**Law of Sines and Cosines Examples With Answers :**

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

**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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