Question #e2f89

Here's how the image looks like

The first important thing you need to notice is that you're dealing with a regular hexagon, which means that all the interior angles are equal to `120""^@`.

So, if `angle(TUP)` is equal to `120""^@` and `TUV` is a straight line, then angle `angle(PUV)` is its complementary angle.

This means that

`angle(TUP) + angle(PUV) = 180""^@`

As a result,

`angle(PUV) = 180 - 120 = 60""^@`

Now, `triangleVUW` is a right triangle, as you can deduce from the fact that `angle(VUW)` is equal to `90^""@`.

Another important thing to notrice here is that `PUWV` is a parallelogram, which means that `bar(UN)` is parralel to `bar(WV)`

If this is the case, then `angle(PVU)` is equal to `90""^@` as well, since you're dealing with two parallel lines intersected by a straight line.

Moreover, `angle(PUV)` is equal to `angle(UVW)` for the same reason.

This means that

`angle(UVW) = angle(PUV) = 60""^@`

Therefore, `angle(VWU)` (angle `x`) will be

`angle(VWU) + angle(VUW) + angle(UVW) = 180""^@`

`angle(VWU) = 180 - 90 - 60 = color(green)(30""^@)`

The interior angles of a regular hexagon are `120^o`
`rArr``color(white)("XXXX")``/_TUP = 120^o`

`/_TUP + /_PUV = 180^o`
`rArr``color(white)("XXXX")``/_PUV = 60^o`

Since `PU || VW`
`rArr``color(white)("XXXX")``/_PUV = /_UVW = 60^0`
`color(white)("XXXX")``color(white)("XXXX")`(opposite angles of a traversal crossing parallel lines [Euclid]

Since Sum of Interior Angles of a Triangle `= 180^o`
`rArr``color(white)("XXXX")``/_WUL + /_UVW + /_VWU = 180^o`

`rArr``color(white)("XXXX")``90^o + 60^o + /_VWU = 180^o`

`rArr``color(white)("XXXX")``/_VWU (= x) = 30^o`