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प्रश्न
In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ? , ∠B = ?, ∠C = ?
उत्तर
sin A = 1 .....[Given]
But, sin 90° = 1
∴ sin A = sin 90°
∴ A = 90°
`sqrt(2)` AC = BC .....[Given]
∴ `"AC"/"BC" = 1/sqrt(2)` .....(i)
∴ sin B = `"AC"/"BC"` ......(ii) [By definition]
∴ sin B = `1/sqrt(2)` .....[From (i) and (ii)]
But, sin 45° = `1/sqrt(2)`
∴ sin B = sin 45°
∴ B = 45°
sin2A + sin2B + sin2C = 2 .....[Given]
∴ `(1)^2 + (1/sqrt(2))^2 + sin^2"C"` = 2
∴ `1 + 1/2 + sin^2"C"` = 2
∴ sin2C = `2 - 3/2`
∴ sin2C = `1/2`
∴ sin C = `1/sqrt(2)`
But, sin 45° = `1/sqrt(2)`
∴ sin C = sin 45°
∴ C = 45°
∴ ∠A = 90°, ∠B = 45°, ∠C = 45°
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