IIT JEE , AIEEE Forum - Mathematics , Trignometry

salonee_hs

In a

ABC

B=

/3 and

C=

/4. Let D divide BC internally in the ratio 1:3, then sin

BAD/sin

CAD=

aysh

hai Salonee, HERE IT GOES>>>

In triangle ABD,
sinB / AD = sinBAD / BD
sinBAD = (sinB x BD) / AD
Similarly,

In triangle CAD,
sinC / AD = sinCAD / CD
sinCAD = (sinC x CD)/AD

THUS,
sinBAD / sinCAD = (sinB / sinC) x (BD / CD)
= (1/6)^(1/2)

PLEASE RATE MY EFFORT...

amar.gupta

good work aysh

chirag

In

ABD and

ACD

using sine rule,

sinBAD/BD=sinABD/AD

therefore, AD=sinABD*BD/sinBAD.......(i)

similarly,in

triangle ACD, sinCAD/CD=sinACD/AD........(ii)

Equating (i) and (ii) for AD

sinBAD*2/

3BD=sinCAD*CD

2

therefore we get sinCAD/sinBAD=1/

6

^{Please rate me...............}

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