Solving Right Triangles: Pythagorean Theorem, Trigonometry - Pre-Calculus
Card 1 of 30
What is $\cos(90^\circ-\theta)$ in terms of $\sin(\theta)$?
What is $\cos(90^\circ-\theta)$ in terms of $\sin(\theta)$?
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$\cos(90^\circ-\theta)=\sin(\theta)$. The cosine of an angle's complement equals the angle's sine.
$\cos(90^\circ-\theta)=\sin(\theta)$. The cosine of an angle's complement equals the angle's sine.
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What is $\sin(90^\circ-\theta)$ in terms of $\cos(\theta)$?
What is $\sin(90^\circ-\theta)$ in terms of $\cos(\theta)$?
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$\sin(90^\circ-\theta)=\cos(\theta)$. The sine of an angle's complement equals the angle's cosine.
$\sin(90^\circ-\theta)=\cos(\theta)$. The sine of an angle's complement equals the angle's cosine.
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State the complementary-angle identity that relates $\cos(\theta)$ and $\sin(90^\circ-\theta)$.
State the complementary-angle identity that relates $\cos(\theta)$ and $\sin(90^\circ-\theta)$.
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$\cos(\theta)=\sin(90^\circ-\theta)$. Cosine of an angle equals sine of its complement.
$\cos(\theta)=\sin(90^\circ-\theta)$. Cosine of an angle equals sine of its complement.
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State the complementary-angle identity that relates $\sin(\theta)$ and $\cos(90^\circ-\theta)$.
State the complementary-angle identity that relates $\sin(\theta)$ and $\cos(90^\circ-\theta)$.
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$\sin(\theta)=\cos(90^\circ-\theta)$. Sine of an angle equals cosine of its complement.
$\sin(\theta)=\cos(90^\circ-\theta)$. Sine of an angle equals cosine of its complement.
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Given $\cos(\beta)=\frac{12}{13}$ and $\beta$ is acute, find $\sin(90^\circ-\beta)$.
Given $\cos(\beta)=\frac{12}{13}$ and $\beta$ is acute, find $\sin(90^\circ-\beta)$.
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$\frac{12}{13}$. Using $\sin(90°-\beta)=\cos(\beta)$.
$\frac{12}{13}$. Using $\sin(90°-\beta)=\cos(\beta)$.
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In a right triangle, if acute angles are $A$ and $B$, what is the relationship between $A$ and $B$?
In a right triangle, if acute angles are $A$ and $B$, what is the relationship between $A$ and $B$?
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$A+B=90^\circ$. The two acute angles in a right triangle are complementary.
$A+B=90^\circ$. The two acute angles in a right triangle are complementary.
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In a right triangle, which ratio equals $\sin(A)$ in terms of $\cos(B)$ when $A$ and $B$ are complementary?
In a right triangle, which ratio equals $\sin(A)$ in terms of $\cos(B)$ when $A$ and $B$ are complementary?
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$\sin(A)=\cos(B)$. In a right triangle, opposite/hypotenuse for one angle equals adjacent/hypotenuse for the other.
$\sin(A)=\cos(B)$. In a right triangle, opposite/hypotenuse for one angle equals adjacent/hypotenuse for the other.
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In a right triangle, which ratio equals $\cos(A)$ in terms of $\sin(B)$ when $A$ and $B$ are complementary?
In a right triangle, which ratio equals $\cos(A)$ in terms of $\sin(B)$ when $A$ and $B$ are complementary?
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$\cos(A)=\sin(B)$. In a right triangle, adjacent/hypotenuse for one angle equals opposite/hypotenuse for the other.
$\cos(A)=\sin(B)$. In a right triangle, adjacent/hypotenuse for one angle equals opposite/hypotenuse for the other.
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Find the complement of $17^\circ$ (the angle that adds to $17^\circ$ to make $90^\circ$).
Find the complement of $17^\circ$ (the angle that adds to $17^\circ$ to make $90^\circ$).
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$73^\circ$. Complementary angles sum to $90°$, so $90°-17°=73°$.
$73^\circ$. Complementary angles sum to $90°$, so $90°-17°=73°$.
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If $\theta$ is acute, what is the complement of $\theta$ written in degrees?
If $\theta$ is acute, what is the complement of $\theta$ written in degrees?
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$90^\circ-\theta$. The complement is the angle that adds to $\theta$ to make $90°$.
$90^\circ-\theta$. The complement is the angle that adds to $\theta$ to make $90°$.
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Find and correct the identity error: $\sin(\theta)=\sin(90^\circ-\theta)$.
Find and correct the identity error: $\sin(\theta)=\sin(90^\circ-\theta)$.
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Correct: $\sin(\theta)=\cos(90^\circ-\theta)$. The error uses sine on both sides; the correct identity uses cosine on the right.
Correct: $\sin(\theta)=\cos(90^\circ-\theta)$. The error uses sine on both sides; the correct identity uses cosine on the right.
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Given $\sin(\alpha)=\frac{3}{5}$ and $\alpha$ is acute, find $\cos(90^\circ-\alpha)$.
Given $\sin(\alpha)=\frac{3}{5}$ and $\alpha$ is acute, find $\cos(90^\circ-\alpha)$.
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$\frac{3}{5}$. Using $\cos(90°-\alpha)=\sin(\alpha)$.
$\frac{3}{5}$. Using $\cos(90°-\alpha)=\sin(\alpha)$.
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If $\cos(\theta)=0.6$ and $\theta$ is acute, what is $\sin(90^\circ-\theta)$?
If $\cos(\theta)=0.6$ and $\theta$ is acute, what is $\sin(90^\circ-\theta)$?
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$0.6$. By the cofunction identity, $\sin(90°-\theta)=\cos(\theta)$.
$0.6$. By the cofunction identity, $\sin(90°-\theta)=\cos(\theta)$.
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If $\sin(\theta)=0.8$ and $\theta$ is acute, what is $\cos(90^\circ-\theta)$?
If $\sin(\theta)=0.8$ and $\theta$ is acute, what is $\cos(90^\circ-\theta)$?
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$0.8$. By the cofunction identity, $\cos(90°-\theta)=\sin(\theta)$.
$0.8$. By the cofunction identity, $\cos(90°-\theta)=\sin(\theta)$.
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Evaluate $\cos(90^\circ-35^\circ)$ by rewriting it as a sine function.
Evaluate $\cos(90^\circ-35^\circ)$ by rewriting it as a sine function.
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$\sin(35^\circ)$. $90°-35°=55°$, and $\cos(55°)=\sin(35°)$ by the cofunction identity.
$\sin(35^\circ)$. $90°-35°=55°$, and $\cos(55°)=\sin(35°)$ by the cofunction identity.
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Evaluate $\sin(90^\circ-20^\circ)$ by rewriting it as a cosine function.
Evaluate $\sin(90^\circ-20^\circ)$ by rewriting it as a cosine function.
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$\cos(20^\circ)$. $90°-20°=70°$, and $\sin(70°)=\cos(20°)$ by the cofunction identity.
$\cos(20^\circ)$. $90°-20°=70°$, and $\sin(70°)=\cos(20°)$ by the cofunction identity.
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What is $\cos(30^\circ)$ using a complementary-angle relationship with $\sin(60^\circ)$?
What is $\cos(30^\circ)$ using a complementary-angle relationship with $\sin(60^\circ)$?
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$\cos(30^\circ)=\sin(60^\circ)$. $30°$ and $60°$ are complementary, so their cosine and sine are equal.
$\cos(30^\circ)=\sin(60^\circ)$. $30°$ and $60°$ are complementary, so their cosine and sine are equal.
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What is $\sin(30^\circ)$ using a complementary-angle relationship with $\cos(60^\circ)$?
What is $\sin(30^\circ)$ using a complementary-angle relationship with $\cos(60^\circ)$?
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$\sin(30^\circ)=\cos(60^\circ)$. $30°$ and $60°$ are complementary, so their sine and cosine are equal.
$\sin(30^\circ)=\cos(60^\circ)$. $30°$ and $60°$ are complementary, so their sine and cosine are equal.
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Identify the cofunction pair: which function equals $\cos(\theta)$ for complementary angles?
Identify the cofunction pair: which function equals $\cos(\theta)$ for complementary angles?
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$\sin(90^\circ-\theta)$. Cosine and sine are cofunctions for complementary angles.
$\sin(90^\circ-\theta)$. Cosine and sine are cofunctions for complementary angles.
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Identify the cofunction pair: which function equals $\sin(\theta)$ for complementary angles?
Identify the cofunction pair: which function equals $\sin(\theta)$ for complementary angles?
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$\cos(90^\circ-\theta)$. Sine and cosine are cofunctions for complementary angles.
$\cos(90^\circ-\theta)$. Sine and cosine are cofunctions for complementary angles.
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Find $\sin(45^\circ)$ by rewriting it as a cosine of a complementary angle.
Find $\sin(45^\circ)$ by rewriting it as a cosine of a complementary angle.
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$\sin(45^\circ)=\cos(45^\circ)=\frac{\sqrt{2}}{2}$. $45°$ is its own complement, so $\sin(45°)=\cos(45°)$.
$\sin(45^\circ)=\cos(45^\circ)=\frac{\sqrt{2}}{2}$. $45°$ is its own complement, so $\sin(45°)=\cos(45°)$.
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Identify the relationship between acute complementary angles $A$ and $B$ in a right triangle.
Identify the relationship between acute complementary angles $A$ and $B$ in a right triangle.
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$A+B=90^\circ$. Complementary angles sum to $90°$ by definition.
$A+B=90^\circ$. Complementary angles sum to $90°$ by definition.
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Identify the missing expression to make a true identity: $\cos(\theta)=\sin(,\underline{\hspace{1.2cm}},)$.
Identify the missing expression to make a true identity: $\cos(\theta)=\sin(,\underline{\hspace{1.2cm}},)$.
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$90^\circ-\theta$. To make $\cos(\theta)=\sin(?)$, the angle must be $\theta$'s complement.
$90^\circ-\theta$. To make $\cos(\theta)=\sin(?)$, the angle must be $\theta$'s complement.
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Find and correct the error: $\sin(\theta)=\cos(90^\circ+\theta)$ for acute $\theta$.
Find and correct the error: $\sin(\theta)=\cos(90^\circ+\theta)$ for acute $\theta$.
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Correct: $\sin(\theta)=\cos(90^\circ-\theta)$. The error uses addition instead of subtraction for complementary angles.
Correct: $\sin(\theta)=\cos(90^\circ-\theta)$. The error uses addition instead of subtraction for complementary angles.
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Which option correctly rewrites $\cos(20^\circ)$ using sine of a complementary angle?
Which option correctly rewrites $\cos(20^\circ)$ using sine of a complementary angle?
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$\cos(20^\circ)=\sin(70^\circ)$. Since $20°+70°=90°$, they are complementary angles.
$\cos(20^\circ)=\sin(70^\circ)$. Since $20°+70°=90°$, they are complementary angles.
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State the complementary-angle identity that rewrites $\sin(\theta)$ using cosine.
State the complementary-angle identity that rewrites $\sin(\theta)$ using cosine.
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$\sin(\theta)=\cos(90^\circ-\theta)$. Complementary angles sum to $90°$, so sine of one equals cosine of the other.
$\sin(\theta)=\cos(90^\circ-\theta)$. Complementary angles sum to $90°$, so sine of one equals cosine of the other.
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State the complementary-angle identity that rewrites $\cos(\theta)$ using sine.
State the complementary-angle identity that rewrites $\cos(\theta)$ using sine.
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$\cos(\theta)=\sin(90^\circ-\theta)$. Complementary angles sum to $90°$, so cosine of one equals sine of the other.
$\cos(\theta)=\sin(90^\circ-\theta)$. Complementary angles sum to $90°$, so cosine of one equals sine of the other.
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What is $\sin(90^\circ-\theta)$ equal to in terms of $\theta$?
What is $\sin(90^\circ-\theta)$ equal to in terms of $\theta$?
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$\sin(90^\circ-\theta)=\cos(\theta)$. Since $\theta$ and $90°-\theta$ are complementary, sine becomes cosine.
$\sin(90^\circ-\theta)=\cos(\theta)$. Since $\theta$ and $90°-\theta$ are complementary, sine becomes cosine.
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What is $\cos(90^\circ-\theta)$ equal to in terms of $\theta$?
What is $\cos(90^\circ-\theta)$ equal to in terms of $\theta$?
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$\cos(90^\circ-\theta)=\sin(\theta)$. Since $\theta$ and $90°-\theta$ are complementary, cosine becomes sine.
$\cos(90^\circ-\theta)=\sin(\theta)$. Since $\theta$ and $90°-\theta$ are complementary, cosine becomes sine.
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In a right triangle with acute complementary angles $A$ and $B$, what is $\sin(A)$ equal to?
In a right triangle with acute complementary angles $A$ and $B$, what is $\sin(A)$ equal to?
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$\sin(A)=\cos(B)$. Since $A+B=90°$, sine of one angle equals cosine of its complement.
$\sin(A)=\cos(B)$. Since $A+B=90°$, sine of one angle equals cosine of its complement.
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