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 7: Line 7:
 
We note that <math>-1</math> <math>\le</math> <math>\sin x</math> <math>\le</math> <math>1</math> and <math>-1</math> <math>\le</math> <math>\cos x</math> <math>\le</math> <math>1</math>.  
 
We note that <math>-1</math> <math>\le</math> <math>\sin x</math> <math>\le</math> <math>1</math> and <math>-1</math> <math>\le</math> <math>\cos x</math> <math>\le</math> <math>1</math>.  
 
Therefore the only way to satisfy this equation is if both <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.
 
Therefore the only way to satisfy this equation is if both <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> (since <math>\cos 0^{\circ}=1</math> and <math>\sin 90^{\circ}=1</math>) and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. It is obvious that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle, and solving for the sides gives us <math>BC=2</math> <math>\Longrightarrow</math>　<math>(C)</math>
+
From this we can easily conclude that <math>2A-B=0^{\circ}</math> and <math>A+B=90^{\circ}</math> (since <math>\cos 0^{\circ}=1</math> and <math>\sin 90^{\circ}=1</math>) and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. It is obvious that <math>\triangle ABC</math> is a <math>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle, and solving for the sides gives us <math>BC=2</math>　<math>\Longrightarrow</math>　<math>(C)</math>

Revision as of 22:03, 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 $-1$ $\le$ $\sin x$ $\le$ $1$ and $-1$ $\le$ $\cos x$ $\le$ $1$. Therefore the only way to satisfy this equation is if both $\cos(2A-B)=1$ 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 conclude that $2A-B=0^{\circ}$ and $A+B=90^{\circ}$ (since $\cos 0^{\circ}=1$ and $\sin 90^{\circ}=1$) and solving this system gives us $A=30^{\circ}$ and $B=60^{\circ}$. It is obvious that $\triangle ABC$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, and solving for the sides gives us $BC=2$　$\Longrightarrow$　$(C)$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_13&oldid=34257"