Question

as shown in the figure below, in right △abc, m∠bac = 90°, ad⊥bc at d. if bd = 3, cd = 2, then tan b = ______. a $\frac{3}{2}$ b $\frac{2}{3}$ c $\frac{sqrt{6}}{2}$ d $\frac{sqrt{6}}{3}$

Explanation:

Step1: Use the geometric - mean theorem

In right - triangle \(ABC\) with \(AD\perp BC\), we know that \(AD^{2}=BD\times CD\). Given \(BD = 3\) and \(CD = 2\), then \(AD=\sqrt{BD\times CD}=\sqrt{3\times2}=\sqrt{6}\).

Step2: Calculate \(\tan B\)

In right - triangle \(ABD\), \(\tan B=\frac{AD}{BD}\). Since \(AD = \sqrt{6}\) and \(BD = 3\), then \(\tan B=\frac{\sqrt{6}}{3}\).

Answer:

D. \(\frac{\sqrt{6}}{3}\)