ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios :

(i) sin 45°

(ii) cos 45°

(iii) tan 45°

In Δ ABC,

⇒ AC² = BC² + AB² (∴ AC is hypotenuse)

⇒ AC² = x² + x²

⇒ AC² = 2x²

⇒ AC = $\sqrt{2\text{x}^2}$

⇒ AC = $\sqrt{2}$ x

As Δ ABC is isosceles right angled triangle, ∠BAC = ∠BCA = $\dfrac{90°}{2}$ = 45°

(i) sin 45°

sin 45° = sin A = sin C

sin A = $\dfrac{Perpendicular}{Hypotenuse}$

= $\dfrac{BC}{AC} = \dfrac{x}{\sqrt2x} = \dfrac{1}{\sqrt2}$

Hence, sin 45° = $\dfrac{1}{\sqrt2}$.

(ii) cos 45°

cos 45° = cos A = cos C

cos A = $\dfrac{Base}{Hypotenuse}$

= $\dfrac{AB}{AC} = \dfrac{x}{\sqrt2x} = \dfrac{1}{\sqrt2}$

Hence, cos 45° = $\dfrac{1}{\sqrt2}$.

(iii) tan 45°

tan 45° = tan A = tan C

tan A = $\dfrac{Perpendicular}{Base}$

= $\dfrac{BC}{AB} = \dfrac{x}{x}$ = 1

Hence, tan 45° = 1.