Question

Let O be the centre of a circle and AC be its diameter. BD is a chord intersecting AC at E. Point A is joined to B and D. If \angle∠BOC = 50^\circ50 ∘ and \angle∠AOD = 110^\circ110 ∘ , then \angle∠BEC = ?

Answer 1

The value of ∠BEC is 80°

Given:

∠BOC = 50°

∠AOD = 110°

To Find:

We need to find the value of ∠BEC

Solution:

In a triangle, the angle bisected at the center is twice the angle subtended at the circumference.

⇒ ∠BOC = 2∠BAC

∠BAC = 50/2

∠BAC = 25

Similarly, ∠AOD = 2∠ABD

∠ABD = 110/2

∠ABD = 55

Now, in triangle ABE, since sum of all angles of a triangle is 180, we have-

25 + 55 + ∠BEA = 180

∠BEA = 180 - 80

∠BEA = 100

Thus, ∠BEC = 180 - 100 = 80

Hence, the value of ∠BEC is 80°

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