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17 Jul 2014, 19:40
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:42) correct
29%
(02:32)
wrong
based on 208
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
geometry_graphics_1.gif [ 6.48 KiB | Viewed 9545 times ]
A. 15º
B. 30º
C. 36º
D. 45º
E. 50º
17 Jul 2014, 20:46
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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=180-y, z = 90-x
step 6: 90-x=180-y
y-x = 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!
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=180-y, z = 90-x
step 6: 90-x=180-y
y-x = 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!
Updated on: 20 Jul 2014, 19:29
Originally posted by SVaidyaraman on 17 Jul 2014, 22:37.
Last edited by SVaidyaraman on 20 Jul 2014, 19:29, edited 1 time in total.
Last edited by SVaidyaraman on 20 Jul 2014, 19:29, edited 1 time in total.
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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.
