ΔABC vuông tại A

=>\(\hat{B}+\hat{C}=90^0\)

=>\(\hat{C}=90^0-58^0=32^0\)

Xét ΔABC vuông tại A có sin B=\(\frac{AC}{BC}=\frac{b}{a}\)

=>\(\frac{b}{72}=\sin58\)

=>\(b=72\cdot\sin58\) ≃61,06(cm)

ΔABC vuông tại A

=>\(b^2+c^2=a^2\)

=>\(c^2=a^2-b^2\)

=>c≃38,15(cm)

Kẻ AH⊥BC tại H

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC\)

=>\(h_{A}=\frac{AB\cdot AC}{BC}\) =38,15*61,06/72≃32,35(cm)

Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot BA\cdot BC\cdot sinABC\)

\(=\dfrac{1}{2}\cdot5\cdot7\cdot sin120=\dfrac{35\sqrt{3}}{4}\)

Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{5^2+7^2-AC^2}{2\cdot5\cdot7}=cos120=\dfrac{-1}{2}\)

=>\(25+49-AC^2=-35\)

=>\(AC^2=25+49+35=109\)

=>\(AC=\sqrt{109}\)

Kẻ AH\(\perp\)BC

=>\(h_A=AH\)

\(S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC\)

=>\(\dfrac{1}{2}\cdot AH\cdot7=\dfrac{35\sqrt{3}}{4}\)

=>\(AH\cdot3,5=\dfrac{35\sqrt{3}}{4}\)

=>\(AH=\dfrac{10\sqrt{3}}{4}=\dfrac{5}{2}\sqrt{3}\)

Xét ΔABC có \(\dfrac{AC}{sinB}=2R\)

=>\(2R=\dfrac{\sqrt{109}}{sin120}=\sqrt{109}\cdot\dfrac{2}{\sqrt{3}}\)

=>\(R=\sqrt{\dfrac{109}{3}}=\dfrac{\sqrt{327}}{3}\)

Bạn xem lại đề nhé, còn thiếu dữ kiện gì nhé

Xét ΔABC có \(cosA=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(=\frac{8^2+10^2-7^2}{2\cdot8\cdot10}=\frac{115}{16\cdot10}=\frac{115}{5\cdot32}=\frac{23}{32}\)

=>\(\sin A=\sqrt{1-\left(\frac{23}{32}\right)^2}=\frac{3\sqrt{55}}{32}\)

\(S_{ABC}=\frac12\cdot b\cdot c\cdot\sin A\)

\(=\frac12\cdot8\cdot10\cdot\frac{3\sqrt{55}}{32}=\frac{80}{64}\cdot3\sqrt{55}=\frac54\cdot3\sqrt{55}=\frac{15\sqrt{55}}{4}\)

Xét ΔABC có \(\frac{BC}{\sin A}=2R\)

=>\(2R=7:\frac{3\sqrt{55}}{32}=7\cdot\frac{32}{3\sqrt{55}}=\frac{224}{3\sqrt{55}}\)

=>\(R=\frac{112}{3\sqrt{55}}\)

\(p=\frac{a+b+c}{2}=\frac{7+8+10}{2}=\frac{25}{2}=12,5\)

S=p*r

=>\(r=\frac{15\sqrt{55}}{4}:12,5=\frac{15\sqrt{55}}{4\cdot12,5}=\frac{15\sqrt{55}}{50}=\frac{3\sqrt{55}}{10}\)

\(S=\frac12\cdot a\cdot h_{a}\)

=>\(\frac12\cdot7\cdot h_{a}=\frac{15\sqrt{55}}{4}\)

=>\(h_{a}=\frac{15\sqrt{55}}{4}:\frac72=\frac{15\sqrt{55}}{4}\cdot\frac27=\frac{15\sqrt{55}}{14}\)

Xét ΔABC có \(m_{a}\) là độ dài đường trung tuyến

nên \(m_{a}^2=\frac{b^2+c^2}{2}-\frac{a^2}{4}=\frac{8^2+10^2}{2}-\frac{7^2}{4}=69,75\)

=>\(m_{a}=\sqrt{69,75}=\sqrt{\frac{279}{4}}=\frac{3\sqrt{31}}{2}\)

Xét ΔABC có \(cosA=\frac{b^2+c^2-a^2}{2\cdot b\cdot c}\)

=>\(8^2+5^2-a^2=2\cdot8\cdot5\cdot cos120=-40\cdot2\cdot\frac12=-40\)

=>\(a^2=64+25+40=129\)

=>\(a=\sqrt{129}\)

\(S_{ABC}=\frac12\cdot b\cdot c\cdot\sin BAC\)

\(=\frac12\cdot8\cdot5\cdot\sin120=20\cdot\frac{\sqrt3}{2}=10\sqrt3\)

=>\(\frac12\cdot h_{A}\cdot a=10\sqrt3\)

=>\(h_{A}=10\sqrt3:\frac{\sqrt{129}}{2}=10\sqrt3\cdot\frac{2}{\sqrt{129}}=\frac{20}{\sqrt{43}}\)

XétΔABC có \(\frac{a}{\sin A}=2R\)

=>\(2R=\sqrt{129}:\sin120=\sqrt{129}:\frac{\sqrt3}{2}=\sqrt{129}\cdot\frac{2}{\sqrt3}=2\cdot\sqrt{43}\)

=>\(R=\sqrt{43}\)