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https://vle.mathswatch.co.uk/images/questions/question2908.png

really need helpsad

prize for who answers this: i will like all your posts.surprise

ask CPhill i liked his posts.smiley

lynx7  Nov 25, 2017

Best Answer 

 #1
avatar+93866 
+3

 

 

\(\angle ODC=\angle OBC=90^\circ\qquad \) A tangent is at right angles to the radius that it intersects with

\(\angle BOD=2*\angle BFD \)               

             When subtended from the same arc (BC), the angle at the  centre = 2* the angle at the circumference

\(\therefore \angle BOD=2*76=152^\circ\)

\(152+90+90+\angle BCD=360 \qquad \text{Angle sum of quadrilateral}\\ 332+\angle BCD=360 \\ \angle BCD=28^\circ\)

Melody  Nov 25, 2017
 #1
avatar+93866 
+3
Best Answer

 

 

\(\angle ODC=\angle OBC=90^\circ\qquad \) A tangent is at right angles to the radius that it intersects with

\(\angle BOD=2*\angle BFD \)               

             When subtended from the same arc (BC), the angle at the  centre = 2* the angle at the circumference

\(\therefore \angle BOD=2*76=152^\circ\)

\(152+90+90+\angle BCD=360 \qquad \text{Angle sum of quadrilateral}\\ 332+\angle BCD=360 \\ \angle BCD=28^\circ\)

Melody  Nov 25, 2017

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