I want to find the $\cos ({\hat{BCD}})$. However, I don't have any idea about how to. Can you assist?

Question

I want to find the $\cos({\hat{BCD}})$. However, I don't have any idea about how to. Can you assist?

I'm too grateful.

Answer 1

$$AC=\sqrt{13+12}=5,$$ which gives $$\measuredangle ACD=90^{\circ}.$$ Id est, $$\cos\measuredangle BCD=\cos\left(90^{\circ}+\arcsin\frac{2\sqrt3}{5}\right)=-\frac{2\sqrt3}{5}.$$

Answer 2

$AC^2 = (2\sqrt 3)^2 + (\sqrt 13)^2 = 25\\ AC = 5$

$5-12-13$ is a Pythagorean triple. ACD is a right triangle.

$\cos \angle BCD = \cos (\angle BCA + 90^\circ) = -\sin \angle BCA$

Answer 3

Hint:

If the angle in $B$ is a right angle, begin with computing the length $AC$, next $\cos\measuredangle BCA$ and $\sin\measuredangle BCA$, then determine $\cos\measuredangle ACD$ with Al Kashi's theorem and finally use addition formula for cosine.

Answer 4

You can find $\angle BCA = \tan^{-1}(2\sqrt{3}/\sqrt{13})$.

You can find the length $AC$ as the hypotenuse of right triangle $ABC$.

Then, you can find $\angle ACD$ from the law of cosines.

Finally, $\angle BCD = \angle BCA + \angle ACD$.