Solve the problem.
-A rotating beacon is located a distance $d$ from a long wall. The distance $d$ is given by $\mathrm { d } = 3 \tan 2 \pi \mathrm { t }$,
where $t$ is the time measured in seconds since the beacon started rotating. Solve the equation for $\mathrm { t }$.
A) $\mathrm { t } = \frac { 1 } { 2 \pi } \tan \frac { \mathrm { d } } { 3 }$
B) $t = 2 \pi \arctan \frac { \mathrm { d } } { 3 }$
C) $t = \frac { 1 } { 2 \pi } \arctan \frac { d } { 3 }$
D) $t = 2 \pi \tan \frac { d } { 3 }$

Question

Quizplus · Accepted Answer

To solve for $t$, we need to isolate $t$ on one side of the equation. Starting with $d = 3 \tan 2 \pi t$, we first divide both sides by 3, getting $\frac{d}{3} = \tan 2 \pi t$. Then, to solve for $t$, we apply the inverse tangent ($\arctan$) to both sides, resulting in $2 \pi t = \arctan \frac{d}{3}$. Finally, dividing both sides by $2 \pi$ gives $t = \frac{1}{2 \pi} \arctan \frac{d}{3}$.

Solve the Problem ddd From a Long Wall ddd Is Given By d=3tan⁡2πt\mathrm { d } = 3 \tan 2 \pi \mathrm { t }d=3tan2πt

Solve the Problem $d$ From a Long Wall $d$ Is Given By $\mathrm { d } = 3 \tan 2 \pi \mathrm { t }$