Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 12B Problems/Problem 13"

(→‎Solution)
(→‎Solution)
Line 5: Line 5:
  
 
== Solution ==
 
== Solution ==
−
We note that the maximum values for both the sine and the cosine function are 1.  
+
We note that the max value for both sine and cosine of any angle is 1.  
−
Therefore, the only way for this equation to be true is if <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if one of these equaled less than 1, the other one would have to be greater than 1, which contradicts our previous statement.
+
Therefore, the only way to satisfy the equation is if <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since if either one of these is less than 1, the other one would have to be greater than 1, which contradicts our previous statement.
−
From this we can easily conclude that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. From this we see that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle with <math>AB=4</math>, <math>AC=2\sqrt{2}</math>, and <math>BC=2</math>　<math>(C)</math>
+
From this we can easily deduce that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. It is obvious from this that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle with <math>AB=4</math>, <math>AC=2\sqrt{2}</math>, and <math>BC=2</math>　<math>(C)</math>

Revision as of 21:38, 6 April 2010

Problem

In $\triangle ABC$, $\cos(2A-B)+\sin(A+B)=2$ and $AB=4$. What is $BC$?

$\textbf{(A)}\ \sqrt{2} \qquad \textbf{(B)}\ \sqrt{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 2\sqrt{2} \qquad \textbf{(E)}\ 2\sqrt{3}$

Solution

We note that the max value for both sine and cosine of any angle is 1. Therefore, the only way to satisfy the equation is if $\cos(2A-B)=1$ and $\sin(A+B)=1$, since if either one of these is less than 1, the other one would have to be greater than 1, which contradicts our previous statement. From this we can easily deduce that $2A-B=0^{\circ}$ and $A+B=90^{\circ}$ and solving this system gives us $A=30^{\circ}$ and $B=60^{\circ}$. It is obvious from this that $\triangle ABC$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with $AB=4$, $AC=2\sqrt{2}$, and $BC=2$　$(C)$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_13&oldid=34251"