Question

In the triangle below, If Sin B =5/13, Find the Tan A

Answer 1

Answer: tgA=12/5

Explanation:

`\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-((5)/(13))^2\\\\cos^2B=1-(5^2)/(13^2) \\\\cos^2B=1-(25)/(169) \\\\cos^2B=(1*169-25)/(169) \\\\cos^2B=(144)/(169)\\\\√(cos^2B) =\sqrt{(144)/(169) } \\\\cosB=б\sqrt{(12^2)/(13^2) } \\\\cosB=б\sqrt{((12)/(13))^2 } \\\\cosB=б(12)/(13) \\\\As,\ 0^0 < B < 90^0\\\\cosB=(12)/(13) \\\\`

`\displaystyle\\tgB=(sinB)/(cosB)\\\\ tgB=((5)/(13) )/((12)/(13) ) \\\\tgB=(5)/(12) \\\\tgA=tg(90^0-tgB)\\\\tgA=ctgB\\\\tgA=(1)/(tgB) \\\\tgA=(1)/((5)/(12) )\\\\ tgA=(12)/(5)`