#
Plz solve this question

Given: AB=AC

$\angle E=100\xb0$

To find: x and y.

$\angle E=100\xb0$

To find: x and y.

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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