# 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.

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