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

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

really need help

ask CPhill i liked his posts.

Nov 25, 2017

#1
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$$\angle ODC=\angle OBC=90^\circ\qquad$$ 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

$$\therefore \angle BOD=2*76=152^\circ$$

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

.
Nov 25, 2017

#1
+97498
+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