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1+2 ) From area we can find radius = 9. From the length of arc we can find angle of AOB (O is the center of angle) Angle = Length of arc / Circumference * 360 degrees --> (7pi/18pi) * 360 = 140 degrees
Angle OBA = 20 degrees Angle OBC = 90 degrees So we can infer that angle ABC = 70 degrees Ans as triangle CBA is isosceles we can infer that x = 40 degrees
Re: In the figure above, two lines are tangent to a circle at points A and
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10 Jun 2015, 01:43
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3
1: insufficient. implies radius is 9. 2: insufficient.
Together: ADB is \(7\pi\), total circumference is \(18\pi\) so if O is the center of the circle, \(\angle AOB = \frac{7}{18}*360=140^{\circ}\). Since \(AO=BO\), \(\triangle AOB\) is isosceles, with \(\angle AOB = 140^{\circ}, \angle ABO = \angle BAO = \frac{180 - 140}{2} = \frac{40}{2} = 20^{\circ}\). Since CB is tangent to the circle, \(\angle CBO = 90^{\circ}\) so \(\angle CBA = \angle CAB =90^{\circ} - 20^{\circ}=70^{\circ}\); therefore, \(x = 40^{\circ}\), so C.
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12 Jun 2015, 09:12
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Hello!!
1. Insufficient 2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.
Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360 320+x=360 Hence x = 40'
Isn't option 2 sufficient to find the angle..Want expert comment.
In the figure above, two lines are tangent to a circle at points A and
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12 Jun 2015, 09:34
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ArnavPaliw wrote:
Hello!!
1. Insufficient 2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.
Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360 320+x=360 Hence x = 40'
Isn't option 2 sufficient to find the angle..Want expert comment.
Hello ArnavPaliw You can't find angle only from arc because circle can be different sizes and it will change the angle If circle really big then this angle can be 1 degree and if circle is very small this arc can be almost all circumference and angle can be for example a 350 degree. Upd: Not in this case. In this case it can be only less than half of circumference. (thanks to the chetan2u for clarification)
But I like your application of formula (n-2)*180 this is more elegant way to solve last step than I used in my solution
_________________
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12 Jun 2015, 14:53
Harley1980 wrote:
ArnavPaliw wrote:
Hello!!
1. Insufficient 2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.
Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360 320+x=360 Hence x = 40'
Isn't option 2 sufficient to find the angle..Want expert comment.
Hello ArnavPaliw You can't find angle only from arc because circle can be different sizes and it will change the angle If circle really big then this angle can be 1 degree and if circle is very small this arc can be almost all circumference and angle can be for example a 350 degree
But I like your application of formula (n-2)*180 this is more elegant way to solve last step than I used in my solution
Hi, you are absolutely correct when you say that the length of the arc will vary as per the size of circle... But this arc can never be almost equal to entire circumference.. it can never be greater than half the circumference. say its equal to half the circumference, the lines will never meet being parallel and greater than circumference will meet on the other side
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1) statement gives us the radius but angle cannot be found by that... insuff 2) statement gives us the length of arc but we know nothing about the radius...in suff combined suff as we can deduce the centre angle from ratio of arc to entire circumference ans C
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12 Jun 2015, 15:03
chetan2u wrote:
Harley1980 wrote:
ArnavPaliw wrote:
Hello!!
1. Insufficient 2. Sufficient: - By knowing the length of arc we can find the angel AOB. Now can't we apply the sum of all the interior angels of a quadrilateral (in this case AOBC) is equal to 360'.
Here we know angle OAC=OBC=90' and angle AOB is 140' (finding the interior angel by length of the arc). So now equation will be 140+90+90+x=360 320+x=360 Hence x = 40'
Isn't option 2 sufficient to find the angle..Want expert comment.
Hello ArnavPaliw You can't find angle only from arc because circle can be different sizes and it will change the angle If circle really big then this angle can be 1 degree and if circle is very small this arc can be almost all circumference and angle can be for example a 350 degree
But I like your application of formula (n-2)*180 this is more elegant way to solve last step than I used in my solution
Hi, you are absolutely correct when you say that the length of the arc will vary as per the size of circle... But this arc can never be almost equal to entire circumference.. it can never be greater than half the circumference. say its equal to half the circumference, the lines will never meet being parallel and greater than circumference will meet on the other side
But I always think that we have two arcs in any case.
In our case we have arc AB with length 7pi and angle 140 degrees And another arc AB with length 11pi and angle 220 degrees
Maybe I miss something but I never met a rule that arc can't be more than half of circumference.
P.S. I'm not saying that I am right, geometry is my weak place. But if it is not too hard, can you please, give some link on that rule?
_________________
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12 Jun 2015, 19:07
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Harley1980 wrote:
thanks for reprimand.
But I always think that we have two arcs in any case.
In our case we have arc AB with length 7pi and angle 140 degrees And another arc AB with length 11pi and angle 220 degrees
Maybe I miss something but I never met a rule that arc can't be more than half of circumference.
P.S. I'm not saying that I am right, geometry is my weak place. But if it is not too hard, can you please, give some link on that rule?
Sorry buddy if i have sounded that way.. you are correct that there will be two arcs between two points on circle and the length of one of them could be even 359 degrees. But this is not possible in this case ... If you have an angle subtended by the tangents at two points on circle, the arc subtended by those points and enclosed within these two points will be less than 180 and ofcourse the other arc may be 359 also as you have explained..
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In the figure above, two lines are tangent to a circle at points A and
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12 Jun 2015, 21:14
chetan2u wrote:
Harley1980 wrote:
thanks for reprimand.
But I always think that we have two arcs in any case.
In our case we have arc AB with length 7pi and angle 140 degrees And another arc AB with length 11pi and angle 220 degrees
Maybe I miss something but I never met a rule that arc can't be more than half of circumference.
P.S. I'm not saying that I am right, geometry is my weak place. But if it is not too hard, can you please, give some link on that rule?
Sorry buddy if i have sounded that way.. you are correct that there will be two arcs between two points on circle and the length of one of them could be even 359 degrees. But this is not possible in this case ... If you have an angle subtended by the tangents at two points on circle, the arc subtended by those points and enclosed within these two points will be less than 180 and ofcourse the other arc may be 359 also as you have explained..
Oh, I get it. I missed constraints that impose tangents. Thanks for clarifying I've changed my post
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The size of the angle depends on two things: (1) INSUFFICIENT: From the rubber band picture on the right, we see that for a circle of fixed size, x can still vary with the length of arc ADB (i.e. x varies with the placement of C).
(2) INSUFFICIENT: We can't rely solely on the pictures above, as the arc length varies with both circle size (see left picture) and with placement of C relative to a circle of fixed size (see right picture). We'll draw two cases to prove that x can vary for a given arc ADB: For a circle with circumference 28π, the arc ADB is 1/4 (= 7π/28π) of the circle, so x is 90°. For a circle with circumference 15π, the arc ADB is nearly half of the circle (= 7π/15π), and the lines tangent to the circle at A and B will be nearly parallel to each other, i.e. x is very small.
(1) AND (2) SUFFICIENT: If the area of the circle is πr^2 = 81π, then r = 9. The circumference of the circle is 2πr = 18. Thus, arc ADB is 7/18 (= 7π/18π) of the circumference of the circle. There is only one way to draw the lines tangent to the circle at A and B, so x can only be one value.
In the figure above, two lines are tangent to a circle at points A and
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24 Mar 2016, 17:29
we can draw 2 radii, from A and B to the origin. and a bisector from C to the origin. we get 2 right triangles with right angles at A and B. angle x is split in 2 similar small angles. to get X, we need to find one of the two similar angles formed at origin COA or COB. knowing this angle, we can apply the triangle principle that the sum of all angles must equal to 180, and since we already know that 1 angle is 90, finding the second one would crack this bad boy.
1. this one tells that the r=9. or OA and OB=9. doesn't help much. Even if we draw a line from A to B, to form another 2 right triangles, we would still not be able to find the values of the angles formed at the origin. so 1 alone is insufficient. A and D - out.
2. ADC is 7pi. well, clearly this is insufficient. if we knew the circumference of the circle, we would be able to find the value of the two angles formed at the origin. B is out.
1+2 => we know area, we know the length of the chord..we can find the angles formed at the origin..C is sufficient..
just for myself calculations: 7pi/18pi = 7/18. multiply by 360 = 360*7/18 = 140. this would be the sum of the two similar angles formed at the origin. now we need to find the value of only 1. 140/2 = 70. 180 - 90 - 70 = 20 -> value of a small x sub-angle. x will be twice this value -> 40.
The attachment 2015-06-09_1333.png is no longer available
Excellent opportunity to explore our two powerful tools when dealing with geometric-related Data Sufficiency problems:
The GEOMETRIC BIFURCATION and THE ALGEBRAIC-GEOMETRIC BIFURCATION
Please study the image attached!
This solution follows the notations and rationale taught in the GMATH method.
Regards, fskilnik.
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Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Nov with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount!
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In the figure above, two lines are tangent to a circle at points A and &nbs
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05 Sep 2018, 14:46
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65% (01:40) correct 
