Geometry Quiz: Proving Applying Laws Of Sines Cosines

What this quiz covers

This quiz focuses on Proving Applying Laws Of Sines Cosines, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $1350\sqrt{3}$ square meters
2. $675\sqrt{3}$ square meters (correct answer)
3. $1350$ square meters
4. $675$ square meters

Question 2

1. $\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(60°) = 45\sqrt{3}$ (correct answer)
2. $\frac{1}{2} \cdot 12 \cdot 15 \cdot \cos(60°) = 45$
3. $\frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(30°) = 45$
4. $\frac{1}{2} \cdot 12 \cdot 15 \cdot \tan(60°) = 90\sqrt{3}$

Question 3

1. Find the altitude to side $XZ$, then use $A = \frac{1}{2} \cdot 16 \cdot h$
2. Find angle $X$ using the Law of Sines, then calculate $A = \frac{1}{2} \cdot 14 \cdot 16 \cdot \cos(X)$
3. Use Heron's formula directly since all three sides are known
4. Find angle $Y$ using the Law of Cosines, then calculate $A = \frac{1}{2} \cdot 10 \cdot 14 \cdot \sin(Y)$ (correct answer)

Question 4

1. $60°$
2. $45°$
3. $30°$ (correct answer)
4. $90°$

Question 5

1. If $\theta=90^\circ$, then $a^2=b^2+c^2-2bc\cos90^\circ=b^2+c^2$. (correct answer)
2. If $\theta=90^\circ$, then $a=b+c$ because right triangles add sides.
3. If $\theta=90^\circ$, then $\sin\theta=1$ so $\tfrac{a}{1}=\tfrac{b}{\sin B}$.
4. If $\theta=90^\circ$, then $a^2=b^2+c^2+2bc\cos90^\circ=b^2+c^2+2bc$.

Question 6

1. Use the Law of Cosines since it generalizes the Pythagorean Theorem: $AB^2=AC^2+BC^2-2(AC)(BC)\cos C$. (correct answer)
2. Use the Pythagorean Theorem because any triangle with an obtuse angle can be split into two right triangles, so $AB^2=AC^2+BC^2$.
3. Use the Law of Sines with $\dfrac{AB}{\sin C}=\dfrac{AC}{\sin C}$ to conclude $AB=AC$ and then find $AB$.
4. Estimate $AB$ from the picture because $\angle C$ looks close to $180^\circ$ so $AB\approx AC+BC$.

Question 7

1. $AD=c\sin\beta$ and $AD=b\sin\gamma$, so $c\sin\beta=b\sin\gamma$. (correct answer)
2. $AD=c\cos\beta$ and $AD=b\cos\gamma$, so $\tfrac{c}{\cos\beta}=\tfrac{b}{\cos\gamma}$.
3. $BD=c\sin\beta$ and $DC=b\sin\gamma$, so $BD=DC$.
4. Since $AD$ is an altitude, $\beta+\gamma=90^\circ$, so $\sin\beta=\cos\gamma$.

Question 8

1. Use $YZ^2=XY^2+XZ^2-2(XY)(XZ)\cos 30^\circ$, which reduces to the Pythagorean Theorem when the included angle is $90^\circ$. (correct answer)
2. Use $YZ=XY+XZ-\cos 30^\circ$ because the cosine term is subtracted from the sum of the sides.
3. Use $YZ^2=XY^2+XZ^2$ because the included angle is given, so the triangle can be treated as right at $X$.
4. Use the Law of Sines: $\frac{YZ}{\sin 30^\circ}=\frac{11}{\sin Y}$ and choose $\angle Y$ by visual estimation.

Question 9

1. Set $17^2=8^2+13^2-2(8)(13)\cos M$, so $\cos M=\frac{8^2+13^2-17^2}{2\cdot 8\cdot 13}$. (correct answer)
2. Set $\frac{17}{\sin M}=\frac{13}{\sin N}$ and choose $\angle N$ by estimating it from the drawing.
3. Assume $\angle M=90^\circ$ because it is the included angle, then use $17^2=8^2+13^2$.
4. Write $\cos M=\frac{17}{8+13}$ since cosine compares the opposite side to the sum of adjacent sides.

Question 10

1. $60°$
2. $30°$
3. $45°$ (correct answer)
4. $135°$

Question 11

1. $10$ units
2. $7.5$ units (correct answer)
3. $15\sqrt{3}$ units
4. $10\sqrt{3}$ units

Question 12

1. $\cos\theta=\dfrac{b^2+c^2-a^2}{2bc}$. (correct answer)
2. $\cos\theta=\dfrac{a^2-b^2-c^2}{2bc}$.
3. $\cos\theta=\dfrac{a}{b+c}$.
4. Since the triangle is not right, $\cos\theta=0$.

Question 13

1. Use the Law of Sines: $\dfrac{BC}{\sin A}=\dfrac{AB}{\sin B}$, then set $B=60^\circ$ because the sides look equal.
2. Use the Law of Cosines: $BC^2=AB^2+AC^2-2(AB)(AC)\cos A$, noting if $A=90^\circ$ then $\cos A=0$. (correct answer)
3. Assume $\angle A=90^\circ$ since $AB=AC$, then use $BC^2=AB^2+AC^2$.
4. Estimate $BC$ by the drawing since equal sides guarantee a right triangle, so $BC\approx 6\sqrt2$.

Question 14

1. Use $CD=AC\sin A$ and $CD=BC\sin B$ to get $AC\sin A=BC\sin B$, then solve for $\sin B$. (correct answer)
2. Use $CD=AC\cos A$ and $CD=BC\cos B$ to get $AC\cos A=BC\cos B$, then solve for $B$.
3. Use the Law of Cosines with $AB^2=AC^2+BC^2-2(AC)(BC)\cos C$, then set $C=90^\circ$ to find $B$.
4. Apply the Law of Sines directly because it is a memorized formula; no altitude reasoning is needed.

Question 15

1. Use the Law of Cosines because it generalizes the Pythagorean Theorem: $BC^2=AB^2+AC^2-2(AB)(AC)\cos A$. (correct answer)
2. Use the Law of Sines because $\dfrac{AB}{\sin A}=\dfrac{AC}{\sin A}$, so $AB=AC$ and then find $BC$.
3. Assume a right triangle at $A$ because the included angle is shown, then use $BC=\sqrt{AB^2+AC^2}$.
4. Estimate $BC$ by comparing how long it looks relative to $AB$ and $AC$ in the diagram.

Question 16

1. Use $BC^2=9^2+12^2-2(9)(12)\cos120^\circ$, then take $BC=\sqrt{9^2+12^2-2(9)(12)\cos120^\circ}$. (correct answer)
2. Use $BC^2=9^2+12^2$ since the largest angle makes it a right triangle.
3. Use $\tfrac{BC}{\sin120^\circ}=\tfrac{12}{\sin120^\circ}$, so $BC=12$.
4. Because the picture would look wide at $A$, estimate $BC\approx 3$.

Question 17

1. Assume a right triangle: $a^2=7^2+10^2$, so $a=\sqrt{149}$.
2. Use $a^2=7^2+10^2-2(7)(10)\cos60^\circ$, so $a=\sqrt{79}$. (correct answer)
3. Use $\tfrac{a}{\sin60^\circ}=\tfrac{7}{\sin60^\circ}$, so $a=7$.
4. Since the diagram looks obtuse at $A$, take $a\approx 17$ by visual estimation.

Question 18

1. Compute $\cos D=\frac{DE^2+DF^2-EF^2}{2(DE)(DF)}$; since this value is $0$, conclude $\angle D$ is right. (correct answer)
2. Since $5+12=13$, the triangle must be right at $D$, so $\angle D=90^\circ$ by inspection.
3. Use the Law of Sines with $\frac{13}{\sin D}$ and infer $\angle D$ from how large 13 is compared to 12.
4. Because the diagram is not drawn to scale, you cannot classify $\angle D$ even with the side lengths given.

Question 19

1. Use $YZ^2=XY^2+XZ^2-2(XY)(XZ)\cos 30^\circ$, which reduces to the Pythagorean Theorem when the included angle is $90^\circ$. (correct answer)
2. Use $YZ=XY+XZ-\cos 30^\circ$ because the cosine term is subtracted from the sum of the sides.
3. Use $YZ^2=XY^2+XZ^2$ because the included angle is given, so the triangle can be treated as right at $X$.
4. Use the Law of Sines: $\frac{YZ}{\sin 30^\circ}=\frac{11}{\sin Y}$ and choose $\angle Y$ by visual estimation.

Question 20

1. Because $CD=AC\sin A$ and $CD=BC\sin B$, so $AC\sin A=BC\sin B$ and the ratio follows by rearranging. (correct answer)
2. Because $CD=AC\sin B$ and $CD=BC\sin A$, so $AC\sin B=BC\sin A$ and the ratio follows.
3. Because $\triangle ABC$ is a right triangle once the altitude is drawn, so the sine ratios are the same in both smaller triangles.
4. Because the diagram appears symmetric about $CD$, so $AC=BC$ and therefore $\dfrac{AC}{\sin B}=\dfrac{BC}{\sin A}$.