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inatriangleABC, angle C =60 degreeand angle A =75 degree. if D is a point on AC such that the area of the triangle BAD is 3timess the area of the triangle BCD find angle ABD

KAILASH APARTMENT , 7 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 7 Years ago

first of all, angle B = 45 degree
by sine law
 
BA/ BC  = sin 60/ sin 75
 which gives 
BA/ BC = \sqrt6 / (\sqrt3  +  1)
 
we know that area  = (1/2) a* b* sin\theta
let \theta =  angle ABD
now
ar(tr  BAD) =  \sqrt3 ar (tr  BCD)
 BA sin \theta / BC  = \sqrt3 sin(45 –\theta)
put the value of BA/ BC in the above equation
 
then solve and you will find that
\theta = 30 degree  = angle ABD

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