Áp dụng hệ quả của định lí cosin, ta có:

 \(\begin{array}{l}\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}};\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\ \Rightarrow \cos A = \frac{{{{13}^2} + {{15}^2} - {{24}^2}}}{{2.13.15}} =  - \frac{7}{{15}};\cos B = \frac{{{{24}^2} + {{15}^2} - {{13}^2}}}{{2.24.15}} = \frac{{79}}{{90}}\\ \Rightarrow \widehat A \approx 117,{8^ \circ },\widehat B \approx 28,{6^o}\\ \Rightarrow \widehat C \approx 33,{6^o}\end{array}\)

Tham khảo:

Đặt \(AB = c,AC = b,BC = a.\)

Ta có: \(a = 152;\widehat A = {180^o} - ({79^o} + {61^o}) = {40^o}\)

Áp dụng định lí sin, ta có:

\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)

Suy ra:

\(\begin{array}{l}AC = b = \frac{{a.\sin B}}{{\sin A}} = \frac{{152.\sin {{79}^o}}}{{\sin {{40}^o}}} \approx 232,13\\AB = c = \frac{{a.\sin C}}{{\sin A}} = \frac{{152.\sin {{61}^o}}}{{\sin {{40}^o}}} \approx 206,82\\R = \frac{a}{{2\sin A}} = \frac{{152}}{{2\sin {{40}^o}}} \approx 118,235\end{array}\)

a) Áp dụng định lí cosin, ta có:

 \(\begin{array}{l}{a^2} = {b^2} + {c^2} - 2bc.\cos A\\ \Leftrightarrow {a^2} = {8^2} + {5^2} - 2.8.5.\cos {120^ \circ } = 129\\ \Rightarrow a = \sqrt {129} \end{array}\)

Áp dụng định lí sin, ta có:

\(\begin{array}{l}\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} \Rightarrow \frac{{\sqrt {129} }}{{\sin {{120}^ \circ }}} = \frac{8}{{\sin B}} = \frac{5}{{\sin C}}\\ \Rightarrow \left\{ \begin{array}{l}\sin B = \frac{{8.\sin {{120}^ \circ }}}{{\sqrt {129} }} \approx 0,61\\\sin C = \frac{{5.\sin {{120}^ \circ }}}{{\sqrt {129} }} \approx 0,38\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\widehat B \approx 37,{59^ \circ }\\\widehat C \approx 22,{41^ \circ }\end{array} \right.\end{array}\)

b) Diện tích tam giác ABC là: \(S = \frac{1}{2}bc.\sin A = \frac{1}{2}.8.5.\sin {120^ \circ } = 10\sqrt 3 \)

c) 

+) Theo định lí sin, ta có: \(R = \frac{a}{{2\sin A}} = \frac{{\sqrt {129} }}{{2\sin {{120}^ \circ }}} = \sqrt {43} \)

+) Đường cao AH của tam giác bằng: \(AH = \frac{{2S}}{a} = \frac{{2.10\sqrt 3 }}{{\sqrt {129} }} = \frac{{20\sqrt {43} }}{{43}}\)

Áp dụng định lí cosin trong tam giác ABC, ta có:

\(\begin{array}{l}{c^2} = {b^2} + {a^2} - 2ab\cos C\\ \Leftrightarrow {c^2} = 26,{4^2} + 49,{4^2} - 2.26,4.49,4\cos {47^ \circ }20'\\ \Rightarrow c \approx 37\end{array}\)

Áp dụng định lí sin, ta có: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\)

\(\begin{array}{l} \Leftrightarrow \frac{{49,4}}{{\sin A}} = \frac{{26,4}}{{\sin B}} = \frac{{37}}{{\sin {{47}^ \circ }20'}}\\ \Rightarrow \sin A = \frac{{49,4.\sin {{47}^ \circ }20'}}{{37}} \approx 0,982 \Rightarrow \widehat A \approx {79^ \circ }\\ \Rightarrow \widehat B \approx {180^ \circ } - {79^ \circ } - {47^ \circ }20' = {53^ \circ }40'\end{array}\)

a² = 8² + 5² - 2.8.5 cos 120⁰ = 64 + 25 + 40 = 129

=> a = √129 ≈ 11, 36cm

Ta có thể tính góc B theo định lí cosin

cosB = = ≈ 0,7936 => = 37⁰48’

Ta cũng có thể tính góc B theo định lí sin :

cosB = = => sinB ≈ 0,6085 => = 37⁰48’

Tính C từ = 180⁰- ( + ) => ≈ 22⁰12’

Gọi \(\widehat{A}:\widehat{B}:\widehat{C}\)lần lượt là a,b,c

Do \(\widehat{A}:\widehat{B}:\widehat{C}=3:4:5\)

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}\)

Mà tổng \(\widehat{A}:\widehat{B}:\widehat{C}=180^o\)(tổng 3 góc trong tam giác)

=>\(\frac{a+b+c}{3+4+5}=\frac{180}{12}=15\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{3}\\\frac{b}{4}\\\frac{c}{5}\end{cases}}=15\)

\(\Rightarrow\hept{\begin{cases}a=45^o\\b=60^o\\c=75^o\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\widehat{A}=45^o\\\widehat{B}=60^o\\\widehat{C}=75^o\end{cases}}\)

MÀ \(\Delta ABC=\Delta A'B'C'\)

\(\Rightarrow\hept{\begin{cases}\widehat{A}=\widehat{A'}\\\widehat{B}=\widehat{B'}\\\widehat{C}=\widehat{C'}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\widehat{A'=45^o}\\\widehat{B'=60^o}\\\widehat{C'}=75^o\end{cases}}\)

Đặt: \(\widehat{A}=3x\Rightarrow\hept{\begin{cases}\widehat{B}=4x\\\widehat{C}=5x\end{cases}}\)

Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow3x+4x+5x=180^o\)

\(\Rightarrow x=15\)

\(\Rightarrow\hept{\begin{cases}\widehat{A'}=\widehat{A}=3x=45^o\\\widehat{B}'=\widehat{B}=4x=60^o\\\widehat{C'}=\widehat{C}=75^o\end{cases}}\)

Vì \(\widehat{A}-\widehat{B}=\widehat{B}-\widehat{C}\) nên \(\widehat{A}-2\widehat{B}+\widehat{C}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}-2\widehat{B}+\widehat{C}=0^0\left(1\right)\\\widehat{A}+\widehat{B}+\widehat{C}=180^0\left(2\right)\end{matrix}\right.\)

Trừ \(\left(2\right)\) cho \(\left(1\right)\), ta được \(3\widehat{B}=180^0\Rightarrow\widehat{B}=60^0\)

\(\Rightarrow\widehat{A}+\widehat{C}=120^0\)

Vậy GTLN của \(\widehat{A}\) là \(119^0\) vì \(\widehat{C}>0\)

$\widehat{ABC}$