Solve the problem.
-A painting 1 meter high and 3 meters from the floor will cut off an angle $\theta$ to an observer, where $\theta = \tan ^ { - 1 } \left( \frac { x } { x ^ { 2 } + 1.6 } \right)$, assuming that the observer is $x$ feet from the wall where the painting is displayed and that the eyes of the observer are $1.6$ meters above the ground (see the figure). Find the value of $\theta$ for $x = 3$. Round to the nearest tenth of a degree.

A) $13.3 ^ { \circ }$
B) $18.3 ^ { \circ }$
C) $15.8 ^ { \circ }$
D) $33.1 ^ { \circ }$

Question

Solve the problem.
-A painting 1 meter high and 3 meters from the floor will cut off an angle $\theta$ to an observer, where $\theta = \tan ^ { - 1 } \left( \frac { x } { x ^ { 2 } + 1.6 } \right)$, assuming that the observer is $x$ feet from the wall where the painting is displayed and that the eyes of the observer are $1.6$ meters above the ground (see the figure). Find the value of $\theta$ for $x = 3$. Round to the nearest tenth of a degree.

A) $13.3 ^ { \circ }$ 
B) $18.3 ^ { \circ }$ 
C) $15.8 ^ { \circ }$ 
D) $33.1 ^ { \circ }$

Quizplus · Accepted Answer

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 $\theta = \tan ^ { - 1 } \left( \frac { x } { x ^ { 2 } + 1.6 } \right)$, assuming that the observer is $x$ feet from the wall where the painting is displayed and that the eyes of the observer are $1.6$ meters above the ground (see the figure). Find the value of $\theta$ for $x = 3$. Round to the nearest tenth of a degree.

A) $13.3 ^ { \circ }$ 
B) $18.3 ^ { \circ }$ 
C) $15.8 ^ { \circ }$ 
D) $33.1 ^ { \circ }$

Solve the Problem θ\thetaθ To an Observer, Where θ=tan⁡−1(xx2+1.6)\theta = \tan ^ { - 1 } \left( \frac { x } { x ^ { 2 } + 1.6 } \right)θ=tan−1(x2+1.6x​)

Solve the Problem $\theta$ To an Observer, Where $\theta = \tan ^ { - 1 } \left( \frac { x } { x ^ { 2 } + 1.6 } \right)$