EXAMPLE PROBLEMS FOR LAW OF SINES AND COSINES

Question 1 :

In a triangle ABC, if sin A/sin C = sin(A − B)/sin(B − C), prove that a², b², c² are in Arithmetic Progression.

Solution :

sin A/sin C = sin(A − B)/sin(B − C)

sin A sin (B - C)  =  sin C sin (A - B)

Since A, B and C are angles of triangle

A + B + C  =  180

A  =  180 - (B + C)

C  =  180 - (A + B)

sin A sin (B - C)  =  sin C sin (A - B)

sin (180 - (B+C)) sin (B-C)  =  sin (180 - (A+B)) sin (A-B)

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

(1/2)[cos (B+C-B+C) - cos (B+C+B-C)]  =  (1/2)[cos (A+B-A+B) - cos (A+B+A-B)]

(1/2)[cos 2C - cos 2B]  =  (1/2)[cos 2B - cos 2A]

cos 2C - cos 2B  =  cos 2B - cos 2A

2 cos 2B  =  cos 2A + cos 2C

2(1-2sin²B)  =  1 - 2sin²A + 1 - 2sin²C

2 - 4sin²B  =  2 - 2sin²A - 2sin²C

4sin²B  =  2sin²A + 2sin²C

2sin²B  =  sin²A + sin²C

Using the law of sin, we get

a/sin A  =  b/sin B  =  c/sin C  =  2R

sin A  =  a/2R, sin B  =  b/2R, sin C  =  c/2R

2sin²B  =  sin²A + sin²C

2(b/2R)^{2  =}  (a/2R)² + (c/2R)²

2b²/4R²  =  (a² + c²)/4R²

2b²  =  (a² + c²)

Hence a², b² and c² are in A.P

Question 2 :

The angles of a triangle ABC, are in Arithmetic Progression and if b : c = √3 : √2, find ∠A.

Solution :

Let A, B and C be the angles of a triangle.

Since the angles are in arithmetic progression, the following relationship holds.

B - A  =  C - B

2B  =  A + C

A + B + C  =  180

2B + B  =  180

3B  =  180, B = 60

Laws of sine :

a/sin A  =  b/sin B  =  c/ sin C  =  2R

a/sin A  =  √3/sin 60  =  √2/sin C

a/sin A  =  √3/(√3/2)  =  √2/sin C

  2  =  √2/sin C

  sin C  =  1/√2

  C  =  45

A + 60 + 45  =  180

A + 105  =  180

A  =  75

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