In triangle ABC, {16 R s Δ A/2 B/2 C/2}{s-c} is equal to

Correct Answer is : 4r1r2r34 r₁ r₂ r₃4r1r2r3

Correct Answer is : 4r1r2r34 r₁ r₂ r₃4r1r2r3

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AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 25 of 160

Marks: +1, -0

In triangle

ABC, \frac{16 R s \Delta \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}}{s-c}

is equal to

Solution:

\Delta ABC, A+B+C=\pi

...(i)

When

A, B, C

are the angles of

\Delta ABC

Let

R

be the circumradius and

s

be semi-perimeter and

\Delta

be the area of

\Delta ABC

[\because \frac{\sin A}{2} = \sqrt{\frac{(s-b)(s-c)}{b c}}

\frac{\sin B}{2} = \sqrt{\frac{(s-a)(s-c)}{a c}}

\frac{\sin C}{2} = \sqrt{\frac{s(s-c)}{a b}}]

and

\left[ r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c} \right]

and

\left[ \Delta = \frac{abc}{4R} \Rightarrow 4R \Delta = abc \right]

\frac{16 R s \Delta \sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}}{s-c} = \frac{(4 R \Delta)(4 s)}{s-c} \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-c)}{a c}} \sqrt{\frac{s(s-c)}{a b}}

= \frac{(a b c)(4 s)}{s-c} \frac{(s-c) \sqrt{ s(s-a)(s-b)(s-c) }}{a b c}

= 4 \Delta s

= \frac{4 s \Delta^{3}}{\Delta^{2}} = \frac{4 s \Delta^{3}}{s(s-a)(s-b)(s-c)}

= 4 \cdot \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c}

= 4 r_{1} r_{2} r_{3}

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