We are given a right-angled triangle $ABC$ with a right angle at $B$. The hypotenuse $AC$ has length $12$ cm and the angle $BCA$ is $45^{\circ}$. Point $D$ is on the line extending $BC$ such that $\angle ACD = 180^{\circ} - 45^{\circ} = 135^{\circ}$. We need to find the length of $BC$ in surd form, the value of $\sin(\angle ACD)$ in surd form and the value of $\tan(\angle ACD)$.

1. Problem Description

We are given a right-angled triangle $ABC$ with a right angle at $B$. The hypotenuse $AC$ has length $12$ cm and the angle $BCA$ is $45^{\circ}$. Point $D$ is on the line extending $BC$ such that $\angle ACD = 180^{\circ} - 45^{\circ} = 135^{\circ}$. We need to find the length of $BC$ in surd form, the value of $\sin(\angle ACD)$ in surd form and the value of $\tan(\angle ACD)$.

2. Solution Steps

a) Find $BC$ in surd form.

In the right-angled triangle $ABC$, we have $\angle BCA = 45^{\circ}$. Since the sum of angles in a triangle is $180^{\circ}$, $\angle BAC = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$. Therefore, $\triangle ABC$ is an isosceles right-angled triangle with $AB = BC$.

Using trigonometry, we have $\cos(45^{\circ}) = \frac{BC}{AC}$. We know that $AC = 12$ cm and $\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$.

Therefore, $BC = AC \cos(45^{\circ}) = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2}$ cm.

b) Find $\sin(\angle ACD)$ in surd form.

We know that $\angle ACD = 135^{\circ}$. Then $\sin(\angle ACD) = \sin(135^{\circ})$.

We can express $\sin(135^{\circ})$ as $\sin(180^{\circ} - 45^{\circ})$.

Since $\sin(180^{\circ} - x) = \sin(x)$, we have $\sin(135^{\circ}) = \sin(45^{\circ})$.

We know that $\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$.

Therefore, $\sin(\angle ACD) = \frac{\sqrt{2}}{2}$.

c) Find $\tan(\angle ACD)$.

We know that $\angle ACD = 135^{\circ}$.

Then $\tan(\angle ACD) = \tan(135^{\circ})$.

We can express $\tan(135^{\circ})$ as $\tan(180^{\circ} - 45^{\circ})$.

Since $\tan(180^{\circ} - x) = -\tan(x)$, we have $\tan(135^{\circ}) = -\tan(45^{\circ})$.

We know that $\tan(45^{\circ}) = 1$.

Therefore, $\tan(\angle ACD) = -1$.

3. Final Answer

a) $BC = 6\sqrt{2}$ cm

b) $\sin(\angle ACD) = \frac{\sqrt{2}}{2}$

c) $\tan(\angle ACD) = -1$