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This is one of our many quizzes on Reference angles and angles in the unit circle. Click on this to access more such competitive quizzes on Geometry.

Reference angles and angles in the unit circle Quiz 2

mathematics

 Find the angle in second quadrant that has a reference angle $\dfrac{\pi }{5}$

 Use the formula ${t}'=t-\pi $ and then substitute the value of the reference angle to find the angle in the third quadrant.

  $\text{In second quadrant},\\ \text{The reference angle is }{t}'=\pi -t \\ \text{So, the angle }t=\pi -{t}' \\ \Rightarrow t=\pi -\dfrac{\pi }{5} \\ \therefore t=\dfrac{4\pi }{5}$

 If the coordinates of the point on the unit circle at an angle of $\dfrac{5\pi }{6}$ are $\left( \alpha ,\beta  \right)$. Then find $\dfrac{\alpha }{\beta }$

 Find the value of $\cos \left( {{150}^{\circ }} \right)$ and $\sin \left( {{150}^{\circ }} \right)$ using reference angle concept and then compare it with $\left( \alpha ,\beta  \right)$ to find the value of $\dfrac{\alpha }{\beta }$

  $\text{We know that, } \\ {{150}^{\circ }}\text{ lies in second quadrant} \\ \text{So, the reference angle is }{{180}^{\circ }}-{{150}^{\circ }}={{30}^{\circ }} \\ \cos \left( {{30}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}\,\text{and }\sin \left( {{30}^{\circ }} \right)=\dfrac{1}{2} \\ \text{As the given angle is in second quadrant}, \\ \text{ cosine is negative and sine is positive } \\ \text{Hence,}\cos \left( {{150}^{\circ }} \right)=-\dfrac{\sqrt{3}}{2}\,\text{and }\sin \left( {{150}^{\circ }} \right)=\dfrac{1}{2} \\ \therefore \left( \alpha ,\beta  \right)=\left( -\dfrac{\sqrt{3}}{2},\dfrac{1}{2} \right) \\ \Rightarrow \dfrac{\alpha }{\beta }=\dfrac{-\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} \\ \Rightarrow \dfrac{\alpha }{\beta }=-\sqrt{3}$

 A certain angle $\varphi $ corresponds to a point on the unit circle at$\left( \dfrac{\sqrt{3}}{2},-\dfrac{1}{2} \right)$. Find the value of $\cos e{{c}^{2}}\varphi -{{\sec }^{2}}\varphi $.

 Take $x=\cos \varphi$ and $y=\sin \varphi $ and then obtain the values of $\cos ec\varphi $ and $\sec \varphi $ using the concept of trigonometry to find the value of $\cos e{{c}^{2}}\varphi -{{\sec }^{2}}\varphi $.

$\text{In a unit circle, } \\ x=\cos \varphi \text{ and }y=\sin \varphi  \\ \cos \varphi =\dfrac{\sqrt{3}}{2}\text{ and sin}\varphi \text{=}\dfrac{-1}{2} \\ \text{We know that, }\cos ec\varphi =\dfrac{1}{\text{sin}\varphi }\text{ and sec}\varphi =\dfrac{1}{\cos \varphi } \\ \Rightarrow \cos ec\varphi =-2;\,\sec \varphi =\dfrac{2}{\sqrt{3}} \\ \therefore \cos e{{c}^{2}}\varphi -{{\sec }^{2}}\varphi ={{\left( -2 \right)}^{2}}-{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}} \\ =4-\dfrac{4}{3} \\ =\dfrac{8}{3}$

 If the coordinates of the point on the unit circle at an angle of$\dfrac{7\pi }{6}$ are $\left( \alpha ,\beta  \right)$. Then find the sum of the roots of ${{\alpha }^{2}}{{x}^{2}}-{{\beta }^{2}}x-\dfrac{1}{2}=0$.

 Using the concept of reference angle , find the value of $\alpha $ and $\beta $ then substitute the values in the given equation to find the sum of the roots of the equation.

 $\dfrac{7\pi }{6}\text{ is in the third quadrant } \\ \text{So , the reference angle is, }\dfrac{7\pi }{6}-\pi =\dfrac{\pi }{6} \\ \text{We know that,}\,\cos \left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}\text{ and }\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2} \\ \text{As the given angle is in the third quadrant}, \\ \text{ both the cosine and sine are negative} \\ \therefore \text{The coordinates }\left( x,y \right)\text{ is }\left( \cos \theta ,\sin \theta  \right)\Rightarrow \left( -\dfrac{\sqrt{3}}{2},-\dfrac{1}{2} \right) \\ \therefore \left( \alpha ,\beta  \right)=\left( -\dfrac{\sqrt{3}}{2},-\dfrac{1}{2} \right) \\ {{\alpha }^{2}}{{x}^{2}}-{{\beta }^{2}}x-\dfrac{1}{2}=0 \\ \Rightarrow 3{{x}^{2}}-x-2=0 \\ \Rightarrow \left( 3x+2 \right)\left( x-1 \right)=0 \\ \text{Hence, the roots are }\dfrac{-2}{3},1 \\ \therefore \text{sum = }\dfrac{1}{3}$

 A certain angle $\theta $ corresponds to a point on the unit circle at $\left( -\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right)$.  If 3 is one of the roots of the equation ${{x}^{2}}{{\tan }^{2}}\theta +px\tan \theta -3=0$, find the value of p.

 Take $x=\cos \theta \text{ and }y=\sin \theta $ to find the value of $\tan \theta $  then substitute it in the given equation to find the value of $p$.

 $\text{In a unit circle, } \\ x=\cos \theta \text{ and }y=\sin \theta \\ \Rightarrow \tan \theta =\dfrac{y}{x} \\ \text{We have, }\left( x,y \right)=\left( -\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2} \right) \\ \therefore \tan \theta =\dfrac{\dfrac{\sqrt{2}}{2}}{-\dfrac{\sqrt{2}}{2}}=-1 \\ {{x}^{2}}{{\tan }^{2}}\theta +px\tan \theta -3=0 \\ \Rightarrow {{x}^{2}}\left( 1 \right)+px\left( -1 \right)-3=0 \\ \Rightarrow {{x}^{2}}-px-3=0 \\ \text{As 3 is root of the given equation,} \\ \Rightarrow {{\left( 3 \right)}^{2}}-p\left( 3 \right)-3=0 \\ \Rightarrow 6-3p=0 \\ \Rightarrow p=2$