Building height $AB=h$, $BC=x$, with $\angle ACB=60^\circ$. Then $ an 60^\circ$ equals—

Question

Building height $AB=h$, $BC=x$, with $\angle ACB=60^\circ$. Then $	an 60^\circ$ equals—

Accepted Answer

$\dfrac{h}{x}$

Answer

$\dfrac{x}{h}$

Answer

$\dfrac{h}{x+42}$

Answer

$\dfrac{x+42}{h}$

Building height AB=hAB=hAB=h, BC=xBC=xBC=x, with ∠ACB=60∘\angle ACB=60^\circ∠ACB=60∘. Then tan⁡60∘\tan 60^\circtan60∘ equals—