We are given a diagram of a triangle ABC, where angle B is a right angle. The length of AC is 12 cm and the angle BCD is 145 degrees. We need to find: a) The length of BC in surd form. b) The value of sin(ACD) in surd form. c) The value of tan(ACD). We are given that sin(45) = $\frac{\sqrt{2}}{2}$ and cos(45) = $\frac{\sqrt{2}}{2}$.

1. Problem Description

We are given a diagram of a triangle ABC, where angle B is a right angle. The length of AC is 12 cm and the angle BCD is 145 degrees. We need to find:

a) The length of BC in surd form.

b) The value of sin(ACD) in surd form.

c) The value of tan(ACD).

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

2. Solution Steps

a) Finding BC:

Since $\angle BCD = 145^\circ$, $\angle BCA = 180^\circ - 145^\circ = 35^\circ$.

This is incorrect. $\angle ACD = 180^\circ - 145^\circ = 35^\circ$, therefore $\angle BCA = 45^\circ$

Since $\angle BCA = 45^\circ$ and $\angle ABC = 90^\circ$, $\angle BAC = 180^\circ - 90^\circ - 45^\circ = 45^\circ$. This means triangle ABC is an isosceles right triangle.

Thus, $BC = AB$.

We know that $AC = 12$ cm. Using the sine formula in triangle ABC,

$\frac{BC}{sin(\angle BAC)} = \frac{AC}{sin(\angle ABC)}$.

So $\frac{BC}{sin(45^\circ)} = \frac{12}{sin(90^\circ)}$.

Therefore, $BC = \frac{12 sin(45^\circ)}{sin(90^\circ)} = 12 \times \frac{\sqrt{2}}{2} = 6\sqrt{2}$ cm.

b) Finding sin(ACD):

Since $\angle BCD = 145^\circ$, $\angle ACD = 180^\circ - \angle BCD = 180^\circ - 145^\circ = 35^\circ$.

So, $\sin(ACD) = \sin(35^\circ)$.

But we are assuming that $\angle BCA = 45^\circ$, so $\angle ACD = 180^\circ - 145^\circ = 35^\circ$.

Since, $\angle BCA = 45^\circ$, then the triangle ABC is isosceles with AB = BC, so, if $AC = 12$, we know that $AC = \sqrt{2} BC$ and thus $BC = \frac{12}{\sqrt{2}} = \frac{12\sqrt{2}}{2} = 6\sqrt{2}$.

Then, $\angle BCA = 45^\circ$, and $\angle ACD = 180^\circ - 145^\circ = 35^\circ$.

Since $\angle ACD = 180^{\circ} - \angle BCD$, therefore $\angle ACD = 180^\circ - 145^\circ = 35^\circ$.

There is no other information that we can use.

We cannot find the exact value of sin(35) in surd form. I suspect that the angle BCD is incorrect. If the question intended for point D to be on line BC, this would be different.

Instead let us assume that angle BCA is a right angle, i.e., 45 degrees.

$\angle ACD = 180 - 145 = 35$. Then we would be calculating sin(35) and tan(35).

I will instead assume that D is on line BC and we are calculating the values for $45^\circ$.

$\angle ACD = 45^\circ$

$\sin(45^\circ) = \frac{\sqrt{2}}{2}$

c) Finding tan(ACD):

Again assume $\angle ACD = 45^\circ$

$tan(45^\circ) = \frac{sin(45^\circ)}{cos(45^\circ)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$.

3. Final Answer

Assuming $\angle BCA = 45^\circ$ (isosceles right angled triangle at B), and point D on line BC, and thus $\angle ACD = 45^\circ$, then:

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

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

c) $tan(ACD) = 1$