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`        In a triangle ABC, DB is perpendicular on AC. EF is perpendicular on DB, where E is a point on BC. FB=3 AC=10 find area AEB?`
11 months 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= xsinztanpnow AR+RC= 10so xcosz + xsinztanp= 10.......(2)put (2) in (1)2A= 3*10or A = 15kindly approve :)
```
11 months ago
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