Giải thích các bước giải:
Ta có:
$\hat B=180^o-\hat A-\hat C=72^o$
Lại có:
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$
$\to \dfrac{a}{\sin30^o}=\dfrac{b}{\sin72^o}=\dfrac{28}{\sin78^o}$
$\to a\approx 14.31, c\approx 27.22, R=14.31$
$\to S_{ABC}=\dfrac12ab\sin C\approx 190.5$
$P_{ABC}=a+b+c=69.53$
Mà $S_{ABC}=\dfrac12r\cdot p\to r\approx 5.48$
Ta có:
$\sin A=\sin30^o=\dfrac12$
$\cos A=\cos30^o=\dfrac{\sqrt{3}}{2}$
$\tan A=\tan30^o=\dfrac{1}{\sqrt{3}}$
$\cot A=\cot 30^o=\sqrt{3}$
$\sin B=\sin72^o\approx 0.95$
$\cos B=\cos72^o\approx 0.31$
$\tan B=\tan72^o\approx 3.08$
$\cot B=\cot 72^o\approx 0.32$
$\sin C=\sin78^o\approx 0.98$
$\cos C=\cos78^o\approx 0.21$
$\tan C=\tan78^o\approx 4.70$
$\cot C=\cot78^o\approx 0.21$
