Question

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Answer 1

Bài 4:

\(sin^2B+cos^2B=1\)

=>\(cos^2B=1-0,4^2=0,84\)

=>\(cosB=\sqrt{0,84}=\sqrt{\dfrac{84}{100}}=\sqrt{\dfrac{21}{25}}=\dfrac{\sqrt{21}}{5}\)

\(sinB=cosC\)

mà \(sinB=0,4\)

nên cosC=0,4

\(cosB=sinC\)

=>\(sinC=\dfrac{\sqrt{21}}{5}\)

\(tanB=\dfrac{sinB}{cosB}=\dfrac{2}{5}:\dfrac{\sqrt{21}}{5}=\dfrac{2}{\sqrt{21}}\)

=>\(cotC=tanB=\dfrac{\sqrt{21}}{2}\)

\(tanC=\dfrac{1}{cotC}=\dfrac{2}{\sqrt{21}}\)

Bài 4:

a: \(sin^2\alpha+cos^2\alpha=1\)

=>\(sin^2\alpha=1-\left(\dfrac{5}{13}\right)^2=1-\dfrac{25}{169}=\dfrac{144}{169}\)

=>\(sin\alpha=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\)

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{12}{13}:\dfrac{5}{13}=\dfrac{12}{5}\)

\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{12}{5}=\dfrac{5}{12}\)

b: \(sin^2\alpha+cos^2\alpha=1\)

=>\(cos^2\alpha=1-\left(\dfrac{12}{13}\right)^2=1-\dfrac{144}{169}=\dfrac{25}{169}\)

=>\(cos\alpha=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}\)

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{12}{13}:\dfrac{5}{13}=\dfrac{12}{5}\)

\(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{5}{13}:\dfrac{12}{13}=\dfrac{5}{12}\)

c: \(tan\alpha\cdot cot\alpha=1\)

=>\(cot\alpha=1:\dfrac{2}{3}=\dfrac{3}{2}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)

=>\(\dfrac{1}{cos^2\alpha}=1+\left(\dfrac{2}{3}\right)^2=\dfrac{13}{9}\)

=>\(cos^2\alpha=\dfrac{9}{13}\)

=>\(cos\alpha=\sqrt{\dfrac{9}{13}}=\dfrac{3}{\sqrt{13}}\)

\(\dfrac{sin\alpha}{cos\alpha}=tan\alpha\)

=>\(sin\alpha=cos\alpha\cdot tan\alpha=\dfrac{3}{\sqrt{13}}\cdot\dfrac{2}{3}=\dfrac{2}{\sqrt{13}}\)