Tính \(\sin M + \cos N\) trong hình bên.

Ta đặt \(AB = m\) thì \(BC = 2m\), suy ra

\(A{C^2} = B{C^2} - A{B^2} = 4{m^2} - {m^2} = 3{m^2} \Rightarrow AC = m\sqrt 3 \).

Ta có \(\sin C = \frac{{AB}}{{BC}} = \frac{m}{{2m}} = \frac{1}{2};\cos C = \frac{{AC}}{{BC}} = \frac{{m\sqrt 3 }}{{2m}} = \frac{{\sqrt 3 }}{2};\)

        \(\tan C = \frac{{AB}}{{AC}} = \frac{m}{{m\sqrt 3 }} = \frac{1}{{\sqrt 3 }};\cot C = \frac{{AC}}{{AB}} = \frac{{m\sqrt 3 }}{m} = \sqrt 3 \).

Ta có \(BH = 6:2 = 3\); \(AB = \sqrt {{4^2} + {3^2}} = 5\). Do đó

\(\begin{array}{*{20}{l}}{}&{\sin B = \frac{{AH}}{{AB}} = \frac{4}{5} = 0,8;}\\{}&{\cos B = \frac{{BH}}{{AB}} = \frac{3}{5} = 0,6;}\\{}&{\tan B = \frac{{AH}}{{AB}} = \frac{4}{3};}\\{}&{\cot B = \frac{{BH}}{{AH}} = \frac{3}{4} = 0,75.}\end{array}\)