Mathematics

In the figure, $$AOB$$ is a straight line. Find the value of $$x$$. Here, find $$\angle AOC,\angle COD$$ and $$\angle BOD$$


SOLUTION
From the figure, we know that $$\angle BOD$$ and $$\angle AOD$$ are linear pair of angles

So we get

$$\angle BOD+\angle AOD={180}^{o}$$

It can also be written as

$$\angle BOD+\angle COD+\angle COA={180}^{o}$$

By substituting the values 

$${ x }^{ o }+{ \left( 2x-19 \right)  }^{ o }+{ \left( 3x+7 \right)  }^{ o }={ 180 }^{ o }$$

$$\Rightarrow x+2x-{ 19 }^{ o } +3x+{ 7 }^{ o }={ 180 }^{ o }$$

$$\Rightarrow 6x-{ 12 }^{ o }={ 180 }^{ o }$$

$$\Rightarrow 6x={ 192 }^{ o }\Rightarrow x={ 32 }^{ o }$$

By substituting the value of $$x$$ we get

$$\angle AOC={ \left( 3x+7 \right)  }^{ o }=3({32}^{o})+{7}^{o}={103}^{o}$$

$$\angle COD={ \left( 2x-19 \right)  }^{ o }=2({32}^{o})-{19}^{o}={45}^{o}$$

$$\angle BOD={x}^{o}={32}^{o}$$

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