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In a triangle ABC, FB is perpendicular on AC. DE is perpendicular on FB, where D is a point on BC. BE=3 AC=10 find area ADB?

Vikas TU
14149 Points
one year ago
Hiii Jitendr
Please type the question clearly , you said that DE is perpendicular to FB , which is not in the figure .
Figure says FE is perpendicular to DB ,
So, type the question clearly .
Good luck
2075 Points
one year ago

hello jitender this was an awesome question tbh. here also we use the formula for area of a triangle as ½ *ab*sinC.
let angle FBE= p, angle EAC= z. let AE=x.
now area of AEB= A= ½ *x*BE*sinAEB. here, BE= 3secp and angle AEB= 90 – p + z (because angle EAC= angle AEF).
so 2A= 3xsecp*sin(90 – p + z)= 3xsecp*cos(z – p)= 3xsecp(coszcosp+sinzsinp)= 3(xcosz+xsinztanp).....(1)
now, drop a perpendicular from E on AC at R. so, AR= xcosz. and ER= xsinz also, angle REC can be easily seen to be equal to p. so, tanp= RC/ER or RC= ERtanp= xsinztanp
now AR+RC= 10
so xcosz + xsinztanp= 10.......(2)
put (2) in (1)
2A= 3*10
or A = 15
kindly approve :)
if u have any further doubts feel free to ask.