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I know this might seem like a really simple question, but I really don't understand where I am going wrong. I am familiar with cyclic quadrilaterals as well as their properties, but this question really isn't making much sense to me - I keep coming out with far fetched answers.

How do I find the size of the angle W? E being the centre of the circle ABCD.

https://i.stack.imgur.com/RK0gd.png

Any help is appreciated!

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3 Answers 3

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First, the quadrilateral ABCD is inscribed in the circle, so the opposite angles $\angle B+\angle D=180^\circ$, therefore $\angle B=20^\circ$.

Then there is a theorem (inscribed angle and central angle) which says that $\angle E=2\angle W$, therefore $\boxed{\angle W=40^\circ}$.

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The angle subtended at the centre of the circle is twice the angle subtended at the circumference.

Thus angle $360-W = 2(160) = 320 \\ W = 360-320 = 40$

Not sure if this is any help but I hope it makes a little more sense, instead of relying on the cyclic quadrilateral to solve the problem, take a look at all the other theorems, this one being one of the most common!

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After erasing some lines, we have the following figures:-

enter image description here

From the left figure, one can find $\angle B$.

From the right figure, one can find $\angle W$.

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