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Quizzes > High School Quizzes > Mathematics

Law of Sines & Cosines Practice Quiz

Sharpen triangle skills with expert practice tips

Difficulty: Moderate
Grade: Grade 10
Study OutcomesCheat Sheet
Paper art promoting Sine Cosine Showdown trivia for high school students.

What does the Law of Sines state for any triangle?
a/sin(A) = b/sin(B) = c/sin(C)
a*sin(A) = b*sin(B) = c*sin(C)
a/sin(B) = b/sin(C) = c/sin(A)
sin(A)/a = sin(B)/b = sin(C)/c
The Law of Sines states that in any triangle, the ratio of a side's length to the sine of its opposite angle is constant. This fundamental relationship is used to solve for unknown sides or angles.
Which formula represents the Law of Cosines for a triangle with sides a, b, c and angle C opposite side c?
c^2 = a^2 + b^2 - ab*cos(C)
c^2 = a^2 + b^2 + 2ab*cos(C)
c^2 = a^2 + b^2 - 2ab*cos(C)
c = a*cos(B) + b*cos(A)
The Law of Cosines is expressed as c^2 = a^2 + b^2 - 2ab*cos(C) when c is opposite angle C. This formula generalizes the Pythagorean theorem to all triangles.
In which situation is the Law of Sines most useful?
Solving for missing sides or angles when two angles and one side are known
Calculating the area of a triangle given its base and height
Determining the circumference of a circle
Finding the midpoint of a triangle's side
The Law of Sines is especially useful when two angles and one side or two sides and a non-included angle are known. It creates a proportion that allows you to solve for the unknown measurements.
Which ratio remains constant in a triangle according to the Law of Sines?
Sine of an angle divided by the side adjacent to it
Side length divided by the sine of its opposite angle
Side length multiplied by the sine of its opposite angle
Cosine of the angle divided by the side length
The Law of Sines establishes that the ratio of a side's length to the sine of its opposite angle remains constant throughout the triangle. This ratio is key to solving many triangle problems.
What value does the Law of Cosines formula reduce to when angle C is 90 degrees?
c^2 = a^2 + b^2 - 2ab
c = sqrt(a^2 + b^2)
c = a + b
c^2 = a^2 + b^2
When angle C is 90°, cos(90°) is 0, so the Law of Cosines simplifies to c^2 = a^2 + b^2. This result is the classic Pythagorean theorem used for right triangles.
Given a triangle with side a = 7, angle A = 30°, and angle B = 45°, what is the value of a/sin(A) using the Law of Sines?
7
14
3.5
28
Using the Law of Sines, a/sin(A) is calculated as 7/sin(30°). Since sin(30°) equals 0.5, the ratio is 7/0.5 = 14. This constant ratio is then used to determine unknown sides or angles.
Given triangle XYZ with sides x = 5, y = 7, and z = 8, which of the following equations correctly uses the Law of Cosines to find angle Y?
8^2 = 5^2 + 7^2 - 2*5*7*cos(Y)
7^2 = 5^2 + 8^2 - 2*5*8*cos(Y)
5^2 = 7^2 + 8^2 - 2*7*8*cos(Y)
7^2 = 5^2 + 8^2 + 2*5*8*cos(Y)
To find angle Y, the Law of Cosines is arranged as 7^2 = 5^2 + 8^2 - 2*5*8*cos(Y). This equation properly relates the sides of the triangle and the cosine of angle Y.
For a triangle with sides 9, 12, and 15, which verification confirms it is a right triangle?
9^2 + 15^2 = 12^2
Calculating cos(angle) equals 1
9/15 = sin(opposite angle)
15^2 = 9^2 + 12^2
A right triangle satisfies the Pythagorean theorem. In this triangle, checking that 15^2 equals 9^2 + 12^2 confirms that the triangle is right-angled.
In the ambiguous case of the Law of Sines (SSA), what potential issue must be considered?
There might be two possible solutions or no solution for the unknown angle or side.
The triangle will always be obtuse.
The triangle must have two right angles.
The side ratios become undefined.
When given two sides and a non-included angle (SSA), the ambiguous case can occur where the triangle may have two valid solutions, one solution, or none at all. Recognizing this possibility is essential to correctly applying the Law of Sines.
Using the Law of Cosines, if triangle ABC has sides a = 10, b = 7, and angle C = 60°, which expression correctly computes side c?
c = 10 + 7 - cos(60°)
c = 10^2 + 7^2 - 2*10*7*sin(60°)
c = sqrt(10^2 + 7^2 - 2*10*7*cos(60°))
c = sqrt(10^2 + 7^2 + 2*10*7*cos(60°))
Substitute the known values into the Law of Cosines: c^2 = 10^2 + 7^2 - 2*10*7*cos(60°). Taking the square root gives the length of side c. This is the correct process for computing the unknown side.
Which of the following steps is necessary when solving for a missing side using the Law of Sines?
Multiply the sum of two sides by the sine of the known angle.
Divide the unknown angle by the known side.
Set up the ratio as the product of the sides and sines of their opposite angles.
Set up the ratio between the known side and sin(its opposite angle) equal to the ratio of the unknown side and sin(its opposite angle).
When applying the Law of Sines, the correct procedure is to equate the ratio of a known side to the sine of its opposite angle with the ratio for the unknown side. This allows for accurate cross-multiplication and solution.
Given triangle PQR with angle P = 50°, angle Q = 60°, and side p = 10, which formula correctly finds side q using the Law of Sines?
q = 10*sin(50)*sin(60)
q = (sin(60)/10)*sin(50)
q = (10*sin(50))/sin(60)
q = (10*sin(60))/sin(50)
Using the proportional relationship from the Law of Sines: p/sin(P) = q/sin(Q), we isolate q as q = (10*sin(60))/sin(50). This correctly uses the given measurements to find the unknown side.
If two sides of a triangle are 8 and 10 with an included angle of 120°, which expression using the Law of Cosines gives the third side?
c = sqrt(8^2 + 10^2 + 2*8*10*cos(120°))
c = (8^2 + 10^2)/cos(120°)
c = sqrt(8^2 + 10^2 - 2*8*10*cos(120°))
c = 8 + 10*cos(120°)
Plugging the given values into the Law of Cosines yields c^2 = 8^2 + 10^2 - 2*8*10*cos(120°). Taking the square root of c^2 produces the length of side c, which is the appropriate method for this problem.
What is the primary difference between the Law of Sines and the Law of Cosines?
The Law of Sines uses ratios of sides, unlike the Law of Cosines which uses sums of angles.
The Law of Sines relates sides to the sines of angles, whereas the Law of Cosines involves the cosine of the included angle.
The Law of Sines can only be used for right triangles, while the Law of Cosines works for any triangle.
There is no difference; they are interchangeable formulas.
The Law of Sines sets up a proportional relationship between a side and the sine of its opposing angle, while the Law of Cosines incorporates the cosine of the included angle to relate all three sides. This distinction makes each law suitable for different given information in a triangle.
An obtuse triangle has sides 13, 14, and 15. Which equation using the Law of Cosines is used to find the angle opposite the 15 unit side?
15^2 = 13^2 + 14^2 - cos(X)
15^2 = 13^2 + 14^2 - 2*13*14*cos(X)
15^2 = 13^2 + 14^2 + 2*13*14*cos(X)
15 = 13 + 14 - 2*13*14*cos(X)
To solve for the angle opposite the side of length 15, the Law of Cosines is set up as 15^2 = 13^2 + 14^2 - 2*13*14*cos(X). This equation directly relates the three sides with the cosine of the unknown angle.
When using the Law of Sines, why is it important to correctly identify the side opposite a given angle?
Because the side opposite an angle is always the largest side.
Because the Law of Sines relates each side to the sine of its opposite angle; a mismatch leads to incorrect results.
Because it determines whether the angle is acute or obtuse.
Because it changes the total sum of the sides in the triangle.
The effectiveness of the Law of Sines depends on pairing each side with the sine of its opposite angle. Incorrect identification disrupts the constant ratio and leads to errors in solving for unknown quantities.
In a triangle with angles 40° and 80°, what is the measure of the third angle?
40°
80°
100°
60°
The sum of the angles in any triangle is 180°. Subtracting the given angles 40° and 80° from 180° leaves 60° for the third angle.
Which scenario is most appropriate for applying the Law of Cosines?
When all three angles are given.
When the triangle is equilateral.
When two sides and the included angle are known.
When two angles and a non-included side are known.
The Law of Cosines is optimal when you know two sides and the included angle of a triangle. This information is sufficient to compute the third side or other angles.
Which of the following is an essential step when solving for an angle using the Law of Cosines?
Multiply by the sine of the known angle.
Use the tangent function to simplify the equation.
Square the angle and then take the square root.
Isolate the cosine term before applying the inverse cosine function.
When solving for an angle with the Law of Cosines, isolating the cosine term is vital. This allows you to correctly apply the inverse cosine function to determine the measure of the angle.
If a triangle has sides 11, 13, and 15, what is the first step in using the Law of Cosines to find the angle opposite the side of length 15?
Calculate the area of the triangle first.
Set up the proportion 15/sin(θ) = 11/sin(α).
Apply the Law of Sines to directly find sin(θ).
Apply the equation 15^2 = 11^2 + 13^2 - 2*11*13*cos(θ) and solve for cos(θ).
The initial step in using the Law of Cosines is to write the equation that relates the three sides with the cosine of the unknown angle. This step allows you to isolate cos(θ) and then calculate the angle using the inverse cosine function.
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Study Outcomes

  1. Understand the fundamental properties of sine and cosine functions.
  2. Apply the law of sines and cosines to solve triangle problems.
  3. Analyze the relationship between angles and sides in various triangle configurations.
  4. Interpret word problems to develop accurate trigonometric equations.
  5. Evaluate and simplify expressions involving sine and cosine functions.

Law of Sines & Cosines Quiz Part 1 Cheat Sheet

  1. Understand the Law of Sines formula - The Law of Sines links each side of a triangle with its opposite angle using a simple ratio: a/sin A = b/sin B = c/sin C. Memorising this formula is like having a secret map to non”right triangles. Dive into the Law of Sines
  2. Recognise when to apply the Law of Sines - This law shines when you know two angles and a side (AAS or ASA) or two sides and a non”included angle (SSA). Spotting these cases quickly will boost your problem‑solving speed. When to use the Law of Sines
  3. Be aware of the ambiguous (SSA) case - With SSA information you might find one, two, or no valid triangles at all. Practice sketching each scenario to build intuition and avoid errors. Master the ambiguous case
  4. Learn the Law of Cosines formulas - Use c² = a² + b² - 2ab cos C (and its cyclic forms) to connect all three sides and an angle. It's the ultimate tool for triangles where you can't use the Law of Sines directly. Explore the Law of Cosines
  5. Use SAS or SSS with the Law of Cosines - When you know two sides and their included angle (SAS) or all three sides (SSS), the Law of Cosines is your go‑to method. It's like a trigonometric Pythagoras that works on any triangle. SAS & SSS in action
  6. See the Pythagorean theorem generalised - Think of the Law of Cosines as a supersized Pythagoras: when C is 90°, the formula collapses to a² + b² = c². This connection helps you appreciate how trigonometry extends basic right‑triangle rules. Pythagorean theorem generalised
  7. Practice combining both laws - Real mastery comes from mixing the Law of Sines and Cosines in multi‑step problems. Tackling varied exercises helps you decide the fastest path to unknown sides or angles. Combined practice exercises
  8. Remember the 180° angle sum - Every triangle's angles add up to 180°, so if you find two angles you can instantly get the third. This handy check keeps your answers on track. Angle sum tip
  9. Use mnemonic devices - Tricks like "SOH‑CAH‑TOA" and "All Students Take Calculus" help you recall sine, cosine, and tangent rules in a flash. Good mnemonics turn long formulas into memorable jingles. Trigonometric mnemonic hacks
  10. Apply these laws in the real world - Navigation, surveying, astronomy and even game development rely on these trigonometric rules. Seeing how they solve real problems makes the concepts stick like glue. Real‑world applications
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