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 "1984 AHSME Problems/Problem 15"
(→‎Solution)
Line 27: Line 27:
 
label("$3x$",(.125,-.06),ESE);
 
label("$3x$",(.125,-.06),ESE);
 
</asy>
 
</asy>
 +
Since the product of the two slopes, $ \tan{2x} $ and $ \tan{-3x} $, is $ -1 $, the lines are perpendicular, and the [[angle]] between them is $ \frac{\pi}{2} $. The angle between them is also $ 2x+3x=5x $, so $ 5x=\frac{\pi}{2} $ and $ x=\frac{\pi}{10} $, or $ 18^\circ, \boxed{\text{A}} $.
  
−
Since the product of the two slopes, <math> \tan{2x} </math> and <math> \tan{-3x} </math>, is <math> -1 </math>, the lines are perpendicular, and the [[angle]] between them is <math> \frac{\pi}{2} </math>. The angle between them is also <math> 2x+3x=5x </math>, so <math> 5x=\frac{\pi}{2} </math> and <math> x=\frac{\pi}{10} </math>, or <math> 18^\circ, \boxed{\text{A}} </math>.
+
NOTE: To show that $ 18^\circ $ is not the only solution, we can also set $ 5x $ equal to another angle measure congruent to $ \frac{\pi}{2} $, such as $ \frac{5\pi}{2} $, yielding another solution as $ \frac{\pi}{2}=90^\circ $, which clearly is a solution to the equation.
−
 
+
==Solution 2==
−
NOTE: To show that <math> 18^\circ </math> is not the only solution, we can also set <math> 5x </math> equal to another angle measure congruent to <math> \frac{\pi}{2} </math>, such as <math> \frac{5\pi}{2} </math>, yielding another solution as <math> \frac{\pi}{2}=90^\circ </math>, which clearly is a solution to the equation.
+
We can simply try the answers. We quickly see that $ 18^\circ$ works, since $sin(36^\circ)sin(54^\circ)=cos(54^\circ)cos(36^\circ)=cos(36^\circ)cos(54^\circ)$ by the identity $cos(x)=sin(90^\circ-c)$
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1984|num-b=14|num-a=16}}
 
{{AHSME box|year=1984|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:49, 1 March 2015

Problem 15

If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$, then one value for $x$ is

$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$

Solution

We divide both sides of the equation by $\cos{2x}\times\cos{3x}$ to get $\frac{\sin{2x}\times\sin{3x}}{\cos{2x}\times\cos{3x}}=1$, or $\tan{2x}\times\tan{3x}=1$.

This looks a lot like the formula relating the slopes of two perpendicular lines, which is $m_1\times m_2=-1$, where $m_1$ and $m_2$ are the slopes. It's made even more relatable by the fact that the tangent of an angle can be defined by the slope of the line that makes that angle with the x-axis.

We can make this look even more like the slope formula by multiplying both sides by $-1$:

$\tan{2x}\times-\tan{3x}=-1$, and using the trigonometric identity $-\tan{x}=\tan{-x}$, we have $\tan{2x}\times\tan{-3x}=-1$.

Now it's time for a diagram:

[asy] unitsize(2.54cm); draw(unitcircle); draw((0,-1.25)--(0,1.25)); draw((-1.25,0)--(1.25,0)); draw((0,0)--(cos(2pi/10),sin(2pi/10))); draw((0,0)--(cos(-3pi/10),sin(-3pi/10))); label("$\tan{-3x}$",(cos(-3pi/10),sin(-3pi/10)),SE); label("$\tan{2x}$",(cos(2pi/10),sin(2pi/10)),NE); label("$2x$",(.125,.03),ENE); label("$3x$",(.125,-.06),ESE); [/asy] Since the product of the two slopes, $ \tan{2x} $ and $ \tan{-3x} $, is $ -1 $, the lines are perpendicular, and the angle between them is $ \frac{\pi}{2} $. The angle between them is also $ 2x+3x=5x $, so $ 5x=\frac{\pi}{2} $ and $ x=\frac{\pi}{10} $, or $ 18^\circ, \boxed{\text{A}} $.

NOTE: To show that $ 18^\circ $ is not the only solution, we can also set $ 5x $ equal to another angle measure congruent to $ \frac{\pi}{2} $, such as $ \frac{5\pi}{2} $, yielding another solution as $ \frac{\pi}{2}=90^\circ $, which clearly is a solution to the equation.

Solution 2

We can simply try the answers. We quickly see that $ 18^\circ$ works, since $sin(36^\circ)sin(54^\circ)=cos(54^\circ)cos(36^\circ)=cos(36^\circ)cos(54^\circ)$ by the identity $cos(x)=sin(90^\circ-c)$

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1984_AHSME_Problems/Problem_15&oldid=68000"