We are given a right triangle $ABC$ with $\angle B = 90^{\circ}$. We are also given that $AC = 12$ cm and $\angle ACB = 45^{\circ}$. We need to find: a) The length of $BC$ in surd form. b) The value of $\sin(\angle ACD)$ in surd form. Note: $\angle ACD$ is supplementary to $45^{\circ}$. c) The value of $\tan(\angle ACD)$.

1. Problem Description

We are given a right triangle $ABC$ with $\angle B = 90^{\circ}$. We are also given that $AC = 12$ cm and $\angle ACB = 45^{\circ}$. We need to find:

a) The length of $BC$ in surd form.

b) The value of $\sin(\angle ACD)$ in surd form. Note: $\angle ACD$ is supplementary to $45^{\circ}$.

c) The value of $\tan(\angle ACD)$.

2. Solution Steps

a) Finding $BC$:

Since $\triangle ABC$ is a right triangle, we can use trigonometric ratios to find $BC$. We know that $\cos(\angle ACB) = \frac{BC}{AC}$.

So, $\cos(45^{\circ}) = \frac{BC}{12}$.

We are given that $\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$.

Therefore, $\frac{\sqrt{2}}{2} = \frac{BC}{12}$.

Multiplying both sides by 12, we get $BC = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2}$ cm.

b) Finding $\sin(\angle ACD)$:

Since $\angle ACB = 45^{\circ}$, and $\angle ACB$ and $\angle ACD$ are supplementary angles, we have $\angle ACD = 180^{\circ} - 45^{\circ} = 135^{\circ}$.

So, $\sin(\angle ACD) = \sin(135^{\circ})$.

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

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

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

c) Finding $\tan(\angle ACD)$:

We need to find $\tan(135^{\circ})$.

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

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