Question

1. what is the tangent ratio of angle bcd? figure 1 options: \\(\frac{10\sqrt{3}}{20}\\), \\(\frac{10}{20\sqrt{3}}\\), \\(\frac{10}{10\sqrt{3}}\\), \\(\frac{10\sqrt{3}}{10}\\)

Explanation:

Step1: Analyze triangle ABD

In right triangle \(ABD\), \(\angle A = 60^\circ\), \(AB = 20\). The side \(AD\) is adjacent to \(\angle A\), and \(BD\) is opposite to \(\angle A\). Using trigonometric ratios, \(\cos 60^\circ=\frac{AD}{AB}\), so \(AD = AB\cos 60^\circ=20\times\frac{1}{2} = 10\). Also, \(\sin 60^\circ=\frac{BD}{AB}\), so \(BD = AB\sin 60^\circ=20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\).

Step2: Analyze triangle BCD

Since \(\angle C = 60^\circ\) and \(BD\perp AC\), triangle \(BCD\) is a right triangle with \(\angle BDC = 90^\circ\). In right triangle \(BCD\), \(\tan\angle BCD=\frac{BD}{CD}\). From the symmetry of the isosceles triangle (since \(\angle A=\angle C = 60^\circ\), triangle \(ABC\) is equilateral, so \(AD = CD = 10\)). So \(CD = 10\) and \(BD = 10\sqrt{3}\). Then \(\tan\angle BCD=\frac{BD}{CD}=\frac{10\sqrt{3}}{10}\).

Answer:

\(\boldsymbol{\frac{10\sqrt{3}}{10}}\) (the last option)