Plz solve this question

Given: AB=AC
$\angle E=100°$
To find: x and y.

In the given figure, BOCE is cyclic quadrilateral.

Therefore

angle BEC + angle BOC = 180^o (because The sum of opposite angles of a cyclic quadrilateral is 180^o )


=>      100^o​ + y = 180^o

​=>      y = 180^o​ - 100^o​ = 80^o

=>      angle BAC = angle BOC (angle y) (Because angles in the same segment are equal)

=>      Therefore angle BAC = 80^o

Now, in triangle ABC, angle BAC = 80^o

​It is given that AB = AC i.e. triangle ABC is an isosceles triangle.

Therefore angle ABC = Angle x (because in a triangle, angles opposite to the two equal sides are equal)          (1)

Therefore using the angle sum property of triangles, we know that the sum of all the angles of a triangle is 180^o.

In triangle ABC,

​=>      angle BAC + angle ABC + angle x =180^o

=>      80^o​ + x + x = 180^o​ (from (1) angle ABC = x )

=>    80^o​ + 2x = 180^o

​=>    2x = ​180^o - 80^o​​ = 100^o

=>    x = 100^o / 2 = 50^o

Hence, x = 50^o and y = 80^o

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