In any triangle ABC, a + b + c = 2s with usual notation, then sin (A/2) =

Correct option is B

Given:

a + b + c = 2s

Formula Used:

​$$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$

​​$$\cos 2\theta = 2\cos^2 \theta - 1$$ 

cosine rule in triangle where a, b and c are the sides of a triangle 

$$c^2 = a^2 + b^2 - 2ab \cdot \cos(\theta)$$​​​

Solution:

a + b + c = 2s

=> a + b = 2s – c

=> a + c = 2s – b

We know that 

​$$c^2 = a^2 + b^2 - 2ab \cdot \cos(\theta)$$

put $$\theta$$ =A​

Cos A = $$\frac{b^2 + c^2– a^2}{2bc}$$​ ___eq(1) 

Now,

​$$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$ 

put $$\theta$$ = A

2sin²($$\frac A2$$​) = 1 – cos A

Now,

put the value of cos(A) from equation(1) 

2sin² (A/2) = $$1 –\frac{b^2 + c^2– a^2}{2bc}$$​​

  $$\implies\frac{2bc -(b^2+c^2-a^2)}{2bc} \\ \ \\ \implies \frac{2bc -(b^2+c^2-a^2)}{2bc} \\ \ \\ \implies \frac{2bc -b^2 -c^2+a^2}{2bc}\\ \ \\ \implies \frac{a^2-(-2bc +b^2+c^2)}{2bc}\\ \ \\ \implies \frac{a^2-(b-c)^2}{2bc}\\ \ \\ \implies \frac{(a+b-c)(a-b+c)}{2bc}$$​​ 

Substitute the semi-perimeter relations based on a + b + c = 2s:

a − b + c = (a + b + c) − 2b = 2s − 2b = 2(s − b)

a + b − c = (a + b + c) − 2c = 2s − 2c = 2(s − c)

Now, 

​$$2\sin^2 (\frac A2) = 2 × \frac{(s – b) (s – c)}{bc}$$​​

​$$\sin^2 (\frac A2) =\frac{ (s – b) (s – c)}{bc}$$​​

$$\sin(\frac A2)$$​ =$$\sqrt{\frac{(s-b)(s-c)}{bc}}$$ 

Exam Hall Method:  

Semi-perimeter s = (a + b + c)/2

For an equilateral triangle: a = b = c and angle A = B = C = 60^∘

Assume an Equilateral Triangle: Let all sides a = b = c = 2

Now, semi perimeter = $$\frac{2+2+2}2$$ = 3

Since A = 60^∘,

​$$\sin(\frac A2)=\sin(\frac{60^∘}{2})=\sin30^∘=\frac12$$​​

from Option (b): $$\sqrt{\frac{(s-b)(s-c)}{bc}} = \sqrt{\frac{(3-2)(3-2)}{2 \times 2}} = \sqrt{\frac 14} = \frac 12$$​​

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