# 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.

$\theta = -210^ {\circ}$

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 $60^ {\circ}$.