Sine Law to Solve Triangle Problems

Problem 1: A triangle ABC has angle A = 106 ^o, angle B = 31 ^o and side a = 10 cm. Solve the triangle ABC by finding angle C and sides b and c.(round answers to 1 decimal place).

Solution to Problem 1:

• Use the fact that the sum of all three angles of a triangle is equal to 180 ^o to write an equation in C.
  
  A + B + C = 180 ^o
  
  
• Solve for C.
  
  C = 180 ^o - (A + B) = 43 ^o
  
  
• Use sine law to write an equation in b.
  
  a / sin(A) = b / sin(B)
  
  
• Solve for b.
  
  b = a sin (B) / sin(A) = (approximately) 5.4 cm
  
  
• Use the sine law to write an equation in c.
  
  a / sin(A) = c / sin(C)
  
  
• Solve for c.
  
  c = a sin (C) / sin(A) = (approximately) 7.1 cm
  
  

Problem 2: The angle of elevation to the top C of a building from two points A and B on level ground are 50 degrees and 60 degrees respectively. The distance between points A and B is 30 meters. Points A, B and C are in the same vertical plane. Find the height h of the building(round your answer to the nearest unit).

Solution to Problem 2:

• We consider triangle ABC. Angle B internal to triangle ABC is equal to
  B = 180 ^o - 60 ^o = 120 ^o
  
  
• In the same triangle, angle C is given by.
  C = 180 ^o - (50 ^o + 120 ^o) = 10 ^o
  
  
• Use sine law to find d.
  d / sin(50) = 30 / sin(10)
  
  
• Solve for d.
  d = 30 *sin(50) / sin(10)
  
  
• We now consider the right triangle.
  sin (60) = h / d
  
  
• Solve for h.
  h = d * sin(60)
  
  
• Substitute d by the expression found above.
  h = 30 *sin(50) * sin(60) / sin(10)
  
  
• Use calculator to approximate h.
  h = (approximately) 115 meters.
  
  

Problem 3: A triangle ABC has side a = 12 cm, side b = 19 cm and angle A = 80 ^o (angle A is opposite side a). Find side c and angles B and C if possible.(round answers to 1 decimal place).

Solution to Problem 3:

• Use sine law to write an equation in sin(B).
  
  a / sin(A) = b / sin(B)
  
  
• Solve for sin(B).
  
  sin (B) = (b / a) sin(A) = (19/12) sin(80) = (approximately) 1.6
  
  
• No real angle B satisfies the equation
  
  sin (B) = 1.6
  
  
• The given problem has no solution.

Problem 4: A triangle ABC has side a = 14 cm, side b = 19 cm and angle A = 32 ^o (angle A is opposite side a). Find side c and angles B and C if possible.(round answers to 1 decimal place).

Solution to Problem 4:

• Use sine law to write an equation in sin(B).
  
  a / sin(A) = b / sin(B)
  
  
• Solve for sin(B).
  
  sin (B) = (b / a) sin(A) = (19/14) sin(32) = (approximately) 0.7192
  
  
• Two angles satisfy the equation sin (B) = 0.7192 and the given problem has two solutions
  
  B1 = 46.0 ^o and B2 = 134 ^o
  
  
• Solution 1: Find angle C1 corresponding to B1
  
  C1 = 180 - B1 - A = 102 ^o
  
  
• Solution 1: Find side c1 corresponding to C1
  
  c1 / sin(C1) = a / sin(A)
  
  c1 = 14 sin(102) / sin(32) = (approximately) 25.8 cm
  
  
• Solution 2: Find angle C2 corresponding to B2
  
  C2 = 180 - B2 - A = 14 ^o
  
  
• Solution 2: Find side c2 corresponding to C2
  
  c2 / sin(C2) = a / sin(A)
  
  c1 = 14 sin(14) / sin(32) = (approximately) 6.4 cm

Exercises:

1. A triangle ABC has angle A = 104 ^o, angle C = 33 ^o and side c = 9 m. Solve the triangle ABC by finding angle B and sides a and b.(round answers to 1 decimal place).

2. Redo problem 2 with the distance between points A and B equal to 50 meters.

Solutions to above exercises

1. B = 43 ^o, a = 16.0 m , b = 11.3 m

2. 191 meters.