The Mysterious Cyclic Quadrilateral...

Since $\angle YXW$ is an inscribed angle it is equal to $\dfrac{1}{2}(100)=50^\circ$ . In a cyclic quadrilateral, opposite angles sum up to $180^\circ$ , so $\angle YZW=180-50=130^\circ$ .

Consider $\triangle YWZ:$ The interior angles of a triangle sum up to $180^\circ$ , so

$\angle YWZ=180-10-130=$ $\boxed{40^\circ}$

There are two ways to solve this problem. They both uses the most important properties of cyclic quadrilaterals. But first, we need to know the measure of YXW to apply make it a cyclic quadrilateral. Since both YOW and YXW opens the name arc, YXW = 1/2 YOW so YXW=50 Then, we can solve this in two ways: The first way is to find the measure of YZW using the property of the opposite angles in a cyclic quadrilateral add up to 180. We then get 130 for angle YZW. Then, using the sum of all angles in a triangle = 180, we get that YWZ=40. The second way is using the property of ZYW=WXZ and YXZ=ZWY in which we get 40 degrees. In both ways, we used the property of cyclic quadrilaterals and applied it to finding the angles inside.

YOW is 100 which means YXW is 50 (circle property) Since YXW is 50, YZW is 180 - 50 = 130 (opposite angles of cyclic quadrilateral equals 180). Draw a dotted line connecting Y and W and see the triangle YWZ. In this triangle, one angle is 10, another is 130. Using angle sum property of triangles, the third angle of triangle YWZ names angle YWZ will be 40 degrees

مجموع زوايا المثلث من الداخل 180 درجه .. وبما ان المثلث متساوي الساقين يصبح زاويتي القاعده متساويتين.. (180-زاويه رأس المثلث 100)=80درجه نقسمها علي 2=40