Question

Answer 1

Bài 1:
a.

$\widehat{C}=90^0-\widehat{B}=90^0-51^0=39^0$

$\frac{AB}{BC}=\cos B=\cos 51^0$

$\Rightarrow BC=\frac{AB}{\cos 51^0}=\frac{c}{\cos 51^0}=\frac{3,8}{\cos 51^0}=6$ (cm)

$\frac{AC}{AB}=\tan B\Rightarrow AC=\tan B.AB=\tan 51^0.c=\tan 51^0.3,8=4,7$ (cm)

b.

$\widehat{B}=90^0-\widehat{C}=90^0-60^0=30^0$

$\frac{AB}{BC}=\sin C\Rightarrow AB=BC\sin C=a.\sin 60^0=11.\sin 60^0=9,5$ (cm)

$\frac{AC}{BC}=\cos C\Rightarrow AC=BC\cos C=a.\cos 60^0=11.\cos 60^0=5,5$ (cm)

Answer 2

Bài 2:

$\cos ^2a=1-\sin ^2a=1-\frac{1}{25}=\frac{24}{25}$

Vì $a$ nhọn nên $\cos a>0$

$\Rightarrow \cos a=\sqrt{\frac{24}{25}}=\frac{2\sqrt{6}}{5}$

$\tan a=\frac{\sin a}{\cos a}=\frac{1}{5}:\frac{2\sqrt{6}}{5}=\frac{\sqrt{6}}{12}$

$\cot a=\frac{1}{\tan a}=\frac{12}{\sqrt{6}}$