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Difference between revisions of "2009 AMC 8 Problems/Problem 19"
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Here, both base angles are <math> 70^\circ</math>, since base angles are congruent. Thus, we can use the equation <math> x+140=180</math>, which simplifies to <math> x=40</math>.
 
Here, both base angles are <math> 70^\circ</math>, since base angles are congruent. Thus, we can use the equation <math> x+140=180</math>, which simplifies to <math> x=40</math>.
  
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Adding up all the cases, we get <math>70+55+40=165</math>, so the answer is <math> \textbf{(D)}\    165</math>.
+
Adding up all the cases, we get <math>70+55+40=165</math>, so the answer is <math> \boxed{\textbf{(D)}\    165}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=18|num-a=20}}
 
{{AMC8 box|year=2009|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:36, 11 November 2014

Problem

Two angles of an isosceles triangle measure $70^\circ$ and $x^\circ$. What is the sum of the three possible values of $x$?

$\textbf{(A)}\ 95 \qquad \textbf{(B)}\ 125 \qquad \textbf{(C)}\ 140 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 180$


Solution

There are 3 cases: where $x^\circ$ is a base angle with the $70^\circ$ as the other angle, where $x^\circ$ is a base angle with $70^\circ$ as the vertex angle, and where $x^\circ$ is the vertex angle with $70^\circ$ as a base angle.

Case 1: $x^\circ$ is a base angle with the $70^\circ$ as the other angle: Here, $x=70$, since base angles are congruent.

Case 2: $x^\circ$ is a base angle with $70^\circ$ as the vertex angle: Here, the 2 base angles are both $x^\circ$, so we can use the equation $2x+70=180$, which simplifies to $x=55$.

Case 3: $x^\circ$ is the vertex angle with $70^\circ$ as a base angle: Here, both base angles are $70^\circ$, since base angles are congruent. Thus, we can use the equation $x+140=180$, which simplifies to $x=40$.

Adding up all the cases, we get $70+55+40=165$, so the answer is $\boxed{\textbf{(D)}\ 165}$.

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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