Precalculus Flashcards: Solving Right Triangles Pythagorean Theorem Trigonometry

What this deck covers

This deck focuses on Solving Right Triangles Pythagorean Theorem Trigonometry, giving you a quick way to review the definitions, rules, and examples that matter most for Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: If $\cos(\theta)=0.35$ and $\theta$ is acute, what is $\sin(90^\circ-\theta)$?

Answer: $\sin(90^\circ-\theta)=0.35$. By the complementary angle identity, $\sin(90°-\theta)=\cos(\theta)$.

Flashcard 2: What is $\sin(90^\circ-\theta)$ equal to in terms of $\theta$?

Answer: $\sin(90^\circ-\theta)=\cos(\theta)$. Since $\theta$ and $90°-\theta$ are complementary, sine becomes cosine.

Flashcard 3: Which option correctly rewrites $\cos(20^\circ)$ using sine of a complementary angle?

Answer: $\cos(20^\circ)=\sin(70^\circ)$. Since $20°+70°=90°$, they are complementary angles.

Flashcard 4: Find $\sin(30^\circ)$ using a complementary-angle identity and a known cosine value.

Answer: $\sin(30^\circ)=\cos(60^\circ)=\frac{1}{2}$. Since $30°$ and $60°$ are complementary, $\sin(30°)=\cos(60°)$.

Flashcard 5: In a right triangle, which ratio equals $\sin(A)$ in terms of $\cos(B)$ when $A$ and $B$ are complementary?

Answer: $\sin(A)=\cos(B)$. In a right triangle, opposite/hypotenuse for one angle equals adjacent/hypotenuse for the other.

Flashcard 6: If $\sin(\theta)=0.8$ and $\theta$ is acute, what is $\cos(90^\circ-\theta)$?

Answer: $0.8$. By the cofunction identity, $\cos(90°-\theta)=\sin(\theta)$.

Flashcard 7: In a right triangle with acute complementary angles $A$ and $B$, what is $\cos(A)$ equal to?

Answer: $\cos(A)=\sin(B)$. Since $A+B=90°$, cosine of one angle equals sine of its complement.

Flashcard 8: Find and correct the error: $\sin(\theta)=\cos(90^\circ+\theta)$ for acute $\theta$.

Answer: Correct: $\sin(\theta)=\cos(90^\circ-\theta)$. The error uses addition instead of subtraction for complementary angles.

Flashcard 9: What is $\sin(90^\circ-\theta)$ in terms of $\cos(\theta)$?

Answer: $\sin(90^\circ-\theta)=\cos(\theta)$. The sine of an angle's complement equals the angle's cosine.

Flashcard 10: In a right triangle with acute complementary angles $A$ and $B$, what is $\sin(A)$ equal to?

Answer: $\sin(A)=\cos(B)$. Since $A+B=90°$, sine of one angle equals cosine of its complement.

Flashcard 11: Find the complement of $17^\circ$ (the angle that adds to $17^\circ$ to make $90^\circ$).

Answer: $73^\circ$. Complementary angles sum to $90°$, so $90°-17°=73°$.

Flashcard 12: Find $\sin(60^\circ)$ using a complementary-angle identity and a known cosine value.

Answer: $\sin(60^\circ)=\cos(30^\circ)=\frac{\sqrt{3}}{2}$. Since $60°$ and $30°$ are complementary, $\sin(60°)=\cos(30°)$.

Flashcard 13: Evaluate $\cos(90^\circ-35^\circ)$ by rewriting it as a sine function.

Answer: $\sin(35^\circ)$. $90°-35°=55°$, and $\cos(55°)=\sin(35°)$ by the cofunction identity.

Flashcard 14: State the complementary-angle identity that rewrites $\sin(\theta)$ using cosine.

Answer: $\sin(\theta)=\cos(90^\circ-\theta)$. Complementary angles sum to $90°$, so sine of one equals cosine of the other.

Flashcard 15: In a right triangle, which ratio equals $\cos(A)$ in terms of $\sin(B)$ when $A$ and $B$ are complementary?

Answer: $\cos(A)=\sin(B)$. In a right triangle, adjacent/hypotenuse for one angle equals opposite/hypotenuse for the other.

Flashcard 16: Identify the cofunction pair: which function equals $\sin(\theta)$ for complementary angles?

Answer: $\cos(90^\circ-\theta)$. Sine and cosine are cofunctions for complementary angles.

Flashcard 17: If $\cos(\theta)=0.6$ and $\theta$ is acute, what is $\sin(90^\circ-\theta)$?

Answer: $0.6$. By the cofunction identity, $\sin(90°-\theta)=\cos(\theta)$.

Flashcard 18: State the complementary-angle identity that relates $\cos(\theta)$ and $\sin(90^\circ-\theta)$.

Answer: $\cos(\theta)=\sin(90^\circ-\theta)$. Cosine of an angle equals sine of its complement.

Flashcard 19: Find and correct the identity error: $\sin(\theta)=\sin(90^\circ-\theta)$.

Answer: Correct: $\sin(\theta)=\cos(90^\circ-\theta)$. The error uses sine on both sides; the correct identity uses cosine on the right.

Flashcard 20: State the complementary-angle identity that relates $\sin(\theta)$ and $\cos(90^\circ-\theta)$.

Answer: $\sin(\theta)=\cos(90^\circ-\theta)$. Sine of an angle equals cosine of its complement.

Flashcard 21: Identify the cofunction pair: which function equals $\cos(\theta)$ for complementary angles?

Answer: $\sin(90^\circ-\theta)$. Cosine and sine are cofunctions for complementary angles.

Flashcard 22: What is $\cos(90^\circ-\theta)$ in terms of $\sin(\theta)$?

Answer: $\cos(90^\circ-\theta)=\sin(\theta)$. The cosine of an angle's complement equals the angle's sine.

Flashcard 23: What is $\cos(30^\circ)$ using a complementary-angle relationship with $\sin(60^\circ)$?

Answer: $\cos(30^\circ)=\sin(60^\circ)$. $30°$ and $60°$ are complementary, so their cosine and sine are equal.

Flashcard 24: Identify the missing expression to make a true identity: $\cos(\theta)=\sin(\,\underline{\hspace{1.2cm}}\,)$.

Answer: $90^\circ-\theta$. To make $\cos(\theta)=\sin(?)$, the angle must be $\theta$'s complement.

Flashcard 25: State the complementary-angle identity that rewrites $\cos(\theta)$ using sine.

Answer: $\cos(\theta)=\sin(90^\circ-\theta)$. Complementary angles sum to $90°$, so cosine of one equals sine of the other.

Flashcard 26: What is $\cos(90^\circ-\theta)$ equal to in terms of $\theta$?

Answer: $\cos(90^\circ-\theta)=\sin(\theta)$. Since $\theta$ and $90°-\theta$ are complementary, cosine becomes sine.

Flashcard 27: Evaluate $\sin(90^\circ-20^\circ)$ by rewriting it as a cosine function.

Answer: $\cos(20^\circ)$. $90°-20°=70°$, and $\sin(70°)=\cos(20°)$ by the cofunction identity.

Flashcard 28: What is $\sin(30^\circ)$ using a complementary-angle relationship with $\cos(60^\circ)$?

Answer: $\sin(30^\circ)=\cos(60^\circ)$. $30°$ and $60°$ are complementary, so their sine and cosine are equal.

Flashcard 29: Find $\cos(30^\circ)$ using a complementary-angle identity and a known sine value.

Answer: $\cos(30^\circ)=\sin(60^\circ)=\frac{\sqrt{3}}{2}$. Since $30°$ and $60°$ are complementary, $\cos(30°)=\sin(60°)$.

Flashcard 30: If $A$ and $B$ are complementary acute angles and $\cos(A)=\frac{12}{13}$, what is $\sin(B)$?

Answer: $\sin(B)=\frac{12}{13}$. Since $A$ and $B$ are complementary, $\cos(A)=\sin(B)$.

Flashcard 31: In a right triangle, if acute angles are $A$ and $B$, what is the relationship between $A$ and $B$?

Answer: $A+B=90^\circ$. The two acute angles in a right triangle are complementary.

Flashcard 32: Identify the missing expression to make a true identity: $\sin(\theta)=\cos(\,\underline{\hspace{1.2cm}}\,)$.

Answer: $90^\circ-\theta$. To make $\sin(\theta)=\cos(?)$, the angle must be $\theta$'s complement.

Flashcard 33: If $A$ and $B$ are complementary acute angles and $\sin(A)=\frac{3}{5}$, what is $\cos(B)$?

Answer: $\cos(B)=\frac{3}{5}$. Since $A$ and $B$ are complementary, $\sin(A)=\cos(B)$.

Flashcard 34: Find $\cos(60^\circ)$ using a complementary-angle identity and a known sine value.

Answer: $\cos(60^\circ)=\sin(30^\circ)=\frac{1}{2}$. Since $60°$ and $30°$ are complementary, $\cos(60°)=\sin(30°)$.

Flashcard 35: If $\theta$ is acute, what is the complement of $\theta$ written in degrees?

Answer: $90^\circ-\theta$. The complement is the angle that adds to $\theta$ to make $90°$.

Flashcard 36: Given $\cos(\beta)=\frac{12}{13}$ and $\beta$ is acute, find $\sin(90^\circ-\beta)$.

Answer: $\frac{12}{13}$. Using $\sin(90°-\beta)=\cos(\beta)$.

Flashcard 37: Identify the relationship between acute complementary angles $A$ and $B$ in a right triangle.

Answer: $A+B=90^\circ$. Complementary angles sum to $90°$ by definition.

Flashcard 38: Find $\sin(45^\circ)$ by rewriting it as a cosine of a complementary angle.

Answer: $\sin(45^\circ)=\cos(45^\circ)=\frac{\sqrt{2}}{2}$. $45°$ is its own complement, so $\sin(45°)=\cos(45°)$.

Flashcard 39: Given $\sin(\alpha)=\frac{3}{5}$ and $\alpha$ is acute, find $\cos(90^\circ-\alpha)$.

Answer: $\frac{3}{5}$. Using $\cos(90°-\alpha)=\sin(\alpha)$.