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This is one of our many quizzes on Unit circle. Click on this to access more such competitive quizzes on Geometry.

Unit circle Quiz 2

mathematics

 If there are two circles with radius 4cm and radius 3cm and a distance of 5cm between their centres, then the maximum number of possible common points between them are:

 Draw circle as per given data by using proper scale.

As per shown in figure: $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/718f992e-8e87-446d-bb31-1e5f8370b4d04032794695806770411.png"/> $ \\ $  There are two common points.

 What part of the circle chart does the point $\left( {\dfrac{{ - \sqrt 3 }}{2},\dfrac{{ - 1}}{2}} \right)$ represent? 

 Use the graph of the unit circle chart and measure according to their coordinate. 

$\left( { - \dfrac{{\sqrt 3 }}{2}, - \dfrac{1}{2}} \right){\text{ lie on the third quadrant of circle chart}}{\text{.}}$

 $\left( {\dfrac{1}{{\sqrt 2 }}, - \dfrac{1}{{\sqrt 2 }}} \right)$ is a terminal point of a unit circle then find angle the angle made by terminal point on x axis.

 Apply terminal point to find out the value of $\theta $.

$ {\text{Given terminal point}}\left( {\dfrac{1}{{\sqrt 2 }}, - \dfrac{1}{{\sqrt 2 }}} \right) \\  {\text{Let angle be }}\theta  \\   \Rightarrow \sin \theta  = \dfrac{P}{H} \\   \Rightarrow \sin \theta  = \dfrac{1}{{\sqrt 2 }} \\   \Rightarrow \theta  = {\sin ^{ - 1}}\dfrac{1}{{\sqrt 2 }} \\   \Rightarrow \theta  = \dfrac{\pi }{4} \\   \Rightarrow \theta  = 45^\circ  \\  $

 If the equation of the unit circle is ${x^2} + {y^2} = 1$. What is the value of $cot\theta $.

 First apply $sin\theta  = \dfrac{P}{H}$ and $cos\theta  = \dfrac{P}{H}$ then divide value of $\cos \theta \,{\text{by}}\,\sin \theta $ to get  the value of $cot\theta $.

 $ {\text{Given equation }}{{\text{x}}^2} + {y^2} = 1 \\  cos\theta {\text{  =  }}\dfrac{B}{H} \\  cos\theta {\text{  =  }}\dfrac{x}{1} \\  \therefore sin\theta {\text{  = }}\dfrac{y}{1} \\  cot\theta {\text{  =  }}\dfrac{x}{y} \\  $

 Find the terminal point of the unit circle at $\theta  = 157^\circ $.

 Apply $\sin 23^\circ  = 0.3907$ and $\cos 23^\circ  = 0.9205$ then further solve for coordinates. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f3084400-511d-454a-add7-c220554dd03a1909249746486755390.png"/> $  \\ {\text{Given angle }-157^\circ \text{ lie in second quadrant}} \\  \therefore {\text{Angle made on the x axis is given by}} \\  \pi -157 = {180^ \circ } - {157^ \circ } \\   \Rightarrow {180^ \circ } - {157^ \circ } = {23^ \circ } \\  \therefore \sin 23^\circ  = \dfrac{P}{H}\left[ {H{\text{ is radius}}} \right] \\   \Rightarrow \dfrac{P}{1} \\   \Rightarrow 0.3907 = P \\  {\text{Similarly }} {{\cos 23^\circ  = }}\dfrac{B}{H} $ $ \\ \Rightarrow 0.9205 = \dfrac{B}{{1^\circ }} \\   \Rightarrow B = 0.9205 \\  \therefore {\text{Terminal point}}\left( {0.9205,0.3907} \right) \\  $