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त्रिभुज

Question
CBSEHHIMAH10010089

आकृति में ∆ODC~∆OBA, angle BOC = 125° और angle CDO = 70° है। angle DOC comma space angle DCO space और space angle OAB ज्ञात कीजिए।

Solution

space space space space space angle DOC space plus space angle BOC space equals space 180 degree
rightwards double arrow space space angle DOC space plus space 125 degree space equals space 180 degree
rightwards double arrow space space angle DOC space equals space 180 degree space minus space 125 degree
rightwards double arrow space space angle DOC space equals space 55 degree
∆DOC में,
angle DOC space plus space angle ODC space plus space angle DCO space equals space 180 degree
[त्रिभुज के तीनों कोणों का योग]
rightwards double arrow space space space space space 55 degree space plus space 70 degree space plus space angle DCO space equals space 180 degree
rightwards double arrow space space space space 125 degree space plus space angle DCO space equals space 180 degree
rightwards double arrow space space space space angle DCO space equals 180 degree space minus space 125 degree
rightwards double arrow space space space space angle DCO space equals space 55 degree
∵ ∆ODC ~ ∆OBA        [ दिया है ]
therefore space space space space space space space angle OCD space equals space angle OAB
rightwards double arrow space space space space space space angle DCO space equals space angle OAB
rightwards double arrow space space space space space space angle OAB space equals space angle DCO
और space space space space space angle DCO space equals space 55 degree
rightwards double arrow space space space space space space angle OAB space equals space 55 degree

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