The pole stands $BC=x$, breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, w

Question

The pole stands $BC=x$, breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, with $\angle BCD=30^\circ$. From $	riangle BCD$, $	an 30^\circ$ equals—

Accepted Answer

$\dfrac{BD}{BC}$

Answer

$\dfrac{BC}{BD}$

Answer

$\dfrac{CD}{BC}$

Answer

$\dfrac{CD}{BD}$

The pole stands BC=xBC=xBC=x, breaks without separation, and the broken piece CDCDCD slopes down to DDD at distance BD=10BD=10BD=10 m, with ∠BCD=30∘\angle BCD=30^\circ∠BCD=30∘. From △BCD\triangle BCD△BCD, tan⁡30∘\tan 30^\circtan30∘ equals—

The pole stands $BC=x$ , breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, with $\angle BCD=30^\circ$ . From $\triangle BCD$ , $\tan 30^\circ$ equals—