Solve the problem.
-A painting 1 meter high and 3 meters from the floor will cut off an angle $\theta$ to an observer, where $\theta = \tan ^ { - 1 } \left( \frac { \mathrm { x } } { \mathrm { x } ^ { 2 } + 1.8 } \right)$, assuming that the observer is $x$ feet from the wall where the painting is displayed and that the eyes of the observer are $1.8$ meters above the ground (see the figure). Find the value of $\theta$ for $x = 2$. Round to the nearest tenth of a degree.

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The Answer of Solve the problem.
-A painting 1 meter high and 3 meters from the floor will cut off an angle $\theta$ to an observer, where $\theta = \tan ^ { - 1 } \left( \frac { \mathrm { x } } { \mathrm { x } ^ { 2 } + 1.8 } \right)$, assuming that the observer is $x$ feet from the wall where the painting is displayed and that the eyes of the observer are $1.8$ meters above the ground (see the figure). Find the value of $\theta$ for $x = 2$. Round to the nearest tenth of a degree.

Solve the Problem θ\thetaθ To an Observer, Where θ=tan⁡−1(xx2+1.8)\theta = \tan ^ { - 1 } \left( \frac { \mathrm { x } } { \mathrm { x } ^ { 2 } + 1.8 } \right)θ=tan−1(x2+1.8x​)

Solve the Problem $\theta$ To an Observer, Where $\theta = \tan ^ { - 1 } \left( \frac { \mathrm { x } } { \mathrm { x } ^ { 2 } + 1.8 } \right)$