Question

In .fig.A,B,C and D are four points on a circle. AC and BD intersect at a point E such that angle BEC = 130 degree and angle ECD = 20 degree. Find angle BAC
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Answer 1

$\Huge\bf\red{Given:-}$

$\bf ∠BEC=130°$

$\bf ∠ECD = 20°$

$\Huge\bf\blue{To \:Find:-}$

$\bf \angle BAC$

$\Huge\bf\green{Solution:-}$

$\bf∠BEC+∠DEC=180°$

${\text {(Linear\:pair)}}$

$\bf \implies 130°+∠DEC=180°$

$\bf \implies ∠DEC=50°$

$\bf In ΔDEC,$

$\bf ∠DEC+∠ECD+∠CDE=180°$

${\text{(Angle sum property)}}$

$\bf \implies 50°+20°+∠CDE=180°$

$\bf \implies ∠CDE=110°$

$\bf ∠BAC=∠CDB (Angle\:in\:the\:same\: segment)$

$\bf \implies ∠BAC=∠CDE=110 °$

$\bf ∴∠BAC=110°$

Answer 2

HERE IS THE ANSWER TO YOUR GIVEN QUESTION