In a Δ ABC. with usual notation, match the items in List - I with the items in List - II and choose the correct option. List - I List - II A) r₁ r₂ {4R - r₁ - r₂/r₁ + r₂ I) b B) {r₂ (r₃ + r₁)}{{r₁ r₂ + r₂ r₃ + r₃ r₁} II) a², b², c² are in A.P. C) a/c = (A-B)/(B-C) III) Δ D) bc ² A/2 IV) R r₁ r₂ r₃ V) s (s - a)

Correct Answer is : A - III , B - I , C - II , D - V

A - IV , B - III , C - I , D - V

A - V , B - IV , C - III , D - II

A - III , B - I , C - II , D - V

A - IV , B - V , C - II , D - I

Correct Answer is : A - III , B - I , C - II , D - V

Test Index

AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 27 of 160

Marks: +1, -0

Solution:

For A

\text{ Consider } r_1 r_2 \sqrt{\frac{4R - r_1 - r_2}{r_1 + r_2}}

=\left[ \frac{\Delta^2}{(s-a)(s-b)} \times \left( \sqrt{ \frac{4R - 4R \cos \frac{C}{2} \left( \sin \frac{A}{2} \cos \frac{B}{2} + \sin \frac{B}{2} \cos \frac{A}{2} \right)}{4R \cos \frac{C}{2} \left( \sin \frac{A}{2} \cos \frac{B}{2} + \sin \frac{B}{2} \cos \frac{A}{2} \right)} } \right) \right]

= \frac{\Delta^2}{(s-a)(s-b)} \sqrt{ \frac{4R \left(1 - \cos^2 \frac{C}{2}\right)}{4R \cos^2 \frac{C}{2}} }

= \frac{\Delta^2}{(s-a)(s-b)} \tan \frac{C}{2}

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

Then

r_1 r_2 \sqrt{\frac{4R - r_1 - r_2}{r_1 + r_2}} = \frac{\Delta^2}{\Delta} = \Delta

For

B

\frac{r_2 (r_3 + r_2)}{\sqrt{r_1 r_2 + r_2 r_3 + r_3 r_1}}

= \frac{ \frac{\Delta}{s-b} \left( \frac{\Delta}{s-c} + \frac{\Delta}{s-a} \right) }{ \sqrt{ \frac{(s-a)+(s-b)+(s-c)}{(s-a)(s-b)(s-c)} } }

= \frac{ \frac{\Delta}{s-b} \cdot \frac{2S - a - c}{(s-a)(s-c)} }{ \sqrt{ \frac{3S - a - b - c}{(s-a)(s-b)(s-c)} } }

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

= b

For C

\frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}

\frac{\sin A}{\sin C} = \frac{\sin(A-B)}{\sin(B-C)}

\sin A \sin (B-C) = \sin(A-B) \sin C

2 \sin A \cos B \sin C = \sin A \sin B \cos C + \sin B \cos A \sin C

This implies

2 \frac{a}{2R} \left( \frac{a^2 + c^2 - b^2}{2ac} \right) \left( \frac{c}{2R} \right)

= \left( \frac{a}{2R} \left( \frac{b}{2R} \right) \left( \frac{a^2 + b^2 - c^2}{2abc} \right) \right) + \left( \frac{b}{2R} \left( \frac{b^2 + c^2 - a^2}{2bc} \right) \left( \frac{c}{2R} \right) \right)

2(a^2 + c^2 - b^2) = a^2 + b^2 - c^2 + b^2 + c^2 - a^2

2a^2 + 2c^2 - 2b^2 = 2b^2

2b^2 = a^2 + c^2

Therefore the correct option is (3)

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