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17 Aug 2020, 23:51
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
29% (02:21) correct
71%
(02:02)
wrong
based on 52
sessions
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Attachment:
File comment: Triangle
Triangle.png [ 22.94 KiB | Viewed 3441 times ]
Triangle.png [ 22.94 KiB | Viewed 3441 times ]
Circle O is inscribed in △ABC. If ∠BAC = 30, then what is the measurement of ∠BOC?
A. 105
B. 115
C. 120
D. 130
E. 150
18 Aug 2020, 01:21
Kudos
Bookmarks
Ans: A (105)
BAC = 30
Therefore: <ABC + <ACB = 180-30 = 150
Line OB and OC bisects <ABC & <ACB respectively
therefore: <OBC + <OCB = (<ABC + <ACB)/2 = 150/2 = 75
Now in triangle OBC-
<OBC + <OCB + <BOC = 180
<BOC = 180- <OBC - <OCB = 180-75 = 105
Thank you = Kudos
BAC = 30
Therefore: <ABC + <ACB = 180-30 = 150
Line OB and OC bisects <ABC & <ACB respectively
therefore: <OBC + <OCB = (<ABC + <ACB)/2 = 150/2 = 75
Now in triangle OBC-
<OBC + <OCB + <BOC = 180
<BOC = 180- <OBC - <OCB = 180-75 = 105
Thank you = Kudos
18 Aug 2020, 05:28
Kudos
Bookmarks
The angular bisectors of a triangle intersect to give the incircle (circle inscribed in a triangle)
OB and OC are the angular bisectors of \(\angle{ABC}\) and \(\angle{ACB}\)
Given that \(\angle{BAC}\) = \(30^o\)
Therefore \(\angle{ACB}\) + \(\angle{ABC}\) = \(180^o - 30^o\) = \(150^o\) (Sum of angles of a triangle = \(180^o\))
\(\angle{OBC}\) = \(\frac{1}{2}\) \(\angle{ABC}\)
\(\angle{OCB}\) = \(\frac{1}{2}\) \(\angle{ACB}\)
Adding we get \(\angle{OBC}\) + \(\angle{OCB}\) = \(\frac{1}{2}\) \(\angle{ABC}\) + \(\angle{ACB}\) = \(\frac{1}{2}\) [\(\angle{ABC}\) + \(\angle{ACB}\)] = \(\frac{1}{2} * 150^o\) = \(75^o\)
In \(\triangle {BOC}\), \(\angle{BOC}\) + \(\angle{OCB}\) + \(\angle{OBC}\) = \(180^o\)
Therefore \(\angle{BOC}\) = \(180^o\) - [\(\angle{OCB}\) + \(\angle{OBC}\)] = \(180^o - 75^o\) = \(105^o\)
Option A
Arun Kumar
OB and OC are the angular bisectors of \(\angle{ABC}\) and \(\angle{ACB}\)
Given that \(\angle{BAC}\) = \(30^o\)
Therefore \(\angle{ACB}\) + \(\angle{ABC}\) = \(180^o - 30^o\) = \(150^o\) (Sum of angles of a triangle = \(180^o\))
\(\angle{OBC}\) = \(\frac{1}{2}\) \(\angle{ABC}\)
\(\angle{OCB}\) = \(\frac{1}{2}\) \(\angle{ACB}\)
Adding we get \(\angle{OBC}\) + \(\angle{OCB}\) = \(\frac{1}{2}\) \(\angle{ABC}\) + \(\angle{ACB}\) = \(\frac{1}{2}\) [\(\angle{ABC}\) + \(\angle{ACB}\)] = \(\frac{1}{2} * 150^o\) = \(75^o\)
In \(\triangle {BOC}\), \(\angle{BOC}\) + \(\angle{OCB}\) + \(\angle{OBC}\) = \(180^o\)
Therefore \(\angle{BOC}\) = \(180^o\) - [\(\angle{OCB}\) + \(\angle{OBC}\)] = \(180^o - 75^o\) = \(105^o\)
Option A
Arun Kumar
