Bài 1: S=$\frac{1}{2}$ bc.sinA=$\frac{1}{2}$ .20.35.sin60o≈303,109
a²=b²+c²-2bc.cosA=>a=√b²+c²-2bc.cosA = √20²+35²-20.35.cos60o = 5√51
+S=$\frac{1}{2}$.a.h <=> 303,109=$\frac{1}{2}$.5√51.h <=> h≈16,977
+$\frac{a}{sinA}$=2R <=> $\frac{5√51}{sin60o}$=2R <=>R=5√17
p=$\frac{a+b++c}{2}$ = $\frac{5√51 + 20+35}{2}$ = $\frac{55+5√51}{2}$
+S=pr<=>303,109=$\frac{55+5√51}{2}$ .r <=> r≈6,683
Bài 2: BC²=AB²+AC²-2.AB.AC.cosA<=>BC=√AB²+AC²-2AB.AC.cosA = √10²+4²-2.10.4.cos60o = 2√19
+p=AB+AC+BC=10+4+2√19 = 14+2√19
+Mình không ra được tanC xin lỗi nha!
Bài 3:
a) BC²=AB²+AC²-2AB.AC.cosA<=>BC=√AB²+AC²-2.AB.AC.cosA = √5²+8²-2.5.8.cos60o = 7
b) S=$\frac{1}{2}$ .AB.AC.sinA=$\frac{1}{2}$ . 5.8.sin60o=10√3
c) AC²=AB²+BC²-2.AB.BC.cosB=>cosB=$\frac{AC²-AB²-BC²}{-2.AB.BC}$=$\frac{8²-5²-7²}{-2.5.7}$ = $\frac{1}{7}$ => B≈81o47'=>B là góc nhọn
d) S=$\frac{1}{2}$ .BC.AH<=>10√3=$\frac{1}{2}$.7.AH <=> AH=$\frac{20√3}{7}$
e) $\frac{BC}{sinA}$ = 2R <=> $\frac{2√19}{sin60o}$ =2R <=> R=$\frac{2√57}{3}$
Xong rồi nhé! ^^
