Math - Angles | Practice Hub

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

$56^\circ$

$66^\circ$

$34^\circ$

$146^\circ$

$156^\circ$

Explanation

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

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

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

$x^\circ=56^\circ$

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

$56^\circ$

$66^\circ$

$34^\circ$

$146^\circ$

$156^\circ$

Explanation

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 choices represents a pair of coterminal angles?

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

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

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

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

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

Explanation

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

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

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

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

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

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

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

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

$49^\circ$

$41^\circ$

$-41^\circ$

$229^\circ$

$140^\circ$

Explanation

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

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

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

$x=49^\circ$

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

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

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

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

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

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

Explanation

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.

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

$49^\circ$

$41^\circ$

$-41^\circ$

$229^\circ$

$140^\circ$

Explanation

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

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

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

$x=49^\circ$

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solve for and .

(Figure not drawn to scale).

$\small x=15^o;\ y=37.5^o$

$\small x=7.5^o;\ y=18.75^o$

$\small x=15^o;\ y=52.5^o$

$\small x=15^o;\ y=7.5^o$

Explanation

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.

$\small 2y=5x$

$\small x=15^o$

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

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

Solve for and .

(Figure not drawn to scale).

$\small x=15^o;\ y=37.5^o$

$\small x=7.5^o;\ y=18.75^o$

$\small x=15^o;\ y=52.5^o$

$\small x=15^o;\ y=7.5^o$

Explanation

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.

$\small 2y=5x$

$\small x=15^o$

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

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

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