Cho tam giác ABC, biết b = 7 , c = 5 , cos A = 3 /5 . Khi đó: a) sin A = 4 /5

a) Đúng

b) Đúng

c) Sai

d) Sai

 

Ta có: \({\sin ^2}A = 1 - {\cos ^2}A = \frac{{16}}{{25}} \Rightarrow \sin A = \frac{4}{5}\) (vì \(\sin A > 0\)).

\(S = \frac{1}{2}bc\sin A = 14\)

\({a^2} = {b^2} + {c^2} - 2bc\cos A = {7^2} + {5^2} - 2.7 \cdot 5 \cdot \frac{3}{5} = 32 \Rightarrow a = 4\sqrt 2 \). \(\frac{a}{{\sin A}} = 2R \Rightarrow R = \frac{a}{{2 \cdot \sin A}} = \frac{{5\sqrt 2 }}{2};p = \frac{{a + b + c}}{2} = 6 + 2\sqrt 2 \)

\(S = pr \Rightarrow r = \frac{S}{p} = \frac{{14}}{{6 + 2\sqrt 2 }} = 3 - \sqrt 2 \).