Question 5
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC=130∘ and ∠ECD=20∘. Find ∠BAC.
Given,
∠BEC=130∘
∠ECD=20∘
∠BEC+∠DEC=180∘ [Linear pair]
130∘+∠DEC=180∘
∠DEC=180∘−130∘=50∘
Now, in ΔDEC
∠DEC+∠ECD+∠CDE=180∘ [Angle sum property of a triangle]
50∘+20∘+∠CDE=180∘
∠CDE=180∘−70∘=110∘
∠CDE=∠BAC [Angles in the same segment]
∴∠BAC=110∘