CBSE BOARD X, asked by mehar1310, 1 month ago

Tangents PA and PB are drawn to a circle with centre O from a point P outside the circle. Prove that
<OPA=<OAM.

please show the working
I'll mark the right answer the brainliest

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Answers

Answered by munmunsgr

Solutions:

L PAN and L PBM
L PAN and L PBMPA =PB
L PAN and L PBMPA =PBPM =PM
L PAN and L PBMPA =PBPM =PML APM = L BPM
L PAN and L PBMPA =PBPM =PML APM = L BPML PAM =~ L PBM
L PAN and L PBMPA =PBPM =PML APM = L BPML PAM =~ L PBM LOAM = x
L PAN and L PBMPA =PBPM =PML APM = L BPML PAM =~ L PBM LOAM = xL POM=90
L PAN and L PBMPA =PBPM =PML APM = L BPML PAM =~ L PBM LOAM = xL POM=90L PAM=90-LOAM
L PAN and L PBMPA =PBPM =PML APM = L BPML PAM =~ L PBM LOAM = xL POM=90L PAM=90-LOAML APM=LOAM
L PAN and L PBMPA =PBPM =PML APM = L BPML PAM =~ L PBM LOAM = xL POM=90L PAM=90-LOAML APM=LOAML APM=L APO=L OPA
L PAN and L PBMPA =PBPM =PML APM = L BPML PAM =~ L PBM LOAM = xL POM=90L PAM=90-LOAML APM=LOAML APM=L APO=L OPAL OPA=L OAM

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