Difference between revisions of "2014 AMC 8 Problems/Problem 9"

(Solution)
(Solution)
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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>
 
==Solution==
 
==Solution==
BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, ADB = 180 - 40 = \boxed{140}.
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BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, ADB = 180 - 40 = <math>\boxed{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}}

Revision as of 08:39, 9 June 2015

Problem

In $\bigtriangleup ABC$, $D$ is a point on side $\overline{AC}$ such that $BD=DC$ and $\angle BCD$ measures $70^\circ$. What is the degree measure of $\angle ADB$?

[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]

$\textbf{(A) }100\qquad\textbf{(B) }120\qquad\textbf{(C) }135\qquad\textbf{(D) }140\qquad \textbf{(E) }150$

Solution

BD = DC, so angle DBC = angle DCB = 70. Then CDB = 40. Since angle ADB and BDC are supplementary, ADB = 180 - 40 = $\boxed{140}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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