Evaluate x and y from the given figure.

In the given figure there are two right angled triangles, △ADC and △BDC.

In △ADC,

∠ACD = 60° and AC = 10 m, CD = x m

cos 60° = $\dfrac{\text{base}}{\text{hypotenuse}}$

$\dfrac{1}{2} = \dfrac{CD}{AC}$

$\dfrac{1}{2} = \dfrac{x}{10}$

x = 5 m.

In △BDC,

BC = $5\sqrt{2}$ m

CD = x = 5 m

sin y° = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{DC}{BC} = \dfrac{5}{5\sqrt{2}} = \dfrac{1}{\sqrt{2}}$

sin y° = $\dfrac{1}{\sqrt{2}}$

sin y° = sin 45°

y° = 45°.

Hence, x = 5 m and y° = 45°.