Toán Lớp 10: Cho a=6, b=13 , B=60° Tính c, C ,A
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Toán Lớp 10: Cho a=6, b=13 , B=60° Tính c, C ,A
Comments ( 1 )
Giải đáp+Lời giải và giải thích chi tiết:
C_1:
Áp dụng định lý sin:
$\dfrac{a}{\sin \widehat{A}}=\dfrac{b}{\sin \widehat{B}}=\dfrac{c}{\sin \widehat{C}}\\ \Rightarrow \sin \widehat{A}=\dfrac{a\sin \widehat{B}}{b} =\dfrac{3\sqrt{3}}{13}\\ \Rightarrow \sin \widehat{A} \approx 23,56^\circ\\ \Rightarrow \widehat{C} =96,44 ^\circ\\ \dfrac{b}{\sin \widehat{B}}=\dfrac{c}{\sin \widehat{C}}\\ \Rightarrow c=\dfrac{c}{b\sin \widehat{C}}{\sin \widehat{B}} \approx 14,92.$
C_2:
Áp dụng định lý cos:
$\cos \widehat{B} =\dfrac{a^2+c^2-b^2}{2ac}\\ \Leftrightarrow \dfrac{1}{2} =\dfrac{6^2+c^2-13^2}{2.6.c}\\ \Leftrightarrow \dfrac{1}{2} =\dfrac{c^2-133}{12c}\\ \Leftrightarrow 2c^2-266=12c\\ \Leftrightarrow 2c^2-12c-266=0\\ \Leftrightarrow c^2-6c-133=0\\ \Leftrightarrow \left[\begin{array}{l} c=3+\sqrt{142} \\ c=3-\sqrt{142}(L)\end{array} \right.\\ \cos \widehat{A} =\dfrac{b^2+c^2-a^2}{2bc} =\dfrac{13^2+(3+\sqrt{142} )^2-6^2}{2.13.(3+\sqrt{142} )}=\dfrac{\sqrt{142}}{13}\\ \Rightarrow \widehat{A} \approx 23,56^\circ\\ \Rightarrow \widehat{C} =96,44 ^\circ.$