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a: AC=căn 5^2+12^2=13cm
sin C=AB/AC=12/13
cos C=5/13
tan C=12/5
cot C=1:12/5=5/12
b: AC=căn 10^2+3^2=căn 109(cm)
sin C=AB/AC=3/căn 109
cos C=BC/AC=10/căn 109
tan C=AB/BC=3/10
cot C=10/3
c: BC=căn 5^2-3^2=4cm
sin C=AB/AC=3/5
cos C=4/5
tan C=3/4
cot C=4/3
a: Xét ΔABC vuông tại A có
\(BC^2=AB^2+AC^2\)
hay AC=12(cm)
b: Xét ΔBAC vuông tại A có
\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AC}{BC}=\dfrac{12}{13}\)
\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{5}{13}\)
\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{AC}{AB}=\dfrac{12}{5}\)
\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{AB}{AC}=\dfrac{5}{12}\)
\(a,\sin\widehat{C}=\dfrac{AB}{BC};\cos\widehat{C}=\dfrac{AC}{BC};\tan\widehat{C}=\dfrac{AB}{AC};\cot\widehat{C}=\dfrac{AC}{AB}\\ b,BC=\sqrt{AB^2+AC^2}=13\left(cm\right)\left(pytago\right)\\ \Rightarrow\sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{12}{13};\cos\widehat{B}=\dfrac{AB}{BC}=\dfrac{5}{13}\\ \tan\widehat{B}=\dfrac{AC}{AB}=\dfrac{12}{5};\cot\widehat{B}=\dfrac{AB}{AC}=\dfrac{5}{12}\)
\(\tan\widehat{B}=\dfrac{AC}{AB}=\dfrac{12}{5}\approx\tan67^022'\\ \Rightarrow\widehat{B}\approx67^022'\\ \Rightarrow\widehat{C}=90^0-67^022'=22^038'\)
a,Sin B=\(\frac{AC}{BC}=\)\(\frac{4}{5}=0.8\)
Cos B=\(\frac{AB}{BC}=\frac{3}{5}=0,6\)
Tan B =\(\frac{AC}{AB}=\frac{4}{3}\)
Cot B=\(\frac{AB}{AC}=\frac{3}{4}=0,75\)
b,Vì sin B = 0,8 => B=53o
=> C=37o(áp dụng hệ quả định lí tổng r tính)
a: \(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AC}{BC}\)
\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{AB}{BC}\)
\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{AC}{AB}\)
\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{AB}{AC}\)
\(\tan B=\sqrt{2}\Leftrightarrow\dfrac{\sin B}{\cos B}=\sqrt{2}\Leftrightarrow\sin B=\sqrt{2}\cos B\\ \sin^2B+\cos^2B=1\Leftrightarrow3\cos^2B=1\\ \Leftrightarrow\cos B=\sqrt{\dfrac{1}{3}}=\dfrac{\sqrt{3}}{3}\\ \Leftrightarrow\sin B=\dfrac{\sqrt{6}}{3}\\ \Leftrightarrow\left\{{}\begin{matrix}\sin C=\cos B=\dfrac{\sqrt{3}}{3}\\\cos C=\sin B=\dfrac{\sqrt{6}}{3}\end{matrix}\right.\\ \cot C=\tan B=\sqrt{3};\tan C=\dfrac{1}{\cot C}=\dfrac{\sqrt{3}}{3}\)
Câu 1:
\(a,=\cos40^0\\ b,=\cot28^0\\ c,=\sin18^0\\ d,=\tan38^0\)
Câu 2:
\(a,\widehat{B}=90^0-52^0=38^0\\ AC=\sin B\cdot BC\approx9,2\left(cm\right)\\ AB=\sin C\cdot BC\approx11,8\left(cm\right)\)