# 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 $\small \angle BAC$        .

$\angle$DEC+ $\angle$BEC = $180 \degree$          (linear pairs)

$\Rightarrow$ $\angle$DEC+ $130 \degree$ = $180 \degree$           ($\angle$ BEC =$130 \degree$)

$\Rightarrow$ $\angle$DEC = $180 \degree$ - $130 \degree$

$\Rightarrow$ $\angle$DEC = $50 \degree$

In $\triangle$ DEC,

$\angle$D+ $\angle$DEC+ $\angle$DCE = $180 \degree$

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

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

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

$\angle$D = $\angle$BAC     (angles in same segment are equal  )

$\angle$BAC  = $110 \degree$

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