In triangle BCD, angle C = 90 degrees, CD = 3, and BD = sqrt(17) . What is tan B?
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\(tan\ B=\dfrac{3}{\sqrt{17-3^2}}=\dfrac{3}{\sqrt{8}}=\dfrac{3\cdot \sqrt8}{8}=\color{blue}\dfrac{3\cdot \sqrt{2}}{4}=1.06066\)
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