text: # LAW OF SINES AND COSINES EXAMPLES
  
  ## About "Law of Sines and Cosines Examples"
  
  Law of Sines and Cosines Examples :
  
  Here we are going to see some example problems based on law of sines and cosines.
  
  ## Law of Sines and Cosines Examples - Examples
  
  Question 1 :
  
  In a triangle ABC, if cos C = sin A / 2 sin B, show that the triangle is isosceles.
  
  Solution :
  
  cos C = sin A / 2 sin B
  
  cos C (2 sin B)  =  sin A
  
  2 sin B cos C  =  sin A
  
  sin (B + C) + sin (B - C)  =  sin A
  
  A + B + C  =  180
  
  B + C  =  180 - A
  
  sin (180 - A) + sin (B - C)  =  sin A
  
  sin A + sin (B - C)  =  sin A
  
  sin (B - C)  =  0
  
  B - C  =  0
  
  B  =  C
  
  Hence the given triangle is isosceles triangle.
  
  Question 2 :
  
  In a triangle ABC, prove that sin B/sinC  =  (c − a cosB)/(b − a cosC)
  
  Solution :
  
  sin B/sinC  =  (c − a cosB)/(b − a cosC)
  
  Laws of sine :
  
  a/sin A  =  b/sin B  =  c/sin C  =  2R
  
  a  =  2R sin A, b = 2R sin B and c = 2R sin C
  
  R.H.S
  
  =  (c − a cosB)/(b − a cosC)
  
  =  (2R sin C - 2R sin A cos B) / (2R sin B - 2R sin A cos C)
  
  =  (sin C - sin A cos B) / (sin B - sin A cos C)
  
  =  (sin C - (1/2) [sin (A+B) + sin (A-B)] / (sin B - (1/2)[sin(A+C) sin (A-C)]
  
  =  (sin C - (1/2) [sin C + sin (A-B)] / (sin B - (1/2)[sinB sin (A-C)]
  
  =  (1/2) [sin C - sin (A-B)]/(1/2)[sin B - sin (A-C)]
  
  =  [sin (A+B) - sin (A-B)]/[sin (A+C) - sin (A-C)]
  
  =  (2 cos A sin B)/(2 cos A sin C)
  
  =  sin B / sin C
  
  Hence proved.
  
  Question 3 :
  
  In a triangle ABC, prove that a cosA + b cosB + c cosC = 2a sinB sinC.
  
  Solution :
  
  L.H.S :
  
  a cosA + b cosB + c cosC
  
  a/sin A  =  b/sin B  =  c/sin C  =  2R
  
  a = 2R sin A, b = 2R sin B  and c = 2R sin C
  
  =  2R sin A cos A + 2R sin B cos B + 2R sin C cos C
  
  =  R[sin 2A + sin 2B + sin 2C]
  
  =  R[2 sin (A + B) cos (A - B) + sin 2C]
  
  =  R[2 sin (180 - C) cos (A - B) + sin 2C]
  
  =  R[2 sin C cos (A - B) + 2 sin C cos C]
  
  =  2R sin C [cos (A - B) + cos C]
  
  =  2R sin C [cos (A-B) + cos (180-(A+B))]
  
  =  2R sin C [cos (A-B) - cos (A+B)]
  
  =  2R sin C [2 sin A sin B]
  
  Instead of 2R sin A, we may apply "a".
  
  =  2a sin A sin B sin C
  
  Hence proved.
  
  After having gone through the stuff given above, we hope that the students would have understood, "Law of Sines and Cosines Examples"
  
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