Consider an acute angle triangle ΔABC with area S. Let the area of its pedal triangle is 'p', satisfies by the relation cosB=2pS and sinB=2√3pS. Then
A
ΔABC is a right angled triangle.
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B
cosA⋅cosB⋅cosC=18
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C
cos(A−C)=1
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D
cos2AcosB+cos2BcosC+cos2CcosA=38
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Solution
The correct options are BcosA⋅cosB⋅cosC=18 Ccos(A−C)=1 Dcos2AcosB+cos2BcosC+cos2CcosA=38 In figure, △DEF is the pedal triangle of △ABC. cosB=2pS,sinB=2√3pS...(1)
We know that, sin2B+cos2B=1 ⇒(2pS)2+(2√3pS)2=1 ⇒S=4p...(2)
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) S=4p ⇒2R2sinA⋅sinB⋅sinC=4×R22sin2A⋅sin2B⋅sin2C ⇒cosA⋅cosB⋅cosC=18...(3)
From eqn(1) and (2), cosB=2pS=2p4p=12 ⇒∠B=π3...(4)
From eqn(3), cosA⋅cosC=14⇒2cosA⋅cosC=12⇒cos(A+C)+cos(A−C)=12⇒cos2π3+cos(A−C)=12⇒cos(A−C)=1⇒A=C