طبق قضیه‌ی فیثاغورس در مثلث $ABC$ داریم: $B{C^2} = {\sqrt {13} ^2} - {2^2} = 13 - 4 = 9 \Rightarrow BC = 3$$\tan A\hat C\,B = \frac{{AB}}{{BC}} = \frac{2}{3} \Rightarrow {\tan ^2}A\hat CB = {(\frac{2}{3})^2} = \frac{4}{9}$$ \to \tan \hat D = \frac{2}{3}$$\sin \hat D = \sin A\hat CB = \frac{2}{{\sqrt {13} }}$$\cos \hat D = \cos A\hat CB = \frac{3}{{\sqrt {13} }}$