Q: 5    In Fig. \small 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that  \small \angle BEC=130^{\circ}  and \small \angle ECD=20^{\circ}. Find \small \angle BAC

        .

Answers (1)

\angleDEC+ \angleBEC = 180 \degree          (linear pairs)

\Rightarrow \angleDEC+ 130 \degree = 180 \degree           (\angle BEC =130 \degree)

\Rightarrow \angleDEC = 180 \degree - 130 \degree

\Rightarrow \angleDEC = 50 \degree

In \triangle DEC,

\angleD+ \angleDEC+ \angleDCE = 180 \degree

\Rightarrow \angle D+50 \degree+20 \degree= 180 \degree

\Rightarrow \angle D+70 \degree= 180 \degree

\Rightarrow \angle D= 180 \degree-70 \degree=110 \degree

  

\angleD = \angleBAC     (angles in same segment are equal  )

 \angleBAC  = 110 \degree

 

 

 

 

 

 

 

 

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