Difference between revisions of "2014 AMC 8 Problems/Problem 9"
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<math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | <math>\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150</math> | ||
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+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/jLnqUOe0HPE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=HP-lBKohxhE ~David | ||
+ | |||
+ | https://youtu.be/j5KrHM81HZ8 ~savannahsolver | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/abSgjn4Qs34?t=3140 | ||
+ | |||
==Solution== | ==Solution== | ||
− | BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, ADB = 180 - 40 = \boxed{\textbf{(D)}~140} | + | Using angle chasing is a good way to solve this problem. <math>BD = DC</math>, so <math>\angle DBC = \angle DCB = 70</math>, because it is an isosceles triangle. Then <math>\angle CDB = 180-(70+70) = 40</math>. Since <math>\angle ADB</math> and <math>\angle BDC</math> are supplementary, <math>\angle ADB = 180 - 40 = \boxed{\textbf{(D)}~140}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=8|num-a=10}} | {{AMC8 box|year=2014|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:32, 16 June 2024
Contents
Problem
In ,
is a point on side
such that
and
measures
. What is the degree measure of
?
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=HP-lBKohxhE ~David
https://youtu.be/j5KrHM81HZ8 ~savannahsolver
Video Solution
https://youtu.be/abSgjn4Qs34?t=3140
Solution
Using angle chasing is a good way to solve this problem. , so
, because it is an isosceles triangle. Then
. Since
and
are supplementary,
.
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.