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ANAM V. RANGA REDDY Grade: Upto college level
`        The vertices of a triangle are (0, 0),(sec θ, tanθ) and (4, -5) (0<θ<∏/2). The minimum possible area is --------`
7 years ago

## Answers : (1)

147 Points
```										Dear ANAM

open determinant
Δ =1/2| [-5secΘ -4tanΘ]|
Δ =1/2| [5secΘ +4tanΘ]|
clearly at Θ =0 both tanΘ  and sec Θ is minimum in there given domain.
so put Θ=0
Δ =5/2
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7 years ago
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