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O is the center

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O is the center [#permalink] New post 17 Mar 2018, 03:48
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Question Stats:

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Attachment:
circle.png
circle.png [ 134.62 KiB | Viewed 810 times ]


O is the center of the circle and ∠AOC is a right angle

Quantity A
Quantity B
OD
BD


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: O is the center [#permalink] New post 17 Mar 2018, 23:07
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Explanation

In this figure, if \(<BDC = 90\) degrees than \(<BDO\) should be \(90\) degrees as well.
This is because angle in a straight line add to \(180\) degree

Similarly, If \(<AOB = 44\)degrees than \(<BOD = 46\) degrees because Given \(AOC\) is a right triangle so \(<AOB\) and \(<BOD\) should add to \(90\)degrees

Now the remaining angle\(<DBO = 44\) degrees
Sum of angles in a triangle add to \(180\) degrees.

From here we can deduce that \(BD>OD\) because side opposite larger angle is larger than side opposite smaller angle in a triangle
option B
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Re: O is the center [#permalink] New post 03 May 2018, 09:47
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Expert's post
Carcass wrote:
Attachment:
circle.png


O is the center of the circle and ∠AOC is a right angle

Quantity A
Quantity B
OD
BD


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Given: ∠AOC = 90°
So, if ∠AOB = 44°, then ∠BOD = 46°

If ∠BOD = 46°, then ∠OBD = 44° (since all 3 angles in ∆OBD must add to 180°

Notice that side OD is opposite the 44° angle (aka ∠OBD),
and side BD is opposite the 46° angle (aka ∠BOD)

Since 46° > 44°, we can conclude that side BD > side OD

Answer: B

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Re: O is the center   [#permalink] 03 May 2018, 09:47
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