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Supplementary and complementary angles Quiz 2

mathematics

 If $\angle QOR={{20}^{\circ }}$. Find the complement of $\angle QPR$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e6423898-0806-4588-8afb-0879393f41de310373531897323404.png"/>

 Using properties of the circle find $\angle PQR$ and then find its complement by taking the sum of complementary angles equal to ${{90}^{\circ }}$.

  $  \angle QOR={{20}^{\circ }} \\ $ Since the angle subtended at the circumference is twice the angle subtended at the center by an arc. $ \\ \Rightarrow 2\angle QPR=\angle QOR \\  \Rightarrow \angle QPR=\dfrac{{20}^{\circ }}{2}={{10}^{\circ }} \\ $ Complement of $\angle QPR={{90}^{\circ }}-{{10}^{\circ }}={{80}^{\circ }}$

 If the angles $x\text{ and }y$ are complementary to each other and the difference between them is  ${{30}^{\circ }}$, then find the value of $\sin x$ and $\cos y$.

 Form linear equations using the given data and solve them simultaneously to get the values of $x$ and $y$. Then make use of the trigonometric values $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$.

  $  x-y={{30}^{\circ }} \\  x+y={{90}^{\circ }} \\  \Rightarrow x={{60}^{\circ }},y={{30}^{\circ }} \\  \sin x=\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} \\  \cos y=\cos 30=\dfrac{\sqrt{3}}{2} \\ $

 Find complement of $\gamma $,  if $\gamma -2$ and $2\gamma +5$ are supplementary to each other.

 As the angles are supplementary so their sum must be ${{180}^{\circ }}$. Solve the linear equation formed. Then take the sum of complementary angles as ${{90}^{\circ }}$ to find the required answer.

$  \left( \gamma -2 \right)+\left( 2\gamma +5 \right)={{180}^{\circ }} \\  \Rightarrow 3\gamma +3={{180}^{\circ }} \\  \Rightarrow 3\gamma ={{177}^{\circ }} \\  \Rightarrow \gamma ={{59}^{\circ }} \\  \therefore \text{Complement of }\gamma =\text{9}{{\text{0}}^{\circ }}-{{59}^{\circ }}={{31}^{\circ }} \\ $

 Find the relation between $w$ and $v$ from the given figure in which $p\parallel q\parallel r$. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c7e881c0-8082-49f8-b23b-e46a7d2b4aa55549125435772109358.png"/>

 Find $\angle u$ using $\angle u+{{45}^{\circ }}={{180}^{\circ }}$ as they are co-interior angles and then find $\angle w$ using $\angle u=\angle w$ as they are corresponding angles. Then find the relation after finding the value of $\angle v$.

 $  \angle u+{{45}^{\circ }}={{180}^{\circ }}\left( \text{Co-interior angles} \right) \\  \Rightarrow \angle u={{135}^{\circ }} \\  \Rightarrow \angle u=\angle w={{135}^{\circ }}\left( \text{Corresponding angles} \right) \\  \text{Again, }\angle v+{{45}^{\circ }}={{90}^{\circ }} \\  \Rightarrow \angle v={{45}^{\circ }} \\  \therefore \angle w+\angle v={{180}^{\circ }} \\ $

 In the given figure $MN\parallel PQ$ and $\angle a={{40}^{\circ }}$. If $c$ is the unequal angle of an isosceles triangle then find the equal angles. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/baaf81d3-7f99-4419-8c78-9fbbc8557aa85175999664341606018.png"/>

 Take $\angle a=\angle b$ as they are corresponding angles and find the value of $\angle b$. Then, find $\angle c$ using the formula of supplementary angles. Next assume the equal angles as $x$ and then apply angle sum property. 

 $  \angle a={{40}^{\circ }} \\ $ $\angle a = \angle b($Corresponding angles$)$ $ \\ \Rightarrow \angle b={{40}^{\circ }} \\ $ $ \angle b+\angle c={{180}^{\circ }}($As seen from figure they are supplementary angles$)$ $ \\ \Rightarrow \angle c={{180}^{\circ }}-{{40}^{\circ }} \\  \Rightarrow \angle c={{140}^{\circ }} \\ $ Let the equal angles of the isosceles triangle be $x$. $ \\ \Rightarrow x+x+{{140}^{\circ }}={{180}^{\circ }}($ Angle Sum Property $)$ $ \\  \Rightarrow 2x={{40}^{\circ }} \\  \Rightarrow x={{20}^{\circ }}$