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=½qos-pos
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Answer
Answer
Answer It is given that OR is perpendicular to PQ
Answer It is given that OR is perpendicular to PQSo that ∠POR = 90°
Answer It is given that OR is perpendicular to PQSo that ∠POR = 90°sum of angle in linear pair always equal to 180°
Answer It is given that OR is perpendicular to PQSo that ∠POR = 90°sum of angle in linear pair always equal to 180°∠POS + ∠SOR + ∠POR = 180°
Answer It is given that OR is perpendicular to PQSo that ∠POR = 90°sum of angle in linear pair always equal to 180°∠POS + ∠SOR + ∠POR = 180°Plug ∠POR = 90°
Answer It is given that OR is perpendicular to PQSo that ∠POR = 90°sum of angle in linear pair always equal to 180°∠POS + ∠SOR + ∠POR = 180°Plug ∠POR = 90°90°+∠SOR + ∠POR = 180°
Answer It is given that OR is perpendicular to PQSo 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°
Answer It is given that OR is perpendicular to PQSo 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)
Answer It is given that OR is perpendicular to PQSo 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°
Answer It is given that OR is perpendicular to PQSo 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
Answer It is given that OR is perpendicular to PQSo 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 raysOP and OR so that
Answer It is given that OR is perpendicular to PQSo 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 raysOP and OR so that ∠QOS − ∠ROS = 90°
Answer It is given that OR is perpendicular to PQSo 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 raysOP and OR so that ∠QOS − ∠ROS = 90°∠ROS = ∠QOS − 90° … (2)
Answer It is given that OR is perpendicular to PQSo 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 raysOP and OR so that ∠QOS − ∠ROS = 90°∠ROS = ∠QOS − 90° … (2)On adding equations (1) and (2), we obtain
Answer It is given that OR is perpendicular to PQSo 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 raysOP and OR so that ∠QOS − ∠ROS = 90°∠ROS = ∠QOS − 90° … (2)On adding equations (1) and (2), we obtain2 ∠ROS = ∠QOS − ∠POS
Answer It is given that OR is perpendicular to PQSo 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 raysOP and OR so that ∠QOS − ∠ROS = 90°∠ROS = ∠QOS − 90° … (2)On adding equations (1) and (2), we obtain2 ∠ROS = ∠QOS − ∠POS∠ROS = 1/2(∠QOS − ∠POS)