Question

Помогите с геометрией

Answer 1

AC=b, <B=β ,<DCB=α
AB=b/sin β
BC=AB*cos β=bcos β/sin β=bctg β
BD=BCtg α= bctg βtg α
DC=BD/sin α= bctg βtg α/sinα=bctg β/cos α S=S(BDA)+S(BDC)+S(CDA)=1/2*BD*AB+1/2*BD*CB+1/2*DC*AC=
=1/2(bctg βtg α* b/sin β+ bctg βtg α* bctg β+ bctg β/cos α*b)=
=b2ctg β(tg α/sin β + ctg βtg α  +1/cos α )

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