LAW OF SINES AND COSINES EXAMPLES WITH ANSWERS

Question 1 :

In a triangle ABC, ∠A = 60°. Prove that b + c = 2a cos (B − C)/2

Solution :

b + c = 2a cos (B − C)/2

R.H.S

=  2a cos (B − C)/2

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

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

here ∠A = 60°

a = 2R sin (60)  =  2R (√3/2)  =  √3R

  =  2(√3 R) cos (B − C)/2  ---(1)

L.H.S

  =  b + c

  =  2R sin B + 2R sin C.

  =  2R [sin B + sin C]

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

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

  =  2R [2 cos A/2 cos (B - C)/2]

  =  2R [2 cos (60/2) cos (B - C)/2]

  =  2R [2 cos 30 cos (B - C)/2]

  =  2R [2 (√3/2)cos (B - C)/2]

  =  2R√3 cos (B - C)/2  ------(2) 

(1)  =  (2)

Hence proved.

Question 2 :

In a triangle ABC, prove the following

(i) a sin (A/2 + B) = (b + c) sin A/2

Solution :

R.H.S

  =  (b + c) sin A/2

  =  (2R sin B + 2R sin C) sin A/2

  =  2R (sin B + sin C) sin A/2

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

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

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

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

  =  2R (sin A cos (B - C)/2)    -------(1)

Finding the value of cos (B - C)/2 :

cos (B - C)/2  =  cos (B - (180 - (A+B)))/2

  =  cos (B-180+A+B)/2

  =  cos (2B-180+A)/2

  =  cos (B-90+A/2) 

  =  cos (90 - (B+A/2)))

cos (B - C)/2  =  sin (B+A/2)

Applying the value of cos (B - C)/2 in (1)

  =  2R sin A sin (B+A/2)

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