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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: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\)
\(\begin{array}{l} \Leftrightarrow \frac{{49,4}}{{\sin A}} = \frac{{26,4}}{{\sin B}} = \frac{{37}}{{\sin {{47}^ \circ }20'}}\\ \Rightarrow \sin A = \frac{{49,4.\sin {{47}^ \circ }20'}}{{37}} \approx 0,982 \Rightarrow \widehat A \approx {79^ \circ }\\ \Rightarrow \widehat B \approx {180^ \circ } - {79^ \circ } - {47^ \circ }20' = {53^ \circ }40'\end{array}\)
Ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\Rightarrow\widehat{A}=75^o\)
* \(\dfrac{BC}{sinA}=\dfrac{AB}{sinC}\Rightarrow AB=\dfrac{BCsinC}{sinA}=a\left(1+\sqrt{3}\right)\)
* \(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\Rightarrow AC=\dfrac{BCsinB}{sinA}=a\left(\dfrac{-6+3\sqrt{2}}{2}\right)\)
\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{18^2+20^2-14^2}{2.18.20}=\dfrac{11}{15}\).
Vậy \(\widehat{A}=42^o50'\).
\(cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{14^2+20^2-18^2}{2.14.20}=\dfrac{17}{20}\).
Vậy \(\widehat{B}=60^o56'\).
Vậy \(\widehat{C}=180^o-\widehat{A}-\widehat{B}=77^o46'\).
Áp dụng định lý cô sin trong tam giác ABC:
\(c^2=a^2+b^2-2abcosC=7^2+23^2-2.7.23.cos130\)\(\cong784cm\).
Vậy \(c=28cm.\)
\(cosA=\dfrac{c^2+b^2-a^2}{2bc}=\dfrac{28^2+23^2-7^2}{2.23.28}=\dfrac{158}{161}\).
\(\Rightarrow\widehat{A}\cong11^o\).
\(\widehat{B}=180^o-\left(\widehat{A}+\widehat{C}\right)=180^o-\left(130^o+11^o\right)=39^o\).
\(\widehat{B}=180^o-\left(40^o+120^o\right)=20^o\).
A C B 35 H
\(AH=AB.sinB=35.sin20^o\cong12cm.\)
\(\widehat{HCA}=180^o-120^o=60^o\).
\(AH=AC.sin60^o\Rightarrow AC=\dfrac{AH}{sin60}=\dfrac{12}{\dfrac{\sqrt{3}}{2}}=8\sqrt{3}\).
Áp dụng định lý Cô-sin:
\(BC=\sqrt{AB^2+AC^2-2.AB.AC.sinA}\)\(=\sqrt{35^2+\left(8\sqrt{3}\right)^2-2.35.8\sqrt{3}.cos40^o}\cong26cm\).
Vậy \(a=26cm;b=8\sqrt{3}cm,\)\(\widehat{B}=20^o\).
