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As shown in the figure above, line segments AB and AC are ta
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17 Jul 2014, 18:40
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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
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Re: As shown in the figure above, line segments AB and AC are ta
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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!



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As shown in the figure above, line segments AB and AC are ta
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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.
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Re: As shown in the figure above, line segments AB and AC are ta
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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 \(2x + x = 90\) \(x = 30\)



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Re: As shown in the figure above, line segments AB and AC are ta
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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
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As shown in the figure above, line segments AB and AC are ta
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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



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Re: As shown in the figure above, line segments AB and AC are ta
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26 Oct 2018, 23:33
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