LAW OF SINES EXAMPLE PROBLEMS

About "Law of Sines Example Problems"

Law of Sines Example Problems :

Here we are going to see some example problems based on law of sines.

Law of Sines - Examples

Question 1 :

In a triangle ABC, prove that

(ii) a(cos B + cos C) = 2(b + c) sin² A/2

Solution :

a(cos B + cos C) = 2(b + c) sin² A

L.H.S

=  a(cos B + cos C)

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) [2cos (B+C)/2 cos (B-C)/2]

=  (2R sin A) [2cos (180-A)/2 cos (B-C)/2]

=  (2R sin A) [2cos (90-(A/2)) cos (B-C)/2]

=  (2R sin A) [2 sin (A/2) cos (B-C)/2]

=  2R (2sin A/2 cos A/2) [2 sin (A/2) cos (B-C)/2]

=  2R sin²A/2 [4cos A/2 cos (B - C)/2]

=  2R sin²A/2 [4cos (180 - (B+C)/2) cos (B - C)/2]

=  2R sin²A/2 [4 sin (B+C)/2 cos (B - C)/2]

=  2R sin²A/2 [2 (2 sin (B+C)/2 cos (B - C)/2)]

=  2R sin²A/2 [2 (sin B + sin C)]

=  2 sin²A/2 [2R sin B + 2R sin C]

=  2 sin²A/2 [b + c]   --> R.H.S

Hence proved.

(iii) (a² − c²) / b² = sin(A − C) / sin(A + C)

Solution :

L.H.S

=  (a² − c²) / b²

=  [(2R sin A)² - (2R sin C)²] / (2R sin B)²

=  [4R² sin²A - 4R² sin²C] / (4R² sin²B)

=  [sin²A - sin²C] / sin²B

Formula for sin2A - sin2B  =  sin (A + B) sin (A - B)

=  sin (A + C) sin (A - C) / sin²B

=  sin (A + C) sin (A - C) / [sin(180 - (A + C)]²

=  sin (A + C) sin (A - C) / [sin(A + C)]²

= sin (A - C)/sin (A + C)

R.H.S

Hence proved.

(iv)  a sin(B − C)/(b² − c²) = b sin(C − A)/c² − a² = c sin(A − B)/(a² − b²)

Solution :

Part 1 :

  =  a sin(B − C)/(b² − c²)

  =  (2R sin A) sin (B - C) / (2R sin B)² - (2R sin C)² 

  =  (2R sin A) sin (B - C) / (4R² sin² B - (4R² sin² C))

  =   a sin (B - C) /2R (sin² B - sin² C)

  =   a sin (B - C) / 2R sin (B+C) sin (B-C)

  =  a/2R sin (B + C)

  =  a/2R sin (180 - A)

  =  a/2R sin A

  =  a / a   =  1    -----------(1)

Part 2 :

b sin(C − A)/c² − a²

  =  (2R sin B) sin (C - A) / (2R sin C)² - (2R sin A)² 

  =  (2R sin B) sin (C - A) / (4R² sin² C - (4R² sin² A))

  =   b sin (C - A) /2R (sin² C - sin² A)

  =   b sin (C - A) / 2R sin (C+A) sin (C-A)

  =  b/2R sin (C + A)

  =  b/2R sin (180 - B)

  =  b/2R sin B

  =  b / b   =  1    -----------(2)

Part 3 :

c sin(A − B)/a² − b²

  =  (2R sin C) sin (A - B) / (2R sin A)² - (2R sin B)² 

  =  (2R sin C) sin (A - B) / (4R² sin² A - (4R² sin² B))

  =   c sin (A - B) /2R (sin² A - sin² B)

  =   c sin (A - B) / 2R sin (A+B) sin (A-B)

  =  c/2R sin (A + B)

  =  c/2R sin (180 - C)

  =  c/2R sin C

  =  c / c   =  1    -----------(3)

(1)  =  (2)  =  (3)

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "Law of Sines Example Problems"

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