Moving back $CD=42$ m to point $D$ gives elevation $\angle ADB = 45^\circ$, so $ an 45^\circ$ equals—

Question

Moving back $CD=42$ m to point $D$ gives elevation $\angle ADB = 45^\circ$, so $	an 45^\circ$ equals—

Accepted Answer

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

Answer

$\dfrac{h}{x}$

Answer

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

Answer

$\dfrac{1}{h}$

Moving back CD=42CD=42CD=42 m to point DDD gives elevation ∠ADB=45∘\angle ADB = 45^\circ∠ADB=45∘, so tan⁡45∘\tan 45^\circtan45∘ equals—