Trigonometry! #28

Since we have the lengths of 2 sides and a cosine of the angle between them, we can use the Law of Cosines to find the length of the 3rd side (which is $a$ ): $a^2 = 9 + 36 - 36*\left(\frac{1}{8} \right) \ \Rightarrow \ \ a = \frac{9 \sqrt{2}}{2}$

Now we'll add in the angle bisector and call $BD \ \ y$ , so $DC = \frac{9 \sqrt{2}}{2} - y$ . Use the Angle Bisector Theorem to find the value of y: $\frac{3}{y} = \frac{6}{9\sqrt{2} / 2 - y} \Rightarrow \ y = \frac{3\sqrt{2}}{2}$ Finally, use Law of Cosines on $\triangle ABD$ , but we'll need to find $cos(\angle BAD)$ to do that. We'll use the Half-Angle Identity: $cos\left(\frac{\angle BAC}{2}\right) = \sqrt{\frac{1 + 1/8}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

Now use Law of Cosines to find $AD$ : $\left(\frac{3\sqrt{2}}{2}\right)^2 = 3^2 + AD^2 - 6AD*cos \left(\angle BAD\right) \\ \frac{9}{2} = 9 + AD^2 - 6AD(\frac{3}{4}) \\ 9 = 18 + 2AD^2 - 9AD \\ 2AD^2 - 9AD + 9 = 0 \\$

Using the Quadratic formula, we see that $AD = \boxed{3}$ .

(Couldn't add in some of the calculations since my solution was getting long :P).

Use Law of Cosines to find $BC$ :

$BC^2=AB^2+AC^2-2(AB)(AC)(\cos \angle BAC)\\ BC^2=3^2+6^2-2(3)(6)(\cos \angle BAC)\\ BC^2=45-36\left(\dfrac{1}{8}\right) \\ BC^2=\dfrac{81}{2}\\ BC=\dfrac{9}{\sqrt{2}}$

Finally, use Stewart's Theorem to find $AD$ :

$AD=\dfrac{\sqrt{(AB)(AC)(AB+AC+BC)(AB+AC-BC)}}{AB+AC}\\ AD=\dfrac{\sqrt{(3)(6)(3+6+\frac{9}{\sqrt{2}})(3+6-\frac{9}{\sqrt{2}})}}{3+6}\\ AD=\boxed{3}$