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Question

# The area of an acute triangle ABC is Δ, the area of its pedal triangle is 'p', where cosB=2pΔ and sinB=2√3pΔ. The value of 8(cos2AcosB+cos2C) is

A
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B
2
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C
3
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D
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Solution

## The correct option is C 3 △DEF is pedal triangle of area p. Area of △ABC=Δ cosB=2pΔ, sinB=2√3pΔ ...(1) ∵sin2B+cos2B=1 ⇒(2pΔ)2+(2√3pΔ)2=1 ⇒Δ=4p ...(2) Δ=12absinC =2R2sinA⋅sinB⋅sinC (∵a=2RsinA,b=2RsinB) Circumradius of a pedal triangle is half the circumradius of the original triangle. Also, angles of the pedal triangle will be 180o−2A,180o−2b,180o−2C Now, area of the pedal triangle will be ∴p=2(R2)2sin2A⋅sin2B⋅sin2C =R22sin2A⋅sin2B⋅sin2C From eqn(2) Δ=4p ⇒2R2sinA⋅sinB⋅sinC=4×R22sin2A⋅sin2B⋅sin2C ⇒cosA⋅cosB⋅cosC=18 ...(3) From eqn(1) and (2), cosB=2pΔ=2p4p=12 ⇒∠B=π3 ...(4) From eqn(3), cosA⋅cosC=14 ∠B=π3⇒∠A+∠C=120∘ ⇒cos(A+C)=−12 ⇒cosA⋅cosC−sinA⋅sinC=−12 ⇒14−sinA⋅sinC=−12 ⇒sinA⋅sinC=34 cos(A−C)=cosA⋅cosC+sinA⋅sinC ⇒cos(A−C)=14+34=1 ⇒A−C=0 ⇒A=C ∵B=π3 ∴A=B=C=π3 8(cos2AcosB+cos2C)=8(14×12+14) =3

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