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

## Answers ( )

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 = BCLet ΔABC be a right angle Δ with ∠B = 90° and ∠BAC = 45°∴∠ACB will also be 45°∵∠BAC = ∠ACB = 45°∴ AB = BCLet AB = xLet 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 xLet 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/hypotenuseLet 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/ACLet 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