Question

1. what is the length of \\(\overline{bd}\\)? figure of a triangle with vertices a, b, c, and d on ac, angle at a and c is 60°, ab = 20, bd is perpendicular to ac options: \\(20\sqrt{3}\\), 10, 20, \\(10\sqrt{3}\\)

Explanation:

Step1: Identify triangle type

In $\triangle ABC$, $\angle A = \angle C = 60^\circ$, so $\triangle ABC$ is equilateral? Wait, no, $\angle B$: sum of angles in triangle is $180^\circ$, so $\angle B = 180 - 60 - 60 = 60^\circ$. Wait, but $BD \perp AC$, so $\triangle ABD$ is a right triangle with $\angle A = 60^\circ$, $\angle ADB = 90^\circ$, so $\angle ABD = 30^\circ$.

Step2: Use trigonometry in $\triangle ABD$

In right triangle $ABD$, $\sin(60^\circ) = \frac{BD}{AB}$. Given $AB = 20$. $\sin(60^\circ) = \frac{\sqrt{3}}{2}$. So $BD = AB \times \sin(60^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$. Wait, no, wait: wait, $\angle A = 60^\circ$, $AB = 20$, $\angle ADB = 90^\circ$. So $\sin(\angle A) = \frac{BD}{AB}$. So $\sin(60^\circ) = \frac{BD}{20}$. So $BD = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$. Wait, but let's check again. Alternatively, in a 30-60-90 triangle, the sides are in ratio $1 : \sqrt{3} : 2$. Wait, if $\angle ABD = 30^\circ$, then the side opposite 30° is $AD$, so $AD = \frac{AB}{2} = 10$, then $BD = AD \times \sqrt{3} = 10\sqrt{3}$. Yes, that's correct.

Answer:

$10\sqrt{3}$ (corresponding to the option with $10\sqrt{3}$)