Let ΔABC be a triangle with incentre at I. Also let P and Q be the feet of perpendiculars from A to BI and CI, respectively. Then which of the following results are correct ?
A
AQCI=sin(C2)cos(B2)sin(A2)
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B
APBI=sin(B2)cos(C2)sin(A2)
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C
AQCI=cos(C2)sin(B2)sin(A2)
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D
APBI+AQCI=√3 if ∠A=30∘
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Solution
The correct option is BAPBI=sin(B2)cos(C2)sin(A2) From ΔAPB, AP=ABsin(B2)
From ΔAIB, BIAB=sin(A2)cos(C2) (A+B+C=π)→(A+B+C2=π2)⇒(∠AIB=π+C2) ⇒APBI=sin(B2)cos(C2)sin(A2)
Similarly, AQCI=sin(C2)cos(B2)sin(A2) ⇒APBI+AQCI=sin(B+C2)sin(A2)=cot(A2) ∵cot(A2)=√3⇒∠A=60∘