# Angle in a circle

• Sep 8th 2010, 09:22 PM
caramelcake
Angle in a circle
Quote:

http://picpanda.com/images/1b3q094l07wnmlq2xua6.jpg
In the diagram, $AB$ is a tangent to the circle with centre $O$. $CDE$ is a straight line and angle $CAB$ = $58°$

Showing all your reasoning, find angle $BOE$
I've found that angle BDE is 122°, but I don't know how to find BOE. I know that BO and OE are the radii, but is there a way to prove that DE and BD are of equal length? And even if I do find angles DBE and DEB, how do I find BOE? I am rather confused. Or is there a way to use the alternate segment theorem?

Any help would be greatly appreciated. Thank you in advance.
• Sep 8th 2010, 10:11 PM
Quote:

Originally Posted by caramelcake
I've found that angle BDE is 122°, but I don't know how to find BOE. I know that BO and OE are the radii, but is there a way to prove that DE and BD are of equal length? And even if I do find angles DBE and DEB, how do I find BOE? I am rather confused. Or is there a way to use the alternate segment theorem?

Any help would be greatly appreciated. Thank you in advance.

There is a theorem which states that the angle formed at the centre of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points.
• Sep 9th 2010, 01:29 AM
Educated

$2 \times \text{angle }BDE = \text{reflex angle }BOE$

I don't know if it's just me, but the way mathaddict said it seemed really complex.
• Sep 9th 2010, 02:47 AM
sa-ri-ga-ma
Consider a point P on the larger segment of the circle with center O.

According to the properties of the circle, $\angle{BOE} = 2\angle{BPE}$

BPED is a cyclic quadrilateral. So $\angle{BDE} + \angle{BPE} = 180^o$

But $\angle{BDE} = 122^o$ So $\angle{BPE} = 58^o, and \angle{BOE} = 116^o$
• Sep 9th 2010, 02:48 AM
sa-ri-ga-ma
Consider a point P on the larger segment of the circle with center O.

According to the properties of the circle, $\angle{BOE} = 2\angle{BPE}$

BPED is a cyclic quadrilateral. So $\angle{BDE} + \angle{BPE} = 180^o$

But $\angle{BDE} = 122^o$ So $\angle{BPE} = 58^o, and \angle{BOE} = 116^o$
• Sep 9th 2010, 06:20 AM
caramelcake
Ah thank you everybody! I didn't spot the theorem in the diagram.
@sa-ri-ga-ma, Why is it necessary to consider the point P?
The method I used in the end:
reflex angle $\angle{BOE} = 2\angle{BDE}$
$= 2 (122^o)$
$= 244^o$
$\angle{BOE} = 360^o - 244^o$
$= 116^o$
• Sep 9th 2010, 07:30 AM
sa-ri-ga-ma
Not necessary at all. I tried to explain in case you did visualize the reflex angle.