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)
m (→‎Solution)
 
(11 intermediate revisions by 4 users not shown)
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
 
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, there is no other way to satisfy this equation other than making both <math>\cos(2A-B)=1</math> and <math>\sin(A+B)=1</math>, since any other way would cause one of these values to become 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> and solving this system gives us <math>A=30^{\circ}</math> and <math>B=60^{\circ}</math>. It is clear that <math>\triangle ABC</math> is a <math>30^{\circ},60^{\circ},90^{\circ}</math> triangle with <math>BC=2</math>　<math>\Longrightarrow</math>　<math>(C)</math>.
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}}
 +
 
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:31, 18 January 2020

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, there is no other way to satisfy this equation other than making both $\cos(2A-B)=1$ and $\sin(A+B)=1$, since any other way would cause one of these values to become greater than 1, which contradicts our previous statement. From this we can easily conclude 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 clear that $\triangle ABC$ is a $30^{\circ},60^{\circ},90^{\circ}$ triangle with $BC=2$　$\Longrightarrow$　$(C)$.

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 AMC 12 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12B_Problems/Problem_13&oldid=114922"