Aditya Gupta
Last Activity: 5 Years 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.
