Answer: tgA=12/5
Step-by-step explanation:
[tex]\displaystyle\\0^0 < B < 90^0\\sin^2B+cos^2B=1\\sin^2B+cos^2B-sin^2B=1-sin^2B\\cos^2B=1-sin^2B\\cos^2B=1-(\frac{5}{13})^2\\\\cos^2B=1-\frac{5^2}{13^2} \\\\cos^2B=1-\frac{25}{169} \\\\cos^2B=\frac{1*169-25}{169} \\\\cos^2B=\frac{144}{169}\\\\\sqrt{cos^2B} =\sqrt{\frac{144}{169} } \\\\cosB=б\sqrt{\frac{12^2}{13^2} } \\\\cosB=б\sqrt{(\frac{12}{13})^2 } \\\\cosB=б\frac{12}{13} \\\\As,\ 0^0 < B < 90^0\\\\cosB=\frac{12}{13} \\\\[/tex]
[tex]\displaystyle\\tgB=\frac{sinB}{cosB}\\\\ tgB=\frac{\frac{5}{13} }{\frac{12}{13} } \\\\tgB=\frac{5}{12} \\\\tgA=tg(90^0-tgB)\\\\tgA=ctgB\\\\tgA=\frac{1}{tgB} \\\\tgA=\frac{1}{\frac{5}{12} }\\\\ tgA=\frac{12}{5}[/tex]
