Answer

Verified

457.5k+ views

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.

Recently Updated Pages

What number is 20 of 400 class 8 maths CBSE

Which one of the following numbers is completely divisible class 8 maths CBSE

What number is 78 of 50 A 32 B 35 C 36 D 39 E 41 class 8 maths CBSE

How many integers are there between 10 and 2 and how class 8 maths CBSE

The 3 is what percent of 12 class 8 maths CBSE

Find the circumference of the circle having radius class 8 maths CBSE

Trending doubts

One cusec is equal to how many liters class 8 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE