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Verified answer
Given :
To Prove :
Solution :
As POQ is a line . So ,
[tex]\longmapsto\tt{\angle{POS}+\angle{ROS}+\angle{ROQ}=180\degree}[/tex]
[tex]\longmapsto\tt{\angle{POS}+\angle{ROS}+90\degree=180\degree}[/tex]
[tex]\longmapsto\tt{\angle{POS}+\angle{ROS}=180\degree-90\degree}[/tex]
[tex]\longmapsto\tt\bf{\angle{POS}+\angle{ROS}=90\degree\:---(1)}[/tex]
Also ,
[tex]\longmapsto\tt{\angle{QOS}=\angle{ROS}+\angle{ROQ}}[/tex]
[tex]\longmapsto\tt{\angle{QOS}=\angle{ROS}+90\degree}[/tex]
[tex]\longmapsto\tt\bf{\angle{QOS}-\angle{ROS}=90\degree\:---(2)}[/tex]
By Equation 1 nd 2 :
[tex]\longmapsto\tt{\angle{POS}+\angle{ROS}=\angle{QOS}-\angle{ROS}}[/tex]
[tex]\longmapsto\tt{\angle{ROS}+\angle{ROS}=\angle{QOS}-\angle{POS}}[/tex]
[tex]\longmapsto\tt\bf{\angle{ROS}=\dfrac{1}{2}(\angle{QOS}-\angle{POS})}[/tex]
HENCE PROVED
Given:-
•OR is perpendicular to PQ
•So that ∠POR = 90°
★Sum of angle in linear pair always equal to 180°.
→∠POS + ∠SOR + ∠POR = 180°
Plug ∠POR = 90°
→90°+∠SOR + ∠POR = 180°
→∠SOR + ∠POR = 90°
→∠ROS = 90° − ∠POS … (1)
→∠QOR = 90°
★Given that OS is another ray lying between rays.
OP and OR so that
→∠QOS − ∠ROS = 90°
→∠ROS = ∠QOS − 90° ...(2)
★On adding equations (1) and (2), we obtain.
→2 ∠ROS = ∠QOS − ∠POS
→∠ROS = 1/2(∠QOS − ∠POS)
Hence verified!