$. The remainder, $0{,}2956^\circ$, multiplied by 60 becomes minutes: $0{,}2956 \times 60 \approx 17{,}738'$, that is, $17'$. The new remainder, $0{,}738'$, multiplied by 60 becomes seconds: $0{,}738 \times 60 \approx 44''$. Therefore

$$1 \text{ rad} \approx 57^\circ 17' 44''.$$

The figure compares 1 rad with $60^\circ$: on equilateral triangle side $r$, the chord measures $r$ and the arc of $60^\circ$ is greater than $r$; the arc of length $r$ ends a little earlier, at $57^\circ 17' 44''$.

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9

Express in radians the measurements of the arcs $a$ and $b$ such that $a - b = 15^\circ$ and $a + b = \dfrac{7\pi}{4}$ rad.

Ver resposta

$a = \dfrac{11\pi}{12}$ rad and $b = \dfrac{5\pi}{6}$ rad

10

Express in degrees the measurements of the arcs $a$, $b$ and $c$ such that $a + b + c = 13^\circ$, $a + b + 2c = \dfrac{\pi}{12}$ rad and $a + 2b + c = \dfrac{\pi}{9}$ rad.

Ver resposta

$a = 4^\circ$, $b = 7^\circ$ e $c = 2^\circ$

Measure of angles

11

Calculate, in degrees, the measure of the angle $a\hat{O}b$ in the figure.

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See the resolution

$\alpha = \dfrac{\ell}{r} = \dfrac{3}{10}$ rad. Converting to degrees:

$$\begin{array}{rcl} \pi \text{ rad} & \longrightarrow & 180^\circ \\ \dfrac{3}{10} \text{ rad} & \longrightarrow & x \end{array} \qquad\Longrightarrow\qquad x = \frac{\frac{3}{10} \times 180^\circ}{\pi} = \frac{54^\circ}{3{,}1416} \approx 17^\circ 11' 19''.$$
12

Calculate the length $\ell$ of the arc $\overset{\frown}{AB}$ defined in a circle of radius $r = 10$ cm by a central angle of $60^\circ$.

See the resolution

Converted to radians, the central angle $a\hat{O}b$ measures $\alpha = \dfrac{\pi}{3}$ rad. Then

$$\alpha = \frac{\ell}{r} \;\Longrightarrow\; \ell = \alpha \cdot r = \frac{\pi}{3} \cdot 10 \approx \frac{31{,}416}{3} \approx 10{,}472 \text{ cm}.$$
13

Calculate the measure of the central angle $a\hat{O}b$ that determines, in a circle of radius $r$, an arc of length $\dfrac{2\pi r}{3}$.

Ver resposta

$\alpha = \dfrac{2\pi}{3}$ rad, or $120^\circ$

14

Calculate the length $\ell$ of the arc $\overset{\frown}{AB}$ defined in a circle of radius 7 cm by a central angle of 4,5 rad.

Ver resposta

$\ell = 31{,}5$ cm

15

Calculate the smallest of the angles formed by the hands of a clock that is ticking: a) 1 h; b) 1 h 15 min; c) 1 h 40 min.

See the resolution

a) The numbers on the dial divide the circumference into 12 equal parts, each of $30^\circ$. At 1 h, the large hand is at 12 and the small one at 1: the hands form a convex angle of $30^\circ$.

b) In 60 minutes the small hand passes through $30^\circ$; in 15 minutes it travels an angle $\alpha$ such that

$$\frac{\alpha}{15} = \frac{30^\circ}{60} \;\Longrightarrow\; \alpha = 7{,}5^\circ = 7^\circ 30'.$$

At 1 h 15 min the big hand is at 3 (at $60^\circ$ from 1), and the small one has advanced $\alpha$ from 1. The angle between them is

$$\theta = 60^\circ - \alpha = 60^\circ - 7^\circ 30' = 52^\circ 30'.$$

c) In 40 minutes the small hand moves through $\beta$ such that $\dfrac{\beta}{40} = \dfrac{30^\circ}{60}$, that is, $\beta = 20^\circ$. The large pointer is at 8, $150^\circ$ from 1 (5 divisions). Then

$$\varphi = 150^\circ + \beta = 150^\circ + 20^\circ = 170^\circ,$$

or even $\varphi = 180^\circ - \gamma = 180^\circ - 10^\circ = 170^\circ$, where $\gamma = 10^\circ$ is what the small pointer needs to reach 2, opposite to 8.

Check out the three answers on interactive clock.

16

Calculate the smallest of the angles formed by the hands of a clock that reads: a) 2 h 40 min; b) 5 h 55 min; c) 6 h 30 min.

Ver resposta

a) $160^\circ$ · b) $152^\circ 30'$ · c) $15^\circ$

Trigonometric cycle

17

Divide the cycle into 12 equal parts, using $A$ as one of the dividing points. Determine the numbers $x \in [0, 2\pi[$ whose images are the dividing points.

See the resolution

Each part measures $\dfrac{1}{12} \cdot 2\pi = \dfrac{\pi}{6}$, and $P$ is the image of $x$ when $\overset{\frown}{AP} = x$. Like this:

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image of $x$ $A$ $P_1$ $P_2$ $B$ $P_3$ $P_4$ $A'$ $P_5$ $P_6$ $B'$ $P_7$ $P_8$
$x$ $0$ $\frac{\pi}{6}$ $\frac{\pi}{3}$ $\frac{\pi}{2}$ $\frac{2\pi}{3}$ $\frac{5\pi}{6}$ $\pi$ $\frac{7\pi}{6}$ $\frac{4\pi}{3}$ $\frac{3\pi}{2}$ $\frac{5\pi}{3}$ $\frac{11\pi}{6}$
18

Divide the cycle into 8 equal parts, using $A$ as one of the dividing points. Determine the set of numbers $x \in [0, 2\pi[$ whose images are the dividing points.

See the answer
image of $x$ $A$ $P_1$ $B$ $P_2$ $A'$ $P_3$ $B'$ $P_4$
$x$ $0$ $\frac{\pi}{4}$ $\frac{\pi}{2}$ $\frac{3\pi}{4}$ $\pi$ $\frac{5\pi}{4}$ $\frac{3\pi}{2}$ $\frac{7\pi}{4}$
19

Indicate the image of each of the numbers in the cycle: a) $\dfrac{3\pi}{4}$; b) $-\dfrac{5\pi}{4}$; c) $11\pi$; d) $-3\pi$; e) $\dfrac{25\pi}{3}$; f) $-\dfrac{19\pi}{6}$.

See the resolution

a) $\dfrac{3\pi}{4} = \dfrac{3}{8} \cdot 2\pi$: we mark from $A$ a route $\overset{\frown}{AP}$ equal to $\dfrac{3}{8}$ of the cycle, in a counterclockwise direction.

b) $-\dfrac{5\pi}{4} = -\dfrac{5}{8} \cdot 2\pi$: from $A$ we mark a route equal to $\dfrac{5}{8}$ of the cycle, clockwise.

c) $11\pi = \pi + 10\pi$. Since $11\pi - \pi$ is a multiple of $2\pi$, the numbers $11\pi$ and $\pi$ have the same image, $A'$.

d) $-3\pi = \pi - 4\pi$. As $(-3\pi) - \pi$ is a multiple of $2\pi$, $-3\pi$ and $\pi$ have the same image, $A'$.

e) $\dfrac{25\pi}{3} = \dfrac{\pi}{3} + \dfrac{24\pi}{3} = \dfrac{\pi}{3} + 8\pi$. Thus $\dfrac{25\pi}{3}$ and $\dfrac{\pi}{3}$ have the same image $P$, obtained by marking a path $\overset{\frown}{AP}$ equal to $\dfrac{1}{6}$ of the cycle, in a counterclockwise direction.

f) $-\dfrac{19\pi}{6} = \dfrac{5\pi}{6} - \dfrac{24\pi}{6} = \dfrac{5\pi}{6} - 4\pi$. So $-\dfrac{19\pi}{6}$ and $\dfrac{5\pi}{6}$ have the same image. As $\dfrac{5\pi}{6} = \dfrac{5}{12} \cdot 2\pi$, the image is the end of the path $\overset{\frown}{AP}$ equal to $\dfrac{5}{12}$ of the cycle, in a counterclockwise direction.

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20

Indicate in the cycle the images of real numbers $\dfrac{\pi}{8}$, $\dfrac{11\pi}{8}$, $-\dfrac{3\pi}{8}$, $-\dfrac{7\pi}{8}$, $\dfrac{13\pi}{6}$, $-\dfrac{15\pi}{2}$, $\dfrac{17\pi}{4}$ and $-\dfrac{31\pi}{4}$.

See the answer

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Note that $\dfrac{13\pi}{6} = \dfrac{\pi}{6} + 2\pi$, $-\dfrac{15\pi}{2} = \dfrac{\pi}{2} - 8\pi$, $\dfrac{17\pi}{4} = \dfrac{\pi}{4} + 4\pi$ and $-\dfrac{31\pi}{4} = \dfrac{\pi}{4} - 8\pi$.

21

Represent the images of sets of numbers in the cycle

$$E = \left\{x \in \mathbb{R} \;\middle|\; x = \frac{\pi}{2} + k\pi,\ k \in \mathbb{Z}\right\} \qquad\text{e}\qquad F = \left\{x \in \mathbb{R} \;\middle|\; x = k\,\frac{\pi}{2},\ k \in \mathbb{Z}\right\}.$$
See the resolution

For $x = \dfrac{\pi}{2} + k\pi$:

  • $k = 0 \Rightarrow x = \dfrac{\pi}{2}$ (image $B$);
  • $k = 1 \Rightarrow x = \dfrac{3\pi}{2}$ (image $B'$);
  • $k = 2 \Rightarrow x = \dfrac{5\pi}{2}$ (repeats $B$).

The set $E$ has as its image the points $B$ and $B'$.

For $x = k \cdot \dfrac{\pi}{2}$:

  • $k = 0 \Rightarrow x = 0$ (image $A$); $\;k = 1 \Rightarrow x = \dfrac{\pi}{2}$ (image $B$);
  • $k = 2 \Rightarrow x = \pi$ (image $A'$); $\;k = 3 \Rightarrow x = \dfrac{3\pi}{2}$ (image $B'$);
  • $k = 4 \Rightarrow x = 2\pi$ (repeats $A$).

The set $F$ has as its image the points $A$, $B$, $A'$ and $B'$.

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22

Represent the images of sets of real numbers in the cycle

$$E = \{x \in \mathbb{R} \mid x = k\pi,\ k \in \mathbb{Z}\} \qquad F = \left\{x \in \mathbb{R} \;\middle|\; x = \frac{k\pi}{3},\ k \in \mathbb{Z}\right\}$$$$G = \left\{x \in \mathbb{R} \;\middle|\; x = \frac{\pi}{3} + k\pi,\ k \in \mathbb{Z}\right\} \qquad H = \left\{x \in \mathbb{R} \;\middle|\; x = \frac{\pi}{4} + k\,\frac{\pi}{2},\ k \in \mathbb{Z}\right\}$$
See the answer

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$E$: points $A$ and $A'$; $F$: the vertices of a regular hexagon with a vertex at $A$; $G$: the images of $\dfrac{\pi}{3}$ and $\dfrac{4\pi}{3}$, diametrically opposite; $H$: the vertices of a square with vertices at $\dfrac{\pi}{4}$, $\dfrac{3\pi}{4}$, $\dfrac{5\pi}{4}$, and $\dfrac{7\pi}{4}$.