Question

In a triangle ABC angle B=60 angle C=45 AND D divides BC internally in the ratio 1:3.?


then
$\frac{sin \: bad}{sin \: cad} \: is \: \: what$
​

Answer 1

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)