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Choose the point on the terminal side of -210°.

  1. (1, $\sqrt{3}$)
  2. (2, 4)
  3. (-$\sqrt{3}$, 3)

The question aims to find the point on the cartesian plane of a given angle on the terminal side.

The question is based on the concept of trigonometric ratios. Trigonometry deals with a right-angle triangle, its sides, and angle with its base.

Expert Answer

The given information about this problem is given as:

\[ \theta = -210^ {\circ} \]

Different points of the terminal side are given and we need to find the correct one. We can use the $\tan$ identity to check the value of the given angle and match it with the given points.

The trigonometric identity is given as:

\[ \tan \theta = \dfrac{ y }{ x } \]

\[ \tan (-210^ {\circ}) = \dfrac{ y }{ x } \]

\[ \dfrac{ y }{ x } = – \dfrac{ \sqrt {3} }{ 3 } \]

a) (1, $\sqrt{3}$)

Here, we replace the values of x and y and simplify them to see if it equals the desired result.

\[ \dfrac{ y }{ x } = \dfrac{ 1 }{ \sqrt {3} } \]

This point is not on the terminal side of $-210^ {\circ}$.

b) (2, 4)

\[ \dfrac{ y }{ x } = \dfrac{ 4 }{ 2 } \]

\[ \dfrac{ y }{ x } = 2 \]

This point is not on the terminal side of $-210^ {\circ}$.

c) ($\sqrt{3}$, 3)

\[ \dfrac{ y }{ x } = \dfrac{ \sqrt {3} }{ 3 } \]

This point lies on the terminal side of $-210^ {\circ}$.

Numerical Result

The point (-$\sqrt{3}$, 3) lies on the terminal side of $-210^ {\circ}$.

Example

Choose the point on the terminal side of $60^ {\circ}$.

– (1, $\sqrt{3}$)

– ($\sqrt {3}$, 1)

– (1, 2)

Calculating the value of the tangent of $60^ {\circ}$, which is given as:

\[ \tan (60^ {\circ} = \dfrac{ y }{ x } \]

\[ \dfrac{ y }{ x } = \sqrt {3} \]

a) (1, $\sqrt{3}$)

\[ \dfrac{ y }{ x } = \dfrac{ 1 }{ \sqrt {3} } \]

This point is not on the terminal side of $60^ {\circ}$.

b) ($\sqrt {3}$, 1)

\[ \dfrac{ y }{ x } = \dfrac{ \sqrt {3} }{ 1 } \]

\[ \dfrac{ y }{ x } = \sqrt {3} \]

This point lies on the terminal side of $60^ {\circ}$.

c) (1, 2)

\[ \dfrac{ y }{ x } = \dfrac{ 1 }{ 2 } \]

This point is not on the terminal side of $60^ {\circ}$.

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