LAW OF SINES AND COSINES 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.

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