Write down the values of the following using the triangle below: ① cos x ② Sine y ③ tan x

In the right-angled triangle $\text{ABD}$, we use the Pythagorean theorem:

$$\text{AB}^2 = \text{AD}^2 + \text{BD}^2$$ $$22^2 = \text{AD}^2 + 11^2$$ $$484 = \text{AD}^2 + 121$$ $$\text{AD}^2 = 484 - 121$$ $$\text{AD}^2 = 363$$ $$\text{AD} = \sqrt{363} = \sqrt{121 \times 3} = 11\sqrt{3}$$

In the right-angled triangle $\text{ACD}$, we use the Pythagorean theorem:

$$\text{AC}^2 = \text{AD}^2 + \text{CD}^2$$ $$\text{AC}^2 = (11\sqrt{3})^2 + 14^2$$ $$\text{AC}^2 = 363 + 196$$ $$\text{AC}^2 = 559$$ $$\text{AC} = \sqrt{559}$$

Now we can find the required trigonometric values:

1. $\cos x$: In the right-angled triangle $\text{ACD}$:

   $$\cos x = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\text{CD}}{\text{AC}}$$ $$\cos x = \frac{14}{\sqrt{559}}$$ $$\boxed{\cos x = \frac{14}{\sqrt{559}}}$$

2. $\sin y$: In the right-angled triangle $\text{ABD}$:

   $$\sin y = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\text{AD}}{\text{AB}}$$ $$\sin y = \frac{11\sqrt{3}}{22}$$ $$\boxed{\sin y = \frac{\sqrt{3}}{2}}$$

3. $\tan x$: In the right-angled triangle $\text{ACD}$:

   $$\tan x = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{AD}}{\text{CD}}$$ $$\tan x = \frac{11\sqrt{3}}{14}$$ $$\boxed{\tan x = \frac{11\sqrt{3}}{14}}$$