Question

1. Cho tam giác ABC có a=13,b=8,c=10

a) Tam giác ABC có góc tù không?

b) Tính R?

c) Tính S tam giác ABC

d) Tính r, ma ,ha

Answer 1

a) Áp dụng định lý Cô sin, ta có :

 \(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{8^2+10^2-13^2}{2.8.10}=\dfrac{-5}{2.8.10}=-\dfrac{1}{32}< 0\)

Nên góc A là góc tù

\(cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{13^2+10^2-8^2}{2.13.10}=\dfrac{205}{2.13.10}=\dfrac{41}{52}>0\)

Nên góc B là góc nhọn

\(cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{13^2+8^2-10^2}{2.13.8}=\dfrac{33}{2.13.8}=\dfrac{33}{208}>0\)

Nên góc B là góc nhọn

b) \(p=\dfrac{a+b+c}{2}=\dfrac{13+8+10}{2}=\dfrac{31}{2}\)

\(S=\sqrt[]{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

\(\Leftrightarrow S=\sqrt[]{\dfrac{31}{2}\left(\dfrac{31}{2}-13\right)\left(\dfrac{31}{2}-8\right)\left(\dfrac{31}{2}-10\right)}\)

\(\Leftrightarrow S=\sqrt[]{\dfrac{31}{2}.\dfrac{5}{2}.\dfrac{15}{2}.\dfrac{11}{2}}=\dfrac{5\sqrt[]{1023}}{4}\)

\(S=\dfrac{abc}{4R}\Rightarrow R=\dfrac{abc}{4S}=\dfrac{13.8.10}{4.\dfrac{5\sqrt[]{1023}}{4}}=\dfrac{208\sqrt[]{1023}}{1023}\)