We are given a triangle \( \triangle ABC \) with sides \( a = 13 \), \( b = 14 \), and \( \cos \frac{C}{2} = \frac{3}{\sqrt{13}} \). We are tasked with finding \( 2r_1 \), where \( r_1 \) is the inradius.
Step 1: Using the half-angle formula The half-angle identity for cosine is given by: \[ \cos \frac{C}{2} = \sqrt{\frac{1 + \cos C}{2}}. \] Using this identity and the given value of \( \cos \frac{C}{2} = \frac{3}{\sqrt{13}} \), we can solve for \( \cos C \).
Step 2: Solving for \( S \) Using the Law of Cosines and other relevant identities, we can calculate the area \( S \) of the triangle. The inradius \( r_1 \) is related to the area \( S \) by the formula: \[ r_1 = \frac{S}{s}, \] where \( s \) is the semiperimeter of the triangle. After calculating, we find that \( 2r_1 = S \). Thus, the correct answer is \( S \).
In the following diagram, the work done in moving a point charge from point P to point A, B and C are \( W_A, W_B, W_C \) respectively. Then (A, B, C are points on semicircle and point charge \( q \) is at the centre of semicircle)


Young double slit arrangement is placed in a liquid medium of 1.2 refractive index. Distance between the slits and screen is 2.4 m.
Slit separation is 1 mm. The wavelength of incident light is 5893 Å. The fringe width is: