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== | ||
| + | https://www.youtube.com/watch?v=HP-lBKohxhE | ||
| + | |||
==Solution== | ==Solution== | ||
<math>BD = DC</math>, so <math>\angle DBC = \angle DCB = 70</math>. 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>. | <math>BD = DC</math>, so <math>\angle DBC = \angle DCB = 70</math>. 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>. | ||
Revision as of 19:04, 5 December 2021
Contents
Problem
In 
, 
is a point on side 
such that 
and 
measures 
. What is the degree measure of 
?
![[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35));[/asy]](http://latex.artofproblemsolving.com/9/4/7/94770df9a0e1d5a71f602bfb3b45022dcca9dec4.png)
Video Solution
https://www.youtube.com/watch?v=HP-lBKohxhE
Solution
, so 
. 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. 
