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

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==Problem==
 
In <math>\bigtriangleup ABC</math>, <math>D</math> is a point on side <math>\overline{AC}</math> such that <math>BD=DC</math> and <math>\angle BCD</math> measures <math>70^\circ</math>. What is the degree measure of <math>\angle ADB</math>?
 
In <math>\bigtriangleup ABC</math>, <math>D</math> is a point on side <math>\overline{AC}</math> such that <math>BD=DC</math> and <math>\angle BCD</math> measures <math>70^\circ</math>. What is the degree measure of <math>\angle ADB</math>?
  
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dot("$C$", C, SE);
 
dot("$C$", C, SE);
 
dot("$D$", D, S);
 
dot("$D$", D, S);
label("$70^\circ$",C,2*dir(180-35));\</asy>
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label("$70^\circ$",C,2*dir(180-35));</asy>
  
 
<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==
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https://www.youtube.com/watch?v=HP-lBKohxhE
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https://youtu.be/j5KrHM81HZ8 ~savannahsolver
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==Solution==
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<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>.
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==See Also==
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{{AMC8 box|year=2014|num-b=8|num-a=10}}
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{{MAA Notice}}

Revision as of 09:15, 27 April 2022

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]

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

Video Solution

https://www.youtube.com/watch?v=HP-lBKohxhE

https://youtu.be/j5KrHM81HZ8 ~savannahsolver

Solution

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

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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