In a triangle ABC angle B=60 angle C=45 AND D divides BC internally in the ratio 1:3.?
then
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now...
BD/DC=1/3...(I)
Now applying sine rule for ∆BAD we can write
sinBAD/sin ADB=BD/AB.....(ii)
noww.....the same rule applying on the ∆ADC
..sinCAD/sinACD=DC/AC
=>sinCAD/sin(180°-ADB)=DC/AC
=>sin CAD/sin ADB=DC/AC.......(iii)
now dividing (ii)÷(iii)....
sinBAD/sinCAD=(BD×AC)/(DC×AB)...,(IV)
now...
AC/AB=sin60°/sin45°.......(v)
now....just combining (I),(IV) and (V)...we get..
sinBAD/sinCAD=sin60°/(3×sin45°)
=>sinBAD/SinCAD=(1/√6)
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