November 20, 2018 November 20, 2018 09:00 AM PST 10:00 AM PST The reward for signing up with the registration form and attending the chat is: 6 free examPAL quizzes to practice your new skills after the chat. November 20, 2018 November 20, 2018 06:00 PM EST 07:00 PM EST What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 486
Location: India
GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49
GPA: 3.3

As shown in the figure above, line segments AB and AC are ta
[#permalink]
Show Tags
17 Jul 2014, 18:40
Question Stats:
68% (02:22) correct 32% (02:20) wrong based on 130 sessions
HideShow timer Statistics
Attachment:
geometry_graphics_1.gif [ 6.48 KiB  Viewed 3670 times ]
As shown in the figure above, line segments AB and AC are tangent to circle O. If line segments BD and DA have the same length, what is angle BAO? (Note: Figure not drawn to scale.) A. 15º B. 30º C. 36º D. 45º E. 50º
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/howtoscore750and750imovedfrom710to189016.html



Intern
Joined: 10 Jun 2014
Posts: 24

Re: As shown in the figure above, line segments AB and AC are ta
[#permalink]
Show Tags
17 Jul 2014, 19:46
I did't think this question was easy, but I finally came up with the solution.
step 1: since AB and AC are tangent to the circle then, angles OBA = OCA =90º step 2: Angle BAO = angle DBA = x and angle BDA = y step 3: OB = OD (both radii of circle O) and therefore triangle BOD is isosceles so angle OBD = angle ODB =z step 4: 2x + y = 180 , y + z =180 , x+z =90 step 5: solve both equations for variable z > z=180y, z = 90x step 6: 90x=180y yx = 90 step 7: y+2x=180 y  x = 90 (1) 3x=90 x = 30
If someone knows a faster way to solve this question please post!



Director
Joined: 17 Dec 2012
Posts: 629
Location: India

As shown in the figure above, line segments AB and AC are ta
[#permalink]
Show Tags
Updated on: 20 Jul 2014, 18:29
It seems to me that if a line drawn from the hypotenuse of a right triangle to the opposite vertex, creates two triangles, such that one of the them is isosceles, the other has to be an isosceles or an equilateral triangle. We can easily arrive at the answer if that is the case.
_________________
Srinivasan Vaidyaraman Sravna Holistic Solutions http://www.sravnatestprep.com
Holistic and Systematic Approach



Manager
Joined: 10 Feb 2014
Posts: 63

Re: As shown in the figure above, line segments AB and AC are ta
[#permalink]
Show Tags
20 Jul 2014, 16:49
a13ssandra wrote: I did't think this question was easy, but I finally came up with the solution.
step 1: since AB and AC are tangent to the circle then, angles OBA = OCA =90º step 2: Angle BAO = angle DBA = x and angle BDA = y step 3: OB = OD (both radii of circle O) and therefore triangle BOD is isosceles so angle OBD = angle ODB =z step 4: 2x + y = 180 , y + z =180 , x+z =90 step 5: solve both equations for variable z > z=180y, z = 90x step 6: 90x=180y yx = 90 step 7: y+2x=180 y  x = 90 (1) 3x=90 x = 30
If someone knows a faster way to solve this question please post! No need to use another variable z in step 3, apply the theorem "exterior angle = sum of opp. interior angle". angle ODB (=OBD) = angle BAO + angle DBA = 2x angle OBD + angle DBA = 90 \(2x + x = 90\) \(x = 30\)



Director
Joined: 17 Dec 2012
Posts: 629
Location: India

Re: As shown in the figure above, line segments AB and AC are ta
[#permalink]
Show Tags
20 Jul 2014, 20:30
SravnaTestPrep wrote: It seems to me that if a line drawn from the hypotenuse of a right triangle to the opposite vertex, creates two triangles, such that one of the them is isosceles, the other has to be an isosceles or an equilateral triangle. We can easily arrive at the answer if that is the case. In the above case 2OBD = BDA We also have OBD+DBO= BDA therefore OBD=DBO and so triangle OBD is equilateral from which we can find BAD=BAO=30 degrees
_________________
Srinivasan Vaidyaraman Sravna Holistic Solutions http://www.sravnatestprep.com
Holistic and Systematic Approach



Intern
Joined: 18 Sep 2013
Posts: 5
WE: Engineering (Manufacturing)

As shown in the figure above, line segments AB and AC are ta
[#permalink]
Show Tags
25 Aug 2015, 02:59
honchos wrote: Attachment: geometry_graphics_1.gif As shown in the figure above, line segments AB and AC are tangent to circle O. If line segments BD and DA have the same length, what is angle BAO? (Note: Figure not drawn to scale.) A. 15º B. 30º C. 36º D. 45º E. 50º Angle OBA = 90 Angle ODB = Angle DBA + Angle DAB  (Exterior angle = Sum of opposite interior angles) We can say ODB is double the angle DBA as the triangle ADB is isosceles ................. Which In turn means Angle OBD is double the angle DBA As the triangle BOD is isosceles Considering above Angle OBA = 90 can only be split in 30 + 60 ....... so the angle OAB = 30



NonHuman User
Joined: 09 Sep 2013
Posts: 8816

Re: As shown in the figure above, line segments AB and AC are ta
[#permalink]
Show Tags
26 Oct 2018, 23:33
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: As shown in the figure above, line segments AB and AC are ta &nbs
[#permalink]
26 Oct 2018, 23:33






