Relation between Exradius ,semiperimeter and circumradius can be given by
( $r_{1}$ is the radius of the circle opposite the angle A)

r_{1} = \frac{2 Δ}{s - a}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

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r_{1} = \frac{Δ}{s - a}, r_{1} = 4 R s i n \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}

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r_{1} = \frac{2 Δ}{(s - a) (s - b)}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

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r_{1} = \frac{Δ (s - a)}{(s - b) (s - c)}, r_{1} = 4 R c o s \frac{A}{2} s i n \frac{B}{2} s i n \frac{C}{2}

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Solution

The correct option is B $r_{1} = \frac{Δ}{s - a}, r_{1} = 4 R s i n \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}$
Relation between Exradius and semiperimeter can be given by, $r_{1} = \frac{Δ}{s - a}$ where $r_{1}$ is the radius of the circle opposite the angle A. and it's helpful to remember the identity $r_{1} = 4 R s i n \frac{A}{2} c o s \frac{B}{2} c o s \frac{C}{2}$

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