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In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = ½ (QOS – POS). Q.5

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Yesterday i was doing the question from class 9th ncert book of math of Lines and Angles chapter of exercise 6.1  What is the easiest way for solving it because i was not able to do this question please help me for solving this question of no.5  In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS = ½ (QOS – POS)

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  1. the question, it is given that (OR ⊥ PQ) and POQ = 180°

    So, POS+ROS+ROQ = 180°

    Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°)

    ∴ POS + ROS = 90°

    Now, QOS = ROQ+ROS

    It is given that ROQ = 90°,

    ∴ QOS = 90° +ROS

    Or, QOS – ROS = 90°

    As POS + ROS = 90° and QOS – ROS = 90°, we get

    POS + ROS = QOS – ROS

    2 ROS + POS = QOS

    Or, ROS = ½ (QOS – POS) (Hence proved).

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