Precalculus Quiz: Applying Laws Of Sines And Cosines

20 questions · exam conditions

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Applying Laws Of Sines And CosinesQuestion 1 of 20

In triangle JKL, $\angle J = 30°$ , $k = 8$ , and $j = 5$ . A student uses the Law of Sines to find $\angle K$ and gets $\sin K = \frac{8 \sin(30°)}{5} = 0.8$ . What should the student consider next?

Since

\sin K = 0.8

, then

\angle K = \arcsin(0.8) \approx 53.1°

There are two possible values:

\angle K \approx 53.1°

\angle K \approx 126.9°

The calculation is impossible because

\sin K > \frac{1}{2}

The triangle is invalid because

j < k

but

\angle J = 30°

is too small

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Precalculus Quiz

Precalculus Quiz: Applying Laws Of Sines And Cosines

Practice Applying Laws Of Sines And Cosines in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

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

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.

All questions

Question 1

Since $\sin K = 0.8$ , then $\angle K = \arcsin(0.8) \approx 53.1°$
There are two possible values: $\angle K \approx 53.1°$ or $\angle K \approx 126.9°$ (correct answer)
The calculation is impossible because $\sin K > \frac{1}{2}$
The triangle is invalid because $j < k$ but $\angle J = 30°$ is too small

Explanation: This is the ambiguous case (SSA) of the Law of Sines. When 0 < sin K < 1, there are potentially two angles: K₁ = arcsin(0.8) ≈ 53.1° and K₂ = 180° - 53.1° = 126.9°. The student must check which value(s) create valid triangles by ensuring the sum of angles equals 180°. Choice A ignores the ambiguous case. Choice C incorrectly suggests the calculation is impossible. Choice D makes an invalid conclusion about the triangle's validity.

Question 2

In triangle $ABC$ , sides $a=8$ and $b=10$ are known, and the included angle $C=60^\circ$ is known. What is the length of side $c$ (opposite $\angle C$ )?

$\sqrt{244}$
$\sqrt{84}$ (correct answer)
$\sqrt{164}$
$\sqrt{36}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a = 8, b = 10, and angle C = 60°, we apply the Law of Cosines: c² = 8² + 10² - 2(8)(10)·cos(60°) = 64 + 100 - 160·(0.5) = 164 - 80 = 84, so c = √84. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice C forgets the crucial -2ab·cos(C) term in the Law of Cosines, computing only a² + b² when the full formula is c² = a² + b² - 2ab·cos(C). Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 3

A surveyor measures a triangular plot of land. From point A, the distance to point B is 200 meters and to point C is 150 meters. The angle at A is $75°$ . If the surveyor needs to find the distance from B to C, which calculation should be used?

$BC = \sqrt{200^2 + 150^2 - 2(200)(150)\cos(75°)}$ (correct answer)
$BC = \sqrt{200^2 + 150^2 + 2(200)(150)\cos(75°)}$
$\frac{BC}{\sin(75°)} = \frac{200}{\sin C}$
$BC = \frac{200 \cdot 150 \cdot \sin(75°)}{200 + 150}$

Explanation: This is a direct application of the Law of Cosines. Given two sides (AB = 200, AC = 150) and the included angle (∠A = 75°), we find the third side BC using

c^2 = a^2 + b^2 - 2ab\cos C

. Choice B incorrectly uses addition instead of subtraction. Choice C attempts the Law of Sines but we don't know angle C. Choice D uses an invalid formula that resembles an area calculation.

Question 4

In triangle $ABC$ , $\angle A=60^\circ$ , $\angle B=45^\circ$ , and side $b=8$ (opposite $\angle B$ ). What is the value of side $a$ (opposite $\angle A$ )?

$4\sqrt{2}$
$8\sqrt{3}$
$4\sqrt{6}$ (correct answer)
$8\sqrt{2}$

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

, and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Knowing

\angle A=60^\circ

\angle B=45^\circ

, and side

b=8

, we use the Law of Sines:

\frac{a}{\sin 60^\circ} = \frac{b}{\sin 45^\circ}

, so

a = 8 \cdot \frac{\sin 60^\circ}{\sin 45^\circ} = 8 \cdot \frac{\sqrt{3}/2}{\sqrt{2}/2} = 8 \cdot \frac{\sqrt{3}}{\sqrt{2}} = 8 \cdot \sqrt{\frac{3}{2}} = 4\sqrt{6}

. Choice C is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice A makes an arithmetic error in calculating

\sin 60^\circ / \sin 45^\circ

, getting a factor of

\sqrt{2}

instead of

\sqrt{3/2}

. For the Law of Sines, always match each angle with its opposite side: angle A with side a, angle B with side b, angle C with side c—using the wrong pairing will give an incorrect answer. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines.

Question 5

In triangle $ABC$ , $a=11$ , $b=9$ , and $\angle C=120^\circ$ (the included angle between sides $a$ and $b$ ). What is the length of side $c$ ?

$\sqrt{301}$ (correct answer)
$\sqrt{103}$
$\sqrt{49}$
$\sqrt{202}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a=11, b=9, and angle C=120°, we apply the Law of Cosines: c²=11² +9² -2(11)(9)·cos(120°)=121+81-198·(-0.5)=202+99=301, so c=√301. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice D uses the wrong sign in the Law of Cosines, subtracting 2ab·cos(C) instead of adding it since cos(120°)<0 makes -2ab·cos(C) positive. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 6

In triangle $ABC$ , $A=45^\circ$ , $B=60^\circ$ , and side $a=10$ (opposite $\angle A$ ). What is the length of side $b$ (opposite $\angle B$ )?

$10\sqrt{\frac{3}{2}}$ (correct answer)
$10\sqrt{2}$
$\frac{10}{\sqrt{3}}$
$10\sqrt{3}$

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

, and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Knowing angle A = 45°, angle B = 60°, and side a = 10, we use the Law of Sines:

\frac{a}{\sin A} = \frac{b}{\sin B}

, so

b = 10 \cdot \sin(60^\circ) / \sin(45^\circ) = 10 \cdot (\sqrt{3}/2) / (\sqrt{2}/2) = 10 \cdot \sqrt{3} / \sqrt{2} = 10 \sqrt{3/2}

. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice D makes an arithmetic error in calculating

\sin(60^\circ)/\sin(45^\circ)

, getting

\sqrt{3}

instead of

\sqrt{3/2}

Question 7

In triangle $ABC$ , $a=9$ , $c=12$ , and $B=60^\circ$ are given. What is the length of side $b$ (opposite $\angle B$ )?

$\sqrt{117}$ (correct answer)
$\sqrt{63}$
$15$
$\sqrt{225}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a = 9, c = 12, and angle B = 60°, we apply the Law of Cosines: b² = a² + c² - 2ac·cos(B) = 81 + 144 - 2(9)(12)·cos(60°) = 225 - 216·0.5 = 225 - 108 = 117, so b = √117. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice D forgets the crucial -2ac·cos(B) term in the Law of Cosines, computing only a² + c². Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 8

In triangle $ABC$ , $\angle A=45^\circ$ , $\angle B=60^\circ$ , and side $a=12$ . What is the length of side $b$ ?

$6\sqrt{6}$ (correct answer)
$6\sqrt{3}$
$12\sqrt{2}$
$8\sqrt{3}$

Explanation: This question tests understanding of the Law of Sines for solving non-right triangles. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C), and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Knowing angle A = 45°, angle B = 60°, and side a = 12, we use the Law of Sines: a/sin(A) = b/sin(B), so b = a·sin(B)/sin(A) = 12·sin(60°)/sin(45°) = 12·(√3/2)/(√2/2) = 12·√3/√2 = 12√3/√2 · √2/√2 = 12√6/2 = 6√6. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice B gives 6√3, which would result from incorrectly simplifying √3/√2 as √3 instead of rationalizing to get √6/2. For the Law of Sines, always match each angle with its opposite side: angle A with side a, angle B with side b, angle C with side c—using the wrong pairing will give an incorrect answer.

Question 9

In triangle $ABC$ , $a=8$ , $b=10$ , and $A=30^\circ$ (SSA configuration). How many triangles satisfy the given conditions?

0
1
2 (correct answer)
Infinitely many

Explanation: This question tests understanding of the Law of Sines for solving non-right triangles. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C), and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Since we know two sides and the non-included angle (SSA configuration with a=8, b=10, A=30°), this requires the Law of Sines because it provides angle-side opposite pairs, but we must check for the ambiguous case. Choice C is correct because with height h = b sin A = 10 sin 30° = 5, and h < a < b (5 < 8 < 10), there are two possible triangles. Choice B identifies only one solution in the SSA ambiguous case when actually two triangles satisfy the given conditions. In the SSA case (two sides and non-included angle), be alert for the ambiguous case: there might be two possible triangles, one triangle, or no triangle, depending on the specific values—always check if a second solution exists.

Question 10

In triangle $ABC$ , $a=9$ , $b=12$ , and $c=15$ are given. What is the measure of angle $C$ (in degrees)?

$30^\circ$
$45^\circ$
$60^\circ$
$90^\circ$ (correct answer)

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). With sides a = 9, b = 12, c = 15, we rearrange the Law of Cosines to solve for the angle: cos(C) = (a² + b² - c²)/(2ab) = (81 + 144 - 225)/(2·9·12) = 0/216 = 0, so C = arccos(0) = 90°. Choice D is correct because it applies the correct law with proper substitution of values and accurate arithmetic, recognizing that this is a 3-4-5 right triangle scaled by 3. Choice C incorrectly gives 60°, which would require cos(C) = 0.5, but our calculation clearly shows cos(C) = 0. When using the Law of Cosines to find an angle, rearrange to cos(C) = (a² + b² - c²)/(2ab), compute the right side, then use arccos to find the angle, checking that the result is between 0° and 180°.

Question 11

In triangle ABC, $a = 12$ , $b = 15$ , and $\angle C = 60°$ . What is the area of triangle ABC?

$45\sqrt{3}$ (correct answer)
$\frac{45\sqrt{3}}{2}$
$90$
$\frac{90\sqrt{3}}{3}$

Explanation: First, find side c using the Law of Cosines:

c^2 = a^2 + b^2 - 2ab\cos C = 144 + 225 - 2(12)(15)\cos(60°) = 369 - 360(\frac{1}{2}) = 189

. So

c = \sqrt{189} = 3\sqrt{21}

. The area is

\frac{1}{2}ab\sin C = \frac{1}{2}(12)(15)\sin(60°) = 90 \cdot \frac{\sqrt{3}}{2} = 45\sqrt{3}

. Choice B uses the wrong area formula. Choice C forgets the

\sin(60°)

factor. Choice D incorrectly simplifies

45\sqrt{3}

Question 12

In triangle $ABC$ , sides $a=8$ and $b=13$ are known, and the included angle $\angle C=120^\circ$ is known. Which law should be used first to solve this triangle?

Law of Sines, because an opposite side-angle pair is given.
Law of Cosines, because two sides and the included angle are given (SAS). (correct answer)
Pythagorean Theorem, because the triangle is right.
Neither; there is not enough information to start.

Explanation: This question tests understanding of when to apply the Law of Cosines versus the Law of Sines for solving triangles. To decide which law to use: Law of Sines when you have angle-side opposite pairs to work with, Law of Cosines when you have two sides and the included angle or three sides with no angles. Since we know two sides (a = 8 and b = 13) and the included angle (C = 120°), this is a SAS configuration, which requires the Law of Cosines because we can directly compute the third side using c² = a² + b² - 2ab·cos(C). Choice B is correct because it correctly identifies the SAS configuration and prescribes the appropriate law. Choice A incorrectly suggests using the Law of Sines, but we don't have an angle-side opposite pair to start with—we'd need to find side c first using the Law of Cosines. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines.

Question 13

In triangle DEF, $d = 15$ , $e = 20$ , and $\angle F = 120°$ . After finding side $f$ using the Law of Cosines, which method would be most accurate for finding $\angle D$ ?

Use Law of Sines: $\sin D = \frac{d \sin F}{f}$
Use Law of Cosines: $\cos D = \frac{e^2 + f^2 - d^2}{2ef}$ (correct answer)
Use the fact that $\angle D = 180° - \angle E - \angle F$
Use Law of Sines: $\sin D = \frac{f \sin F}{d}$

Explanation: While both Law of Sines and Law of Cosines could work, the Law of Cosines is more accurate when we know all three sides because it avoids the ambiguous case and rounding errors that can occur with inverse sine. The Law of Cosines gives a unique answer for the angle. Choice A uses the correct Law of Sines formula but is less accurate. Choice C requires finding ∠E first. Choice D has the Law of Sines formula backwards.

Question 14

In triangle PQR, $\angle P = 45°$ , $\angle Q = 70°$ , and side $r = 20$ (opposite to angle R). What is the length of side $p$ (opposite to angle P)?

$\frac{20\sin(70°)}{\sin(45°)}$
$\frac{20\sin(65°)}{\sin(45°)}$
$\frac{20\sin(45°)}{\sin(70°)}$
$\frac{20\sin(45°)}{\sin(65°)}$ (correct answer)

Explanation: When you encounter a triangle with two angles and one side given, you're dealing with a Law of Sines problem. This law states that in any triangle, the ratio of each side to the sine of its opposite angle is constant:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

. First, find the missing angle. Since angles in a triangle sum to 180°, angle R = 180° - 45° - 70° = 65°. Now you have all three angles: P = 45°, Q = 70°, R = 65°, and side r = 20. To find side p (opposite angle P), set up the Law of Sines proportion:

\frac{p}{\sin P} = \frac{r}{\sin R}

. Substituting the known values:

\frac{p}{\sin 45°} = \frac{20}{\sin 65°}

. Solving for p:

p = \frac{20\sin 45°}{\sin 65°}

. Let's examine why the other answers are incorrect. Answer A uses

\sin 70°

in the numerator and

\sin 45°

in the denominator, which would actually give you the ratio

\frac{r}{p}

rather than p itself. Answer B incorrectly uses

\sin 70°

(angle Q) instead of

\sin 45°

(angle P) in the numerator. Answer C places

\sin 70°

in the denominator instead of

\sin 65°

, which would be trying to relate sides p and q instead of p and r. Study tip: Always identify all three angles first, then carefully match each side with its opposite angle in your Law of Sines setup. Double-check that you're solving for the correct variable by ensuring the unknown appears in the numerator of your final expression.

Question 15

In triangle $ABC$ , sides $a=8$ and $b=15$ and the included angle $C=90^\circ$ are given. Which law should be used first to solve this triangle for side $c$ ?

Law of Sines
Law of Cosines (correct answer)
Pythagorean theorem only (Law of Sines/Cosines cannot be used)
Neither law; there is not enough information

Explanation: This question tests understanding of when to apply the Law of Cosines for solving non-right triangles. To decide which law to use: Law of Sines when you have angle-side opposite pairs to work with, Law of Cosines when you have two sides and the included angle or three sides with no angles. Since we know two sides (a = 8, b = 15) and the included angle (C = 90°), this is a SAS configuration, which requires the Law of Cosines because we have the two sides and the angle between them. Choice B is correct because it correctly identifies that the Law of Cosines should be used for this SAS configuration. Choice C incorrectly suggests using only the Pythagorean theorem and claims the Laws cannot be used, but the Law of Cosines actually reduces to the Pythagorean theorem when C = 90° since cos(90°) = 0, making c² = a² + b² - 2ab·cos(90°) = a² + b² - 0 = a² + b². The Law of Cosines reduces to the Pythagorean theorem when the angle is 90° (since cos(90°) = 0, the -2ab·cos(C) term vanishes), making it a generalization that works for all triangles.

Question 16

In triangle $ABC$ , $a=9$ , $b=12$ , and the included angle $C=120^\circ$ are given. What is the length of side $c$ ?

$\sqrt{117}$
$\sqrt{333}$ (correct answer)
$15$
$\sqrt{225}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a = 9, b = 12, and angle C = 120°, we apply the Law of Cosines: c² = a² + b² - 2ab·cos(C) = 9² + 12² - 2(9)(12)·cos(120°) = 81 + 144 - 216·(-1/2) = 225 - (-108) = 225 + 108 = 333, so c = √333. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic, remembering that cos(120°) = -1/2. Choice D gives the intermediate result c² = 225 instead of including the -2ab·cos(C) term, forgetting that cos(120°) is negative. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 17

In triangle $ABC$ , $a=13$ , $b=14$ , and the included angle is $\angle C=60^\circ$ . What is the value of side $c$ ?

$\sqrt{183}$ (correct answer)
$\sqrt{379}$
$\sqrt{365}$
$\sqrt{169}$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). Given sides a=13, b=14, and angle C=60°, we apply the Law of Cosines: c² =13² +14² -2(13)(14)·cos(60°)=169+196-364·(0.5)=365-182=183, so c=√183. Choice A is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice C forgets the crucial -2ab·cos(C) term in the Law of Cosines, computing only 13² +14²=365 when the full formula is c²=13² +14² -2(13)(14)·cos(60°). Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. Remember that the Law of Cosines generalizes the Pythagorean theorem: when the angle is 90°, cos(90°) = 0 makes the formula reduce to a² + b² = c², but for any other angle, you must include the -2ab·cos(C) term.

Question 18

In triangle $ABC$ , side $a=9$ , side $b=12$ , and side $c=15$ . What is the measure of $\angle C$ (in degrees)?

$60^\circ$
$45^\circ$
$90^\circ$ (correct answer)
$120^\circ$

Explanation: This question tests understanding of the Law of Cosines for solving non-right triangles. The Law of Cosines states that c² = a² + b² - 2ab·cos(C) for any triangle (with analogous formulas for the other sides), generalizing the Pythagorean theorem, and is used when you know two sides and the included angle (SAS) or all three sides (SSS). With sides a = 9, b = 12, c = 15, we rearrange the Law of Cosines to solve for the angle: cos(C) = (a² + b² - c²)/(2ab) = (81 + 144 - 225)/(2·9·12) = 0/216 = 0, so C = arccos(0) = 90°. Choice C is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice D might result from a sign error, computing cos(C) = -0.5 instead of 0, which would give 120°. When using the Law of Cosines to find an angle, rearrange to cos(C) = (a² + b² - c²)/(2ab), compute the right side, then use arccos to find the angle, checking that the result is between 0° and 180°.

Question 19

In triangle $ABC$ , $a=8$ , $b=10$ , and angle $A=30^\circ$ are given (SSA, ambiguous case). How many triangles satisfy the given conditions?

0
1
2 (correct answer)
Infinitely many

Explanation: This question tests understanding of when to apply the Law of Sines for solving non-right triangles, specifically the ambiguous SSA case. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C), and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Since we know a = 8, b = 10, and angle A = 30° (SSA configuration), we first find sin(B) using the Law of Sines: sin(B) = b·sin(A)/a = 10·sin(30°)/8 = 10·(0.5)/8 = 5/8 = 0.625. Since sin(B) = 0.625 < 1 and B could be either arcsin(0.625) ≈ 38.7° or 180° - 38.7° ≈ 141.3°, and both give valid triangles (since A + B < 180° in both cases), there are two possible triangles. Choice C is correct because it correctly identifies that the SSA configuration with these specific values yields two valid triangles. Choice B incorrectly suggests only one triangle exists, missing the second solution in the ambiguous case. In the SSA case (two sides and non-included angle), be alert for the ambiguous case: there might be two possible triangles, one triangle, or no triangle, depending on the specific values—always check if a second solution exists.

Question 20

In triangle $ABC$ , $A=45^\circ$ , $B=45^\circ$ , and $c=10$ (opposite $\angle C$ ). What is the length of side $a$ (opposite $\angle A$ )?

$10$
$5\sqrt{2}$ (correct answer)
$10\sqrt{2}$
$5$

Explanation: This question tests understanding of the Law of Sines for solving non-right triangles. The Law of Sines states that in any triangle, the ratio of each side to the sine of its opposite angle is constant: a/sin(A) = b/sin(B) = c/sin(C), and is used when you know two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). Knowing angle A = 45°, angle B = 45°, and side c = 10, we first find C = 180° - 45° - 45° = 90°, then use the Law of Sines: a/sin(A) = c/sin(C), so a = 10·sin(45°)/sin(90°) = 10·(√2/2)/1 = 5√2. Choice B is correct because it applies the correct law with proper substitution of values and accurate arithmetic. Choice C uses the wrong angle-side pairing in the Law of Sines, matching angle A with side c instead of its opposite side a. Key to choosing the right law: identify what's given—if you have two sides and the included angle (SAS) or three sides (SSS), use Law of Cosines; if you have two angles and a side (AAS/ASA) or two sides and a non-included angle (SSA), use Law of Sines. For the Law of Sines, always match each angle with its opposite side: angle A with side a, angle B with side b, angle C with side c—using the wrong pairing will give an incorrect answer.