a) In triangle ABC, angle C = (3 x angle A) If a = 27, c = 48, find length of line b.
b) (Separate question) In triangle ABC, angle A = 2 x angle B. If BC = 9, AC=6, and D is a point on BC such that AD bisects angle A, find the length AD.
These questions are supposedly about sine law, cosine law, solutions of triangles, and ratio and proportion in triangles. I can't seem to find how to apply those to the questions. Thank you so much for all your help.
The first problem.
Let $\measuredangle A=\alpha$.
Thus, by law of sines we obtain:
$$\frac{\sin{3\alpha}}{48}=\frac{\sin\alpha}{27}$$ or $$\frac{3-4\sin^2\alpha}{16}=\frac{1}{9}$$ and from here we can get a value of $\alpha$ and $$\measuredangle B=180^{\circ}-4\alpha.$$ I got $$\alpha=\arcsin\frac{\sqrt{11}}{6},$$ $$\cos\measuredangle B=-\cos4\arcsin\frac{\sqrt{11}}{6}=\frac{113}{162}$$ and by law of cosines we obtain: $$AC=\sqrt{27^2+48^2-2\cdot27\cdot48\cdot\frac{113}{162}}=35.$$
The second problem.
Let $\measuredangle B=\beta.
Thus, $\measuredangle C=180^{\circ}-3\beta$ and by law of sines we obtain: $$\frac{\sin\beta}{6}=\frac{\sin2\beta}{9}$$ or $$\frac{1}{2}=\frac{2\cos\beta}{3}$$ or $$\cos\beta=\frac{3}{4}.$$ Now, by law of sines for $\Delta ADC$ we obtain: $$\frac{AD}{\sin3\beta}=\frac{6}{\sin2\beta}$$ or $$\frac{AD}{3-4\sin^2\beta}=\frac{3}{\cos\beta},$$ which gives $$AD=\frac{3(4\cos^2\beta-1)}{\cos\beta}=\frac{3\left(4\left(\frac{3}{4}\right)^2-1\right)}{\frac{3}{4}}=5.$$