Exercises from the chapter 3 — Angles
Exercises and resolutions
The calculator below does, step by step, the calculations with degrees, minutes and seconds for exercises 29 to 33. Try solving it first and use it to check.
Simplify the following measurements: a) $30^{\circ}70^{\prime}$ · b) $45^{\circ}150^{\prime}$ · c) $65^{\circ}39^{\prime}123^{\prime\prime}$ · d) $110^{\circ}58^{\prime}300^{\prime\prime}$ · e) $30^{\circ}56^{\prime}240^{\prime\prime}$
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a) $31^{\circ}10^{\prime}$ · b) $47^{\circ}30^{\prime}$ · c) $65^{\circ}41^{\prime}3^{\prime\prime}$ · d) $111^{\circ}3^{\prime}$ · e) $31^{\circ}$
Determine the sums: a) $30^{\circ}40^{\prime} + 15^{\circ}35^{\prime}$ · b) $10^{\circ}30^{\prime}45^{\prime\prime} + 15^{\circ}29^{\prime}20^{\prime\prime}$
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a) $46^{\circ}15^{\prime}$ · b) $26^{\circ}0^{\prime}5^{\prime\prime}$
Determine the differences: a) $20^{\circ}50^{\prime}45^{\prime\prime} - 5^{\circ}45^{\prime}30^{\prime\prime}$ · b) $31^{\circ}40^{\prime} - 20^{\circ}45^{\prime}$ · c) $90^{\circ}15^{\prime}20^{\prime\prime} - 45^{\circ}30^{\prime}50^{\prime\prime}$ · d) $90^{\circ} - 50^{\circ}30^{\prime}45^{\prime\prime}$
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a) $15^{\circ}5^{\prime}15^{\prime\prime}$ · b) $10^{\circ}55^{\prime}$ · c) $44^{\circ}44^{\prime}30^{\prime\prime}$ · d) $39^{\circ}29^{\prime}15^{\prime\prime}$
Determine the products: a) $2 \times (10^{\circ}35^{\prime}45^{\prime\prime})$ · b) $5 \times (6^{\circ}15^{\prime}30^{\prime\prime})$
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a) $21^{\circ}11^{\prime}30^{\prime\prime}$ · b) $31^{\circ}17^{\prime}30^{\prime\prime}$
Determine the divisions: a) $(46^{\circ}48^{\prime}54^{\prime\prime}) \div 2$ · b) $(31^{\circ}32^{\prime}45^{\prime\prime}) \div 3$ · c) $(52^{\circ}63^{\prime}42^{\prime\prime}) \div 5$
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a) $23^{\circ}24^{\prime}27^{\prime\prime}$ · b) $10^{\circ}30^{\prime}55^{\prime\prime}$ · c) $10^{\circ}36^{\prime}44{,}4^{\prime\prime}$
Determine the value of $x$ in the following cases:
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a) $20^{\circ}$ · b) $55^{\circ}$ · c) $60^{\circ}$ · d) $23^{\circ}$ · e) $25^{\circ}$
$\overrightarrow{Oa}$ and $\overrightarrow{Ob}$ are two opposite collinear rays. $\overrightarrow{Oc}$ is any ray. Are angles $a\hat{O}c$ and $c\hat{O}b$ adjacent? Are they supplementary?
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They are adjacent and supplementary.
Prove: If two angles are vertex-opposed, then they are congruent.
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It is the theorem of angles opposite the vertex, demonstrated above: $x + z = 180^{\circ}$ and $y + z = 180^{\circ}$ give $x = y$.
Determine the value of $x$ in the following cases:
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a) $25^{\circ}$ · b) $30^{\circ}$
Determine the value of $\alpha$ in the following cases:
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a) $(2x - 10^{\circ}) + (x + 40^{\circ}) = 90^{\circ} \Rightarrow x = 20^{\circ}$, then $\alpha = 60^{\circ}$.
b) Opposite at the vertex: $3x - 15^{\circ} = x + 35^{\circ} \Rightarrow x = 25^{\circ}$, each measures $60^{\circ}$; $\alpha$ is supplementary: $\alpha = 120^{\circ}$.
c) $x + y = 2x - y \Rightarrow x = 2y$; and $(4x - 2y) + (x + y) = 180^{\circ}$. Hence $y = 20^{\circ}$, $x = 40^{\circ}$ and $\alpha = 4x - 2y = 120^{\circ}$.
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a) $\alpha = 60^{\circ}$ · b) $\alpha = 120^{\circ}$ · c) $\alpha = 120^{\circ}$
If $\overrightarrow{OP}$ is angle bisector of $A\hat{O}B$, determine $x$ in the cases:
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a) $x = 15^{\circ}$ · b) $x = 10^{\circ}$
Exercise 40 — Classify as true (V) or false (F)
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Two consecutive angles are adjacent.
Falso. They may have common internal points (one “inside” the other).
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Two adjacent angles are consecutive.
Verdadeiro. It is part of the definition of adjacent.
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Two adjacent angles are opposed by the vertex.
Falso. o.p.v angles they have no common side.
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Two angles opposite at the vertex are adjacent.
Falso. They are not even consecutive.
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Two angles opposite the vertex are consecutive.
Falso. The sides of one are opposite to those of the other; no side is common.
Exercise 41 — Classify as true (V) or false (F)
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Two supplementary angles are adjacent.
Falso. Being supplementary only depends on the measures; angles can be anywhere.
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Two complementary angles are adjacent.
Falso. Idem: see exercise 42.
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Two adjacent angles are complementary.
Falso. The sum can be any value less than $360^{\circ}$.
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The measuring angles $10^{\circ}$, $20^{\circ}$ and $60^{\circ}$ are complementary.
Falso. The definition is for two angles (and besides, the sum is $90^{\circ}$ just by coincidence).
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The measuring angles $30^{\circ}$, $60^{\circ}$ and $90^{\circ}$ are supplementary.
Falso. There are also three angles.
Are the angles in the following figure complementary? Are they adjacent?
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They are complementary ($60^{\circ} + 30^{\circ} = 90^{\circ}$), but they are not adjacent.
Calculate the value of $x$ in the case below, where $m(r\hat{O}s) = 90^{\circ}$.
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$x = 10^{\circ}$, because $2x + 3x + 4x = 90^{\circ}$.
The sum of two adjacent angles is $120^{\circ}$. Calculate the measure of each angle, knowing that the measure of one of them is the difference between three times the other and $40^{\circ}$.
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$40^{\circ}$ e $80^{\circ}$.
Calculate the complement of the following angles: a) $25^{\circ}$ · b) $47^{\circ}$ · c) $37^{\circ}25^{\prime}$
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a) $65^{\circ}$ · b) $43^{\circ}$ · c) $52^{\circ}35^{\prime}$
Calculate the supplement of the following angles: a) $72^{\circ}$ · b) $141^{\circ}$ · c) $93^{\circ}15^{\prime}$
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a) $108^{\circ}$ · b) $39^{\circ}$ · c) $86^{\circ}45^{\prime}$
Given a measuring angle $x$, indicate: a) its complement; b) its supplement; c) twice its complement; d) half of your supplement; e) triple your supplement; f) the seventh part of the complement; g) the fifth part of the supplement; h) the complement of its third part; i) triple the supplement of its fifth part.
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a) $90^{\circ} - x$ · b) $180^{\circ} - x$ · c) $2(90^{\circ} - x)$ · d) $\dfrac{180^{\circ} - x}{2}$ · e) $3(180^{\circ} - x)$ · f) $\dfrac{90^{\circ} - x}{7}$ · g) $\dfrac{180^{\circ} - x}{5}$ · h) $90^{\circ} - \dfrac{x}{3}$ · i) $3\left(180^{\circ} - \dfrac{x}{5}\right)$
ExemploHow to translate problems
Call the angle $x$. So the complement is $90^{\circ} - x$ and the supplement is $180^{\circ} - x$. Translate the sentence into an equation. In exercise 52, “the angle exceeds its complement in $76^{\circ}$” becomes $x - (90^{\circ} - x) = 76^{\circ}$, therefore $2x = 166^{\circ}$ and $x = 83^{\circ}$.
Give the measure of the angle that is twice its complement.
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$60^{\circ}$
Determine the measure of the angle equal to three times its complement.
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$67^{\circ}30^{\prime}$
Calculate the angle that is four times its complement.
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$72^{\circ}$
Calculate an angle, knowing that a quarter of your supplement is worth $36^{\circ}$.
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$36^{\circ}$
What is the angle that exceeds its complement by $76^{\circ}$?
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$83^{\circ}$ (resolved in the example above).
What is the angle that exceeds your supplement by $66^{\circ}$?
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$123^{\circ}$
Determine an angle, knowing that its supplement exceeds the angle itself by $70^{\circ}$.
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$55^{\circ}$
What is the angle that added to triple its complement gives $210^{\circ}$?
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$30^{\circ}$
An angle exceeds its complement by $48^{\circ}$. Determine the supplement of this angle.
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$111^{\circ}$ (angle measures $69^{\circ}$).
The supplement of an angle exceeds this angle by $120^{\circ}$. Determine the angle.
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$30^{\circ}$
The third’s complement of an angle exceeds the complement of that angle by $30^{\circ}$. Determine the angle.
See the resolution of the book
Angle: $x$; angle complement: $90^{\circ} - x$; complement of the third part: $90^{\circ} - \dfrac{x}{3}$.
$$\left(90^{\circ} - \frac{x}{3}\right) - (90^{\circ} - x) = 30^{\circ} \ \Longrightarrow\ \frac{2x}{3} = 30^{\circ} \ \Longrightarrow\ x = 45^{\circ}.$$Ver resposta
$45^{\circ}$
The supplement of the triple of the half’s complement of an angle is equal to the triple of the complement of that angle. Determine the angle.
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$80^{\circ}$
The supplement of the complement of an angle exceeds one-third of the complement of the double of that angle in $85^{\circ}$. Determine the angle.
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$15^{\circ}$
Two angles are supplementary and the ratio between the complement of one and the supplement of the other, in that order, is $\dfrac{1}{8}$. Determine these angles.
See the resolution of the book
Let $x$ and $y$ be the measurements. One’s complement: $90^{\circ} - x$; supplement of the other: $180^{\circ} - y$.
$$x + y = 180^{\circ} \quad\text{e}\quad \frac{90^{\circ} - x}{180^{\circ} - y} = \frac{1}{8} \ \Longrightarrow\ 720^{\circ} - 8x = 180^{\circ} - y.$$With $y = 180^{\circ} - x$: $720^{\circ} - 8x = x$, then $x = 80^{\circ}$ and $y = 100^{\circ}$.
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$80^{\circ}$ e $100^{\circ}$
Two angles are in the relationship $\dfrac{4}{9}$. Being $130^{\circ}$ its sum, determine the complement of the smallest.
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$50^{\circ}$ (angles measure $40^{\circ}$ and $90^{\circ}$).
Determine two supplementary angles, knowing that one of them is triple the other.
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$135^{\circ}$ e $45^{\circ}$
Two angles are supplementary. One of them is the complement of the fourth part of the other. Calculate these angles.
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$60^{\circ}$ e $120^{\circ}$
The ratio between two supplementary angles is equal to $\dfrac{2}{7}$. Determine the complement of the minor.
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$50^{\circ}$ (angles measure $40^{\circ}$ and $140^{\circ}$).
Determine the complement of an angle, knowing that the ratio between the angle and its complement is equal to $\dfrac{5}{4}$.
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$40^{\circ}$ (angle measures $50^{\circ}$).
The complement of an angle is to its supplement as 2 to 7. Calculate the measure of the angle.
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$54^{\circ}$
Triple the complement of an angle, increased by $50^{\circ}$, is equal to the supplement of the angle. Determine the measure of the angle.
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$70^{\circ}$
Determine the measures of two supplementary angles, knowing that twice one of them, added to the seventh part of the other, results in $100^{\circ}$.
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$40^{\circ}$ e $140^{\circ}$
The sum of an angle and the third part of its complement results in $46^{\circ}$. Determine the supplement of this angle.
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$156^{\circ}$ (angle measures $24^{\circ}$).
Determine two complementary angles such that twice one, increased by a third of the other, is equal to a right angle.
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$36^{\circ}$ e $54^{\circ}$
In the figure, the angle $x$ measures one-sixth of the angle $y$, plus half of the angle $z$. Calculate the angle $y$.
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$x$ and $z$ are opposite by the vertex, so $z = x$. From the condition, $x = \dfrac{y}{6} + \dfrac{x}{2}$, that is, $y = 3x$. As $x + y = 180^{\circ}$, comes $4x = 180^{\circ}$, $x = 45^{\circ}$ and $y = 135^{\circ}$.
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$y = 135^{\circ}$
The angles $\alpha$ and $\beta$ are opposite at the vertex. The first is expressed in degrees by $9x - 2$ and the second by $4x + 8$. Determine these angles.
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$16^{\circ}$ e $16^{\circ}$ ($x = 2$).
Five rays start from the same point $V$, forming five angles that cover the entire plane and are proportional to the numbers 2, 3, 4, 5 and 6. Calculate the largest angle.
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$108^{\circ}$
Show that the angle bisectors of two angles opposite the vertex are opposite rays. (Use the “angle bisectors of x and y” button in the o.p.v. angles figure.)
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The angle bisectors form an angle of $x/2 + (180^{\circ} - x) + x/2 = 180^{\circ}$; two rays with the same origin that form an angle of $180^{\circ}$ are opposite.
Show that the angle bisectors of two adjacent and supplementary angles form right angles.
See the resolution of the book
Hypothesis: $r\hat{O}s$ and $s\hat{O}t$ adjacent and supplementary; $\overrightarrow{Ox}$ and $\overrightarrow{Oy}$ their angle bisectors. Thesis: $x\hat{O}y$ is straight.
Let $a$ be the measure of $r\hat{O}x$ and $x\hat{O}s$, and $b$ be the measure of $s\hat{O}y$ and $y\hat{O}t$. Then
$$a + a + b + b = 180^{\circ} \ \Longrightarrow\ 2a + 2b = 180^{\circ} \ \Longrightarrow\ a + b = 90^{\circ},$$therefore $x\hat{O}y$ is straight.
Show that the angle bisectors of two adjacent and complementary angles form an angle of $45^{\circ}$.
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$a + a + b + b = 90^{\circ} \Rightarrow a + b = 45^{\circ}$.
Two adjacent angles add up to $136^{\circ}$. What is the measure of the angle formed by your angle bisectors?
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$68^{\circ}$
The angle bisectors of two consecutive angles form an angle of $52^{\circ}$. If one of them measures $40^{\circ}$, what is the measurement of the other?
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$64^{\circ}$ or $144^{\circ}$, depending on whether the angles are adjacent or not.