Question #4536d

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Question #4536d

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Answer 1 · 2017-04-07T23:10:06

$θ \approx 230^{\circ}$ $theta ~~ 230^@$ or $θ \approx 4 radians$ $theta ~~ 4" radians"$

Explanation:

We know that the angle must be in the Third Quadrant, because the signs of both of the components are negative, therefore, we must add $180^{\circ}$ $180^@$ or $π radians$ $pi" radians"$ to the inverse tangent function:

$θ = {tan}^{- 1} (\frac{y}{x}) + 180^{\circ}$ $theta = tan^-1(y/x)+180^@$ or $θ = {tan}^{- 1} (\frac{y}{x}) + π radians$ $theta = tan^-1(y/x)+pi" radians"$

$θ = {tan}^{- 1} (\frac{- 4}{- 3}) + 180^{\circ}$ $theta = tan^-1((-4)/-3)+180^@$ or $θ = {tan}^{- 1} (\frac{- 4}{- 3}) + π radians$ $theta = tan^-1((-4)/-3)+pi" radians"$

$θ \approx 230^{\circ}$ $theta ~~ 230^@$ or $θ \approx 4 radians$ $theta ~~ 4" radians"$

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Question #4536d

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