COM Q2) Find the value of cos 45º geometrically. ​

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COM
Q2) Find the value of cos 45º geometrically.

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Gabriella 1 month 2021-08-20T11:59:12+00:00 1 Answer 0 views 0

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    2021-08-20T12:00:13+00:00

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BC

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = x

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²)

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²) = √(x² + x²) (∵ AB = BC = x)

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²) = √(x² + x²) (∵ AB = BC = x) =√(2x²)

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²) = √(x² + x²) (∵ AB = BC = x) =√(2x²)∴ AC = √2 x

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²) = √(x² + x²) (∵ AB = BC = x) =√(2x²)∴ AC = √2 xNow we know that cosФ = base/hypotenuse

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²) = √(x² + x²) (∵ AB = BC = x) =√(2x²)∴ AC = √2 xNow we know that cosФ = base/hypotenuse∴ cos∠BAC = AB/AC

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²) = √(x² + x²) (∵ AB = BC = x) =√(2x²)∴ AC = √2 xNow we know that cosФ = base/hypotenuse∴ cos∠BAC = AB/ACor, cos45° = x/√2 x (∵ ∠BAC = 45°)

    Let ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xThen BC = x (∵AB = BC)By Pythagorean Theorem,AC = √(AB² + BC²) = √(x² + x²) (∵ AB = BC = x) =√(2x²)∴ AC = √2 xNow we know that cosФ = base/hypotenuse∴ cos∠BAC = AB/ACor, cos45° = x/√2 x (∵ ∠BAC = 45°)∴ cos45° = 1/√2

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