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sin a=12/13
cos^2a=1-(12/13)^2=25/169
=>cosa=5/13
tan a=12/13:5/13=12/5
cot a=1:12/5=5/12
sin b=căn 3/2
cos^2b=1-(căn 3/2)^2=1/4
=>cos b=1/2
tan b=căn 3/2:1/2=căn 3
cot b=1/căn 3
Bài 1:
a) Ta có:
\(tanB=\dfrac{AC}{AB}\Rightarrow\dfrac{AC}{AB}=\dfrac{5}{2}\)
\(\Rightarrow AC=\dfrac{AB\cdot5}{2}=\dfrac{6\cdot5}{2}=15\)
b) Áp dụng Py-ta-go ta có:
\(BC^2=AB^2+AC^2=6^2+15^2=261\)
\(\Rightarrow BC=\sqrt{261}=3\sqrt{29}\)
Bài 2:
\(\left\{{}\begin{matrix}sinM=sin40^o\approx0,64\Rightarrow cosN\approx0,64\\cosM=cos40^o\approx0,77\Rightarrow sinN\approx0,77\\tanM=tan40^o\approx0,84\Rightarrow cotN\approx0,84\\cotM=cot40^o\approx1,19\Rightarrow tanN\approx1,19\end{matrix}\right.\)
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)
hay \(\cos\alpha=\dfrac{4}{5}\)
Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)
\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)
\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)
\(=\dfrac{141}{25}\)
c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)
\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)
\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)
1.
\(cosa=\sqrt{1-sin^2a}=\frac{4}{5}\)
\(tana=\frac{sina}{cosa}=\frac{3}{4}\)
2.
\(1+tan^2x=\frac{1}{cos^2x}\Rightarrow cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{3}{5}\)
\(sinx=\sqrt{1-cos^2x}=\frac{4}{5}\)
3.
\(sina=\sqrt{1-cos^2a}=\frac{2\sqrt{2}}{3}\)
\(tana=\frac{sina}{cosa}=2\sqrt{2}\)
\(cota=\frac{1}{tana}=\frac{\sqrt{2}}{4}\)
1) \(tan\alpha=\dfrac{2}{3}\)
Mà: \(tan\alpha\cdot cot\alpha=1\)
\(\Rightarrow cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\dfrac{2}{3}}=\dfrac{3}{2}\)
Và: \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)
\(\Rightarrow cos^2\alpha=\dfrac{1}{1+tan^2\alpha}\)
\(\Rightarrow cos\alpha=\sqrt{\dfrac{1}{1+tan^2\alpha}}=\sqrt{\dfrac{1}{1+\left(\dfrac{2}{3}\right)^2}}=\dfrac{3\sqrt{13}}{13}\)
Lại có:
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)
\(\Rightarrow sin\alpha=tan\alpha\cdot cos\alpha=\dfrac{2}{3}\cdot\dfrac{3\sqrt{13}}{13}=\dfrac{2\sqrt{13}}{13}\)