A. B, C and D are points on the circumference of a circle, centre O. FDE is a tangent to the circle. (a) Show that $y - x = 90$. You must give a reason for each stage of your working. Dylan was asked to give some possible values for x and y. He said, "y could be 200 and x could be 110, because 200 - 110 = 90". (b) Is Dylan correct? You must give a reason for your answer.

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A. B, C and D are points on the circumference of a circle, centre O - Edexcel - GCSE Maths - Question 14 - 2018 - Paper 2

Question 14

A. B, C and D are points on the circumference of a circle, centre O. FDE is a tangent to the circle. (a) Show that $y - x = 90$. You must give a reason for each s... show full transcript

Worked Solution & Example Answer:A. B, C and D are points on the circumference of a circle, centre O - Edexcel - GCSE Maths - Question 14 - 2018 - Paper 2

Step 1

Show that $y - x = 90$

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Answer

To show that $y - x = 90$ , we start by noting that the angle $y$ at point E is an angle subtended by the tangent FDE and the radius OE. Since the angle between a radius and a tangent at the point of contact is always $90$ degrees, we have:

The angle EFD is $90^{ ext{o}}$ because FDE is tangent to the circle.
Therefore, we can express this relationship as:

$y = x + 90$
Rearranging gives:

$y - x = 90$ .

Step 2

Is Dylan correct?

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Answer

No, Dylan is not correct. For the angles x and y, the sum of angles in a triangle must be less than $180^{ ext{o}}$ . Given

$y = 200^{ ext{o}}$ and $x = 110^{ ext{o}}$ , this gives:

$y - x = 200 - 110 = 90$

However, if we consider the triangle formed by these angles, the total must be less than $180^{ ext{o}}$ , which is not the case here. Hence, Dylan's values for x and y are not possible.

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