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In fig .1  PQ is a tangent at a point C to a circle with center O . If AB is diameter and 

\angle CAB = 30 , find \: \: \angle PCA


 

 

 

 

 
 
 
 
 

Answers (1)

Given \angle CAB \\\\ draw \rightarrow join \: \: OC \\\\ \angle CAB = \angle OCA \\\\ (OA = OC = radisu and equall side have equal opposite angle )

\angle OCA = 30 \degree ---- (1) \\\\ we \: \: know \: \: PQ \perp OC (PQ is tangent and point C is the point of contact ) 

\angle PCO = 90 \\\\ \angle PCO = \angle OCA + \angle PCA = 90 \\\\ 30 + \angle PCA = 90 \\\\ \angle PCA = 60

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Ravindra Pindel

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