In the given figure, $\displaystyle \angle B =90^{\circ}$ and $\displaystyle \angle ADB=x^{\circ}$, then find

$\displaystyle \cos^{2} C^{\circ}+\sin^{2} C^{\circ}$.

In $\triangle CAB$,

$CA = 20$ and $AB = 12$

Using Pythagoras theorem,

$CA^2 = AB^2 + BC^2$

$20^2 = 12^2 + BC^2$

$BC = 16$

$\cos^2 C + \sin^2 C = \left (\dfrac{BC}{AC}\right )^2 + \left (\dfrac{AB}{AC}\right )^2$

= $\left (\dfrac{16}{20}\right )^2 + \left (\dfrac{12}{20}\right )^2$

= $\dfrac{256 + 144}{400}$

= $\dfrac{400}{400}$

= $1$