Precalculus Flashcards: All Circles Are Similar

What this deck covers

This deck focuses on All Circles Are Similar, 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: Use the Law of Cosines: find $c$ if $a=5$, $b=5$, and included angle $C=60^\circ$.

Answer: $c=\sqrt{5^2+5^2-2\cdot^5\cdot^5\cos 60^\circ}=5$. Isosceles with $60°$ angle forms equilateral triangle.

Flashcard 2: Find $A$ if $\sin A=\frac{1}{2}$ and $A$ is acute.

Answer: $30^\circ$. $\sin 30°=\frac{1}{2}$ from special right triangle.

Flashcard 3: State the Law of Sines for triangle $ABC$ using sides $a,b,c$ opposite angles $A,B,C$.

Answer: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Relates each side to the sine of its opposite angle.

Flashcard 4: In any triangle, which side is opposite angle $A$ when using standard notation?

Answer: $a$. Standard notation pairs lowercase sides with uppercase angles.

Flashcard 5: Use the Law of Sines: if $a=10$, $A=30^\circ$, and $B=60^\circ$, what is $b$?

Answer: $b=10\frac{\sin 60^\circ}{\sin 30^\circ}=10\sqrt{3}$. Apply $\frac{a}{\sin A}=\frac{b}{\sin B}$ and simplify.

Flashcard 6: Identify the included angle for sides $b$ and $c$ in triangle $ABC$ (standard opposite-side notation).

Answer: $A$. Angle $A$ is between sides $b$ and $c$ in standard notation.

Flashcard 7: What is the triangle angle sum rule used after finding two angles?

Answer: $A+B+C=180^\circ$. Sum of interior angles in any triangle equals $180°$.

Flashcard 8: Use the Law of Sines: if $a=12$, $A=90^\circ$, and $B=30^\circ$, what is $b$?

Answer: $b=12\frac{\sin 30^\circ}{\sin 90^\circ}=6$. Apply sine ratio with $\sin 90^\circ=1$ and $\sin 30^\circ=\frac{1}{2}$.

Flashcard 9: What is the area formula using two sides and the included angle, for sides $b,c$ and angle $A$?

Answer: $K=\frac{1}{2}bc\sin A$. Uses half the product of two sides times sine of included angle.

Flashcard 10: Find the magnitude of the resultant of two forces $10$ and $10$ with included angle $60^\circ$.

Answer: $R=\sqrt{10^2+10^2+2\cdot10\cdot10\cos60^\circ}=10\sqrt{3}$. Use cosine formula with angle between forces.

Flashcard 11: Which triangle data type is directly solvable using the Law of Sines without extra steps: $AAS$, $ASA$, $SAS$, or $SSS$?

Answer: $AAS$ or $ASA$. Both have two angles and one side, perfect for sine ratio.

Flashcard 12: Use the Law of Sines: if $a=8$, $A=45^\circ$, and $B=30^\circ$, what is $b$ (exact form)?

Answer: $b=8\frac{\sin 30^\circ}{\sin 45^\circ}=4\sqrt{2}$. Apply sine ratio and simplify using $\sin 30^\circ=\frac{1}{2}$, $\sin 45^\circ=\frac{\sqrt{2}}{2}$.

Flashcard 13: State the Law of Cosines formula for angle $A$ using sides $a,b,c$.

Answer: $\cos A=\frac{b^2+c^2-a^2}{2bc}$. Rearranges Law of Cosines to solve for angle's cosine.

Flashcard 14: What triangle information type is the standard use case for the Law of Sines: $AAS, ASA, SAA, SSA,$ or $SSS$?

Answer: $AAS, ASA, \text{or } SAA$. Two angles and a side allow unique triangle solution.

Flashcard 15: State the Law of Sines for a triangle with sides $a,b,c$ opposite angles $A,B,C$.

Answer: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Relates ratios of sides to sines of opposite angles.

Flashcard 16: State the Law of Cosines formula that solves for side $a$ in triangle $ABC$.

Answer: $a^2=b^2+c^2-2bc\cos A$. Relates side $a$ to the other sides and angle $A$.

Flashcard 17: Which triangle data type is the classic ambiguous case for the Law of Sines: $SSA$, $SAS$, or $SSS$?

Answer: $SSA$. Two sides and non-included angle can yield 0, 1, or 2 triangles.

Flashcard 18: Classify the triangle by angles if $a=3$, $b=4$, and $c=5$ (largest side is $5$).

Answer: $\text{Right, since }5^2=3^2+4^2$. Pythagorean theorem confirms $90°$ angle.

Flashcard 19: Find the missing angle if $A=35^\circ$ and $B=75^\circ$ in a triangle.

Answer: $C=70^\circ$. Triangle angles sum to $180^\circ$.

Flashcard 20: State the Law of Cosines formula for side $a$ in terms of $b,c,$ and included angle $A$.

Answer: $a^2=b^2+c^2-2bc\cos A$. Generalizes Pythagorean theorem with cosine correction term.

Flashcard 21: In the $SSA$ case, what is the first step to test for $0$, $1$, or $2$ triangles when angle $A$ is known?

Answer: Compute $h=b\sin A$. Height $h$ determines if side $a$ can reach the opposite side.

Flashcard 22: In the $SSA$ case with acute $A$, what condition gives two triangles?

Answer: $h<b$ and $h<a<b$ where $h=b\sin A$. Side $a$ can swing to two positions when $A$ is acute.

Flashcard 23: Use the Law of Sines: find $b$ if $A=45^\circ$, $B=30^\circ$, and $a=8$.

Answer: $b=\frac{8\sin 30^\circ}{\sin 45^\circ}=4\sqrt{2}$. Calculate: $b=\frac{8(0.5)}{\frac{\sqrt{2}}{2}}=4\sqrt{2}$.

Flashcard 24: Find $C$ if $A=35^\circ$ and $B=65^\circ$ in a triangle.

Answer: $80^\circ$. Apply angle sum: $C=180°-35°-65°=80°$.

Flashcard 25: In the $SSA$ case with given $A$, $a$, and $b$, what condition gives no triangle?

Answer: $a<h=b\sin A$. Side $a$ is too short to reach the opposite side.

Flashcard 26: What triangle information type is the standard use case for the Law of Cosines: $SAS, SSS,$ or $ASA$?

Answer: $SAS\text{ or }SSS$. Two sides with included angle or three sides determine triangle.

Flashcard 27: State the altitude relation for the $SSA$ case with known $A,a,$ and adjacent side $b$.

Answer: $h=b\sin A$. Height from vertex to opposite side uses sine of angle.

Flashcard 28: For $SSA$ with $A=30^\circ$, $a=4$, $b=10$, how many triangles are possible?

Answer: $0\text{ triangles}$. Since $a<b\sin A=5$, no triangle exists.

Flashcard 29: State the Law of Sines form that directly solves for $a$ given $A,b,$ and $B$.

Answer: $a=\frac{b\sin A}{\sin B}$. Cross-multiply Law of Sines to isolate desired side.

Flashcard 30: State the Law of Cosines formula that solves for angle $A$ given sides $a,b,c$.

Answer: $\cos A=\frac{b^2+c^2-a^2}{2bc}$. Rearranges to solve for the cosine of angle $A$.

Flashcard 31: Identify the ambiguous case in triangle solving that can yield $0,1,$ or $2$ solutions.

Answer: $SSA$. Two sides and non-included angle may have multiple solutions.

Flashcard 32: For $SSA$ with $A=30^\circ$, $a=6$, $b=10$, how many triangles are possible?

Answer: $2\text{ triangles}$. Since $b\sin A<a<b$, two triangles possible.