# Find the values of $\theta $ and $p$, if the equation \[xcos\theta + ysin\theta = p\] is the normal form of the line \[\;\sqrt {3x} + y + 2 = 0\].

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Hint: In the above given question, it is important to find out the quadrant in which$\theta $lies. This can be known with the help of the values of the trigonometric functions $\cos \theta $ and $\sin \theta $ so obtained. Once the quadrants are known, further values can be easily calculated.

We have the given equation of the normal form as \[xcos\theta + ysin\theta = p\]

The equation of the given line is \[\sqrt 3 \;x + y + 2 = 0\]

This equation can be reduced as \[ - \sqrt 3 x - y = 2\]

Now, on dividing both sides by we obtain,

\[ - \dfrac{{\sqrt 3 }}{2}x - \dfrac{1}{2}y = \dfrac{2}{2}\]

$ \Rightarrow \left\{ { - \dfrac{{\sqrt 3 }}{2}} \right\}x + \left\{ { - \dfrac{1}{2}} \right\}y = 1$ … (1)

On comparing equation (1) to \[xcos\theta + ysin\theta = p\],

We obtain \[cos\theta = - \dfrac{{\sqrt 3 }}{2}\],\[sin\theta = - \dfrac{1}{2}\]and\[\;p = 1\]

Since the value of \[sin\theta \;\] and \[cos\theta \;\] are both negative,

So, $\theta $ is in the third quadrant.

\[\;\therefore \theta = \pi + \dfrac{\pi }{6}{\text{ }} = \dfrac{{7\pi }}{6}\]

Thus, the respective values of $\theta $ and $p$ are $\dfrac{{7\pi }}{6}$ and $1$.

Note: Whenever we face such types of problems the key point is to have a good grasp of the equations of lines. We need to equate the equation obtained after appropriate manipulations to the given equation of the normal form of the line to obtain the required solution. Also, the knowledge of trigonometric functions is needed, in which quadrants the sign of trigonometric functions is positive or negative.

We have the given equation of the normal form as \[xcos\theta + ysin\theta = p\]

The equation of the given line is \[\sqrt 3 \;x + y + 2 = 0\]

This equation can be reduced as \[ - \sqrt 3 x - y = 2\]

Now, on dividing both sides by we obtain,

\[ - \dfrac{{\sqrt 3 }}{2}x - \dfrac{1}{2}y = \dfrac{2}{2}\]

$ \Rightarrow \left\{ { - \dfrac{{\sqrt 3 }}{2}} \right\}x + \left\{ { - \dfrac{1}{2}} \right\}y = 1$ … (1)

On comparing equation (1) to \[xcos\theta + ysin\theta = p\],

We obtain \[cos\theta = - \dfrac{{\sqrt 3 }}{2}\],\[sin\theta = - \dfrac{1}{2}\]and\[\;p = 1\]

Since the value of \[sin\theta \;\] and \[cos\theta \;\] are both negative,

So, $\theta $ is in the third quadrant.

\[\;\therefore \theta = \pi + \dfrac{\pi }{6}{\text{ }} = \dfrac{{7\pi }}{6}\]

Thus, the respective values of $\theta $ and $p$ are $\dfrac{{7\pi }}{6}$ and $1$.

Note: Whenever we face such types of problems the key point is to have a good grasp of the equations of lines. We need to equate the equation obtained after appropriate manipulations to the given equation of the normal form of the line to obtain the required solution. Also, the knowledge of trigonometric functions is needed, in which quadrants the sign of trigonometric functions is positive or negative.

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