Law of Sines & Triangles
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This flashcard set covers the Law of Sines, triangle properties, and step-by-step methods for solving various triangle cases. It is designed for students learning trigonometry.
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Cards · 52
- Basic formula of the Law of Sines
- a / sin(A) = b / sin(B) = c / sin(C)
- Alternative reciprocal form of the Law of Sines
- sin(A) / a = sin(B) / b = sin(C) / c
- Law of Sines relation to circumradius R
- a / sin(A) = b / sin(B) = c / sin(C) = 2R, where R is the radius of the circumscribed circle.
- Triangle angle sum theorem
- The sum of interior angles in any Euclidean triangle satisfies A + B + C = 180 degrees.
- When can the Law of Sines directly solve a triangle?
- When you are given AAS (Angle-Angle-Side), ASA (Angle-Side-Angle), or SSA (Side-Side-Angle).
- Why cannot the Law of Sines immediately solve SAS or SSS cases?
- There is no known side-and-opposite-angle pair to set up a solvable single-variable ratio.
- Solving AAS: Step 1 when given A, B, and a
- Compute angle C directly using C = 180 - (A + B).
- Solving AAS: Step 2 to find side b when given A, B, and a
- Use b = a * sin(B) / sin(A).
- Solving AAS: Step 3 to find side c when given A, B, and a
- Use c = a * sin(C) / sin(A).
- Solving ASA: Step 1 when given A, c, and B
- Determine angle C using C = 180 - (A + B).
- Solving ASA: Step 2 to find side a given A, c, and B
- Apply side formula a = c * sin(A) / sin(C).
- Solving ASA: Step 3 to find side b given A, c, and B
- Apply side formula b = c * sin(B) / sin(C).
- Definition of the Ambiguous Case (SSA)
- Given two sides and a non-included acute/obtuse angle, there may be 0, 1, or 2 valid triangles.
- SSA with acute angle A: condition for 0 solutions when a < h
- If a < b * sin(A), side a is too short to reach the opposite side, yielding no valid triangle.
- SSA with acute angle A: condition for exactly 1 right triangle
- If a = b * sin(A), exactly one right-angled triangle exists where angle B = 90 degrees.
- SSA with acute angle A: condition for 2 valid triangles
- If b * sin(A) < a < b, two valid triangles exist (one acute B, one obtuse B).
- SSA with acute angle A: condition for 1 unique triangle when a >= b
- If a >= b, only one triangle is possible because the obtuse alternative would not close.
- SSA with obtuse angle A: condition for 0 solutions
- If angle A >= 90 degrees and a <= b, no triangle can be formed.
- SSA with obtuse angle A: condition for 1 solution
- If angle A >= 90 degrees and a > b, exactly one obtuse triangle is formed.
- Finding secondary angle B2 in ambiguous SSA case
- Calculate B2 = 180 - B1, then check if A + B2 < 180 degrees.
- Area of a triangle using sine formula
- Area = (1/2) * a * b * sin(C) = (1/2) * b * c * sin(A) = (1/2) * a * c * sin(B).
- How to isolate sin(B) in SSA case given a, b, A
- sin(B) = (b * sin(A)) / a.
- Significance of sin(B) > 1 during SSA calculation
- No real angle exists for sin(B) > 1, proving no triangle can be constructed.
- Given A = 30 deg, B = 45 deg, a = 10. Find angle C.
- C = 180 - 30 - 45 = 105 degrees.
- Given A = 30 deg, B = 45 deg, a = 10. Find side b.
- b = 10 * sin(45) / sin(30) = 10 * (sqrt(2)/2) / (1/2) = 10 * sqrt(2) approx 14.14.
- Given A = 60 deg, a = 6, b = 2*sqrt(3). Find sin(B).
- sin(B) = (2*sqrt(3) * sin(60)) / 6 = (2*sqrt(3) * sqrt(3)/2) / 6 = 3 / 6 = 0.5.
- Given sin(B) = 0.5 with A = 60 deg, a = 6, b = 2*sqrt(3). Find B.
- B = 30 degrees (B = 150 is rejected because 60 + 150 = 210 > 180).
- Given A = 40 deg, B = 60 deg, c = 5. Step 1 to solve.
- Calculate angle C: C = 180 - (40 + 60) = 80 degrees.
- Given A = 40 deg, B = 60 deg, C = 80 deg, c = 5. Find side a.
- a = 5 * sin(40) / sin(80) approx 5 * 0.6428 / 0.9848 approx 3.26.
- Given A = 40 deg, B = 60 deg, C = 80 deg, c = 5. Find side b.
- b = 5 * sin(60) / sin(80) approx 5 * 0.8660 / 0.9848 approx 4.40.
- Given A = 120 deg, a = 4, b = 5. How many triangles exist?
- Zero triangles exist because A is obtuse and side a is smaller than side b.
- Given A = 30 deg, a = 4, b = 10. How many triangles exist?
- Zero, because height h = 10 * sin(30) = 5, and a = 4 < h.
- Given A = 30 deg, a = 5, b = 10. How many triangles exist?
- Exactly one right triangle, because h = 10 * sin(30) = 5 = a.
- Given A = 30 deg, a = 7, b = 10. How many triangles exist?
- Two triangles, because h = 5 < a < 10.
- Given A = 30 deg, a = 12, b = 10. How many triangles exist?
- Exactly one triangle, because a > b.
- Altitude h from vertex C to side c in terms of angle A
- h = b * sin(A).
- Altitude h from vertex C to side c in terms of angle B
- h = a * sin(B).
- Derivation basis of the Law of Sines
- Equating two expressions for triangle altitude: h = b * sin(A) = a * sin(B) gives a/sin(A) = b/sin(B).
- Can the Law of Sines be applied to right triangles?
- Yes; when C = 90 deg, c/sin(90) = c, reducing the law to basic SOH definitions.
- Step to find missing angle when two sides and one opposite angle are given
- Calculate sin(B) = (b * sin(A)) / a, then apply arcsin to find angle B.
- Why check angle sum after using arcsin in SSA?
- To determine whether the obtuse supplement (180 - B) leaves room for a third positive angle.
- Given A = 45 deg, B = 45 deg, c = 8. Find angle C.
- C = 180 - (45 + 45) = 90 degrees.
- Given A = 45 deg, C = 90 deg, c = 8. Find side a.
- a = 8 * sin(45) / sin(90) = 8 * (sqrt(2)/2) / 1 = 4 * sqrt(2) approx 5.66.
- Relationship between greatest side and greatest angle
- In any triangle, the longest side is always opposite the angle with the greatest measure.
- Relationship between smallest side and smallest angle
- In any triangle, the shortest side is always opposite the angle with the smallest measure.
- Circumradius formula from Law of Sines
- R = a / (2 * sin(A)) = b / (2 * sin(B)) = c / (2 * sin(C)).
- Given a = 14, A = 30 deg. Find the diameter of the circumscribed circle.
- Diameter = 2R = a / sin(A) = 14 / sin(30) = 14 / 0.5 = 28.
- Given b = 10, B = 90 deg. Find circumradius R.
- 2R = 10 / sin(90) = 10, so circumradius R = 5.
- Given a = 6, b = 8, C = 30 deg. Find triangle area.
- Area = 0.5 * 6 * 8 * sin(30) = 24 * 0.5 = 12 square units.
- How rounding intermediate angles can introduce error
- Rounding an angle early distorts subsequent sine values; store exact values in memory.
- Domain limitation of arcsine when solving for angles
- Arcsin returns angles in [-90, 90] degrees; triangle angles must fall strictly in (0, 180) degrees.
- Role of the Law of Sines in navigation and triangulation
- Determines an unknown distance to an object by measuring two angles from a known baseline.