Angles - Math

Card 0 of 60

What angle is supplementary to $131^\circ$ ?

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Supplementary angles add up to $180^\circ$ . That means:

$x+131^\circ=180^\circ$

$x=180^\circ-131^\circ$

$x=49^\circ$

Solve for and .

(Figure not drawn to scale).

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The angles containing the variable all reside along one line, therefore, their sum must be .

$\small $5x+4x+3x=180^o$$

$\small $12x=180^o$$

$\small $x=15^o$$

Because and are opposite angles, they must be equal.

$\small 2y=5x$

$\small $x=15^o$$

$\small $2y=5(15^o$$)=75^o$$

$\small $y=\frac{75^o$$}{2}$=37.5^o$$

Which one of these is positive in quadrant III?

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The pattern for positive functions is All Student Take Calculus. In quandrant I, all trigonometric functions are positive. In quadrant II, sine is positive. In qudrant III, tangent is positive. In quadrant IV, cosine is positive.

Solve for .

(Figure not drawn to scale).

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The angles are supplementary, therefore, the sum of the angles must equal .

$\small $(4n+22^o$$)+(8n-10^o$$)=180^o$$

$\small $4n+22^o$$+8n-10^o$$=180^o$$

$\small $12n+12^o$$=180^o$$

$\small $12n=168^o$$

$\small $n=14^o$$

Are and supplementary angles?

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Since supplementary angles must add up to , the given angles are indeed supplementary.

Are and complementary angles?

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Complementary angles add up to . Therefore, these angles are complementary.

Which of the following angles is supplementary to $15^\circ$ ?

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When two angles are supplementary, they add up to $180^\circ$ .

For this problem, we can set up an equation and solve for the supplementary angle:

$x^\circ+15^\circ=180^\circ$

$x^\circ=180^\circ-15^\circ$

$x^\circ=165^\circ$

What angle is complementary to $34^\circ$ ?

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Two complementary angles add up to $90^\circ$ .

Therefore, $34^\circ+x^\circ=90^\circ$ .

$x^\circ=90^\circ-34^\circ$

$x^\circ=56^\circ$

Which of the following angles is complementary to $32^\circ$ ?

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Two complementary angles add up to $90^\circ$ .

$90^\circ=32^\circ+x^\circ$

$90^\circ-32^\circ=x^\circ$

$58^\circ=x^\circ$

What angle is supplementary to $142^\circ$ ?

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When two angles are supplementary, they add up to $180^\circ$ .

$142^\circ+x=180^\circ$

Solve for :

$142^\circ+x=180^\circ$

$x=180^\circ-142^\circ$

$x=38^\circ$

Find a coterminal angle for .

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Coterminal angles are angles that, when drawn in the standard position, share a terminal side. You can find these angles by adding or subtracting 360 to the given angle. Thus, the only angle measurement that works from the answers given is .

Which of the following angles is coterminal with $\frac{2\pi}{7}$ ?

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For an angle to be coterminal with $\theta$ , that angle must be of the form $\theta + 2\pi N$ for some integer - or, equivalently, the difference of the angle measures multiplied by $\frac{1}{2\pi}$ must be an integer. We apply this test to all five choices.

$\frac{9\pi}{7}$ :

$\left ( $\frac{1}{2\pi }$ \right ) \left ( $\frac{2\pi}{7}$ - $\frac{9\pi}{7}$ \right )$

$=\left ( $\frac{1}{2\pi }$ \right )\left (- $\frac{7\pi}{7}$ \right ) =\left ( $\frac{1}{2\pi }$ \right ) \pi = $\frac{1}{2 }$$

$\frac{12\pi}{7}$ :

$\left ( $\frac{1}{2\pi }$ \right ) \left ( $\frac{2\pi}{7}$ - $\frac{12\pi}{7}$ \right )$

$=\left ( $\frac{1}{2\pi }$ \right )\left (- $\frac{10\pi}{7}$ \right ) =- $\frac{5}{7}$$

$-$\frac{2\pi}{7}$$ :

$\left ( $\frac{1}{2\pi }$ \right ) \left [$\frac{2\pi}{7}$ -\left ( -$\frac{2\pi}{7}$ \right ) \right ]$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{2\pi}{7}$ + $\frac{2\pi}{7 }$\right )$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{4\pi}{7}$ \right ) = $\frac{2}{7}$$

$-$\frac{9\pi}{7}$$ :

$\left ( $\frac{1}{2\pi }$ \right ) \left [$\frac{2\pi}{7}$ -\left ( -$\frac{9\pi}{7}$ \right ) \right ]$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{2\pi}{7}$ + $\frac{9\pi}{7 }$\right )$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{11\pi}{7}$ \right ) = $\frac{11}{14 }$$

$-$\frac{12\pi}{7}$$ :

$\left ( $\frac{1}{2\pi }$ \right ) \left [$\frac{2\pi}{7}$ -\left ( -$\frac{12\pi}{7}$ \right ) \right ]$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{2\pi}{7}$ + $\frac{12\pi}{7 }$\right )$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{14\pi}{7}$ \right ) = \left ( $\frac{1}{2\pi }$ \right ) \cdot2\pi = 1$

$-$\frac{12\pi}{7}$$ is the correct choice, since only that choice passes our test.

Which of the following angles is coterminal with $\frac{5\pi}{9}$ ?

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For an angle to be coterminal with $\theta$ , that angle must be of the form $\theta + 2\pi N$ for some integer - or, equivalently, the difference of the angle measures multiplied by $\frac{1}{2\pi}$ must be an integer. We apply this test to all four choices.

$\frac{41\pi}{9}$ :

$\left ( $\frac{1}{2\pi }$ \right ) \left ($\frac{41\pi}{9}$- $\frac{5\pi}{9}\right)$

$=\left( $\frac{1}{2\pi }$ \right)\left( $\frac{36\pi}{9}$ \right) =\left( $\frac{1}{2\pi }$ \right)4 \pi =2$

$\frac{23\pi}{9}$ :

$\left( $\frac{1}{2\pi }$ \right) \left($\frac{23\pi}{9}$- $\frac{5\pi}{9}\right )$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{18\pi}{9}$ \right ) =\left ( $\frac{1}{2\pi }$ \right )2 \pi =1$

$-$\frac{13\pi}{9}$$ :

$\left ( $\frac{1}{2\pi }$ \right ) \left (-$\frac{13\pi}{9}$- $\frac{5\pi}{9}\right)$

$=\left( $\frac{1}{2\pi }$ \right)\left( $\frac{-18\pi}{9}$ \right) =\left( $\frac{1}{2\pi }$ \right)\left(- 2 \pi \right)=-1$

$-$\frac{31\pi}{9}$$ :

$\left( $\frac{1}{2\pi }$ \right) \left(-$\frac{31\pi}{9}$- $\frac{5\pi}{9}\right )$

$=\left ( $\frac{1}{2\pi }$ \right )\left ( $\frac{-36\pi}{9}$ \right ) =\left ( $\frac{1}{2\pi }$ \right )\left (- 4 \pi \right )=-2$

All four choices pass the test, so all four angles are coterminal with $\frac{5\pi}{9}$ .

Which of the following choices represents a pair of coterminal angles?

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For two angles to be coterminal, they must differ by $2\pi N$ for some integer - or, equivalently, the difference of the angle measures multiplied by $\frac{1}{2\pi}$ must be an integer. We apply this test to all five choices.

$\frac{17\pi }{2}$ \textrm{ and } $\frac{23\pi }{2}$ :

$\frac{1}{2\pi }$ \cdot \left ( $\frac{23\pi }{2}$ - $\frac{17\pi }{2}$ \right ) = $\frac{1}{2\pi }$ \cdot $\frac{6\pi }{2}$ = $\frac{3 }{2}$

$\frac{17\pi }{3}$ \textrm{ and } $\frac{23\pi }{3}$ :

$$\frac{1}{2\pi }$ \cdot \left ( $\frac{23\pi }{3}$ - $\frac{17\pi }{3}$ \right ) = $\frac{1}{2\pi }$ \cdot $\frac{6\pi }{3}$ = 1$

$\frac{17\pi }{4}$ \textrm{ and } $\frac{23\pi }{4}$ :

$\frac{1}{2\pi }$ \cdot \left ( $\frac{23\pi }{4}$ - $\frac{17\pi }{4}$ \right ) = $\frac{1}{2\pi }$ \cdot $\frac{6\pi }{4}$ = $\frac{3 }{4}$

$\frac{17\pi }{5}$ \textrm{ and } $\frac{23\pi }{5}$ :

$\frac{1}{2\pi }$ \cdot \left ( $\frac{23\pi }{5}$ - $\frac{17\pi }{5}$ \right ) = $\frac{1}{2\pi }$ \cdot $\frac{6\pi }{5}$ = $\frac{3 }{5}$

$\frac{17\pi }{6}$ \textrm{ and } $\frac{23\pi }{6}$ :

$\frac{1}{2\pi }$ \cdot \left ( $\frac{23\pi }{6}$ - $\frac{17\pi }{6}$ \right ) = $\frac{1}{2\pi }$ \cdot $\frac{6\pi }{6}$ = $\frac{1 }{2}$

The only angles that pass the test - and are therefore coterminal - are $\frac{17\pi }{3}$ \textrm{ and } $\frac{23\pi }{3}$ .

$\textup{Which of the following angle measures, given in radians, is coterminal with an}$

$\textup{an angle of 90 degrees?}$

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$90\textup{ degrees }=\frac{\pi }{2}$:: \textup{ Coterminal angles share the same position on the unit circle.}$

$\textup{Therefore, coterminal angles can be found by adding or subtracting multiples of }2\pi$ .

$\frac{\pi }{2}$+2\pi=\frac{5\pi }{2}$

What angle is supplementary to $131^\circ$ ?

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Supplementary angles add up to $180^\circ$ . That means:

$x+131^\circ=180^\circ$

$x=180^\circ-131^\circ$

$x=49^\circ$

Solve for and .

(Figure not drawn to scale).

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The angles containing the variable all reside along one line, therefore, their sum must be .

$\small $5x+4x+3x=180^o$$

$\small $12x=180^o$$

$\small $x=15^o$$

Because and are opposite angles, they must be equal.

$\small 2y=5x$

$\small $x=15^o$$

$\small $2y=5(15^o$$)=75^o$$

$\small $y=\frac{75^o$$}{2}$=37.5^o$$

Which one of these is positive in quadrant III?

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The pattern for positive functions is All Student Take Calculus. In quandrant I, all trigonometric functions are positive. In quadrant II, sine is positive. In qudrant III, tangent is positive. In quadrant IV, cosine is positive.

Solve for .

(Figure not drawn to scale).

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The angles are supplementary, therefore, the sum of the angles must equal .

$\small $(4n+22^o$$)+(8n-10^o$$)=180^o$$

$\small $4n+22^o$$+8n-10^o$$=180^o$$

$\small $12n+12^o$$=180^o$$

$\small $12n=168^o$$

$\small $n=14^o$$

Are and supplementary angles?

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Since supplementary angles must add up to , the given angles are indeed supplementary.