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Grade: 11

                        

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

3 years ago

Answers : (1)

Arun
24742 Points
							
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
3 years ago
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