Difference between revisions of "2009 AMC 8 Problems/Problem 19"

Revision as of 23:52, 1 May 2022

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 $\boxed{\textbf{(D)}\ 165}$ .

Video Solution

https://youtu.be/FDgcLW4frg8?t=902

~ pi_is_3.14

Video Solution 2

https://youtu.be/iB0dcpVREE8 Soo, DRMS, NM

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 32: / Line 32: @@
 ==Video Solution 2==
 https://youtu.be/iB0dcpVREE8 Soo, DRMS, NM
 ==See Also==
 {{AMC8 box|year=2009|num-b=18|num-a=20}}
 {{MAA Notice}}

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