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In the figure above, point O is the center of the circle and

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Re: In the figure above, point O is the center of the circle and  [#permalink]

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New post 06 May 2018, 08:53
1
dave13 wrote:
VeritasPrepKarishma wrote:
tonebeeze wrote:
In the figure attached, point O is the center of the circle and OC = AC = AB. What is the value of x (in degrees)?

a. 40
b. 36
c. 34
d. 32
e. 30


A small diagram helps:
Attachment:
Ques1.jpg


In the figure, you see (180 - 2x) + y = 180 (straight angle) so 2x = y

Also, the moment you see the circle and its two radii, mark them equal and the corresponding angles equal. (the red angle = blue angle)
x + 180 - 2y = y
5x = 180 (from above, y = 2x)
x = 36



VeritasPrepKarishma hello :)

cant understand the logic behind this (180 - 2x) + y = 180 (straight angle) so 2x = y

if you are deducting 2x from 180 then why are you adding y ? there are 2x and 2y right and how 2x = y :?

why triangle OAB an isosceles and not an equilateral :?

please explain:)

have a great wekkend :)


In triangle OAC, OC = AC
That is why angle COA = angle OAC = x
So angle OCA = 180 - x - x = 180 - 2x

Also AC = BC, so in triangle ACB,
angle ACB = angle CBA = y

Angles OCA and ACB form a straight angle so
180 - 2x + y = 180

In triangle OAB, OA = OB = radius of the circle. But these are not equal to AB (so not equilateral)
Note that AB is actually equal to OC (and AC). OC is less than the radius of the circle.
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Re: In the figure above, point O is the center of the circle and  [#permalink]

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New post 07 Oct 2018, 11:25
From triangle OAC,
<AOC = <OAC = x degree. (since OC = AC)

And ext. <ACB = <AOC + <OAC = 2x degree.

Now from triangle ACB,

AC = AB , therefore <ABC = <ACB = 2x degree and <CAB = (180 - 4x) degree.

Now considering triangle OAB,
<BOA + <OAB + <ABO = 180
x + [x+180-4x] + 2x = 180

Where and why am I going wrong on this ?
GMAT Club Bot
Re: In the figure above, point O is the center of the circle and &nbs [#permalink] 07 Oct 2018, 11:25

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In the figure above, point O is the center of the circle and

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