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Difference between revisions of "1982 AHSME Problems/Problem 23"

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\textbf{(E)}\ \text{none of these} </math>
 
\textbf{(E)}\ \text{none of these} </math>
  
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== Solution 1 (Sines and Cosines) ==
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== Solution 1 (Law of Sines and Law of Cosines) ==
 
In <math>\triangle ABC,</math> let <math>a=n,b=n+1,c=n+2,</math> and <math>\angle A=\theta</math> for some positive integer <math>n.</math> We are given that <math>\angle C=2\theta,</math> and we need <math>\cos\theta.</math>
 
In <math>\triangle ABC,</math> let <math>a=n,b=n+1,c=n+2,</math> and <math>\angle A=\theta</math> for some positive integer <math>n.</math> We are given that <math>\angle C=2\theta,</math> and we need <math>\cos\theta.</math>
  
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We apply the Law of Cosines to solve for <math>\cos\angle A:</math> <cmath>\cos\theta=\frac{b^2+c^2-a^2}{2bc}=\frac{n+5}{2(n+2)}.</cmath>
+
We apply the Law of Cosines to solve for <math>\cos\theta:</math> <cmath>\cos\theta=\frac{b^2+c^2-a^2}{2bc}=\frac{n+5}{2(n+2)}.</cmath>
−
 
+
We apply the Law of Sines to <math>\angle A</math> and <math>\angle C:</math> <cmath>\frac{\sin\theta}{a}=\frac{\sin(2\theta)}{c}.</cmath>
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Let the brackets denote areas. We write <math>[ABC]</math> in terms of <math>\sin\angle A</math> and <math>\sin\angle C,</math> respectively: <cmath>[ABC]=\frac12 bc\sin\theta=\frac12 ab\sin(2\theta).</cmath>
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By the Double-Angle Formula <math>\sin(2\theta)=2\sin\theta\cos\theta,</math> we simplify and rearrange to solve for <math>\cos\theta:</math> <cmath>\cos\theta=\frac{c}{2a}=\frac{n+2}{2n}.</cmath>
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Recall that <math>\sin(2\theta)=2\sin\theta\cos\theta</math> holds for all <math>\theta.</math> Equating the last two expressions and then simplifying, we have <cmath>\cos\theta=\frac{c}{2a}=\frac{n+2}{2n}.</cmath>
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We equate the expressions for <math>\cos\theta:</math> <cmath>\frac{n+5}{2(n+2)}=\frac{n+2}{2n},</cmath>
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Equating the expressions for <math>\cos\theta,</math> we get <cmath>\frac{n+5}{2(n+2)}=\frac{n+2}{2n},</cmath>
 
 
from which <math>n=4.</math> By substitution, the answer is <math>\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.</math>
 
from which <math>n=4.</math> By substitution, the answer is <math>\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
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== Solution 2 (Cosines Only) ==
+
== Solution 2 (Law of Cosines Only) ==
 
This solution uses the same variable definitions as Solution 1 does. Moreover, we conclude that <math>\cos\theta=\frac{n+5}{2(n+2)}</math> from the second paragraph of Solution 1.
 
This solution uses the same variable definitions as Solution 1 does. Moreover, we conclude that <math>\cos\theta=\frac{n+5}{2(n+2)}</math> from the second paragraph of Solution 1.
  
−
We apply the Law of Cosines to solve for <math>\cos\angle C:</math> <cmath>\cos(2\theta)=\frac{a^2+b^2-c^2}{2ab}=\frac{n-3}{2n}.</cmath>
+
We apply the Law of Cosines to solve for <math>\cos(2\theta):</math> <cmath>\cos(2\theta)=\frac{a^2+b^2-c^2}{2ab}=\frac{n-3}{2n}.</cmath>
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By the Double-Angle Formula <math>\cos(2\theta)=2\cos^2\theta-1,</math> we have <cmath>2\left(\frac{n+5}{2(n+2)}\right)^2-1=\frac{n-3}{2n},</cmath> from which <math>n=-3,-\frac12,4.</math> Recall that <math>n</math> is a positive integer, so <math>n=4.</math> By substitution, the answer is <math>\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.</math>
+
By the Double-Angle Formula <math>\cos(2\theta)=2\cos^2\theta-1,</math> we set up an equation for <math>n:</math> <cmath>\frac{n-3}{2n}=2\left(\frac{n+5}{2(n+2)}\right)^2-1,</cmath> from which <math>n=-3,-\frac12,4.</math> Recall that <math>n</math> is a positive integer, so <math>n=4.</math> By substitution, the answer is <math>\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Latest revision as of 22:29, 15 September 2021

Problem

The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is

$\textbf{(A)}\ \frac{3}{4}\qquad \textbf{(B)}\ \frac{7}{10}\qquad \textbf{(C)}\ \frac{2}{3}\qquad \textbf{(D)}\ \frac{9}{14}\qquad \textbf{(E)}\ \text{none of these}$

Solution 1 (Law of Sines and Law of Cosines)

In $\triangle ABC,$ let $a=n,b=n+1,c=n+2,$ and $\angle A=\theta$ for some positive integer $n.$ We are given that $\angle C=2\theta,$ and we need $\cos\theta.$

We apply the Law of Cosines to solve for $\cos\theta:$ \[\cos\theta=\frac{b^2+c^2-a^2}{2bc}=\frac{n+5}{2(n+2)}.\] We apply the Law of Sines to $\angle A$ and $\angle C:$ \[\frac{\sin\theta}{a}=\frac{\sin(2\theta)}{c}.\] By the Double-Angle Formula $\sin(2\theta)=2\sin\theta\cos\theta,$ we simplify and rearrange to solve for $\cos\theta:$ \[\cos\theta=\frac{c}{2a}=\frac{n+2}{2n}.\] We equate the expressions for $\cos\theta:$ \[\frac{n+5}{2(n+2)}=\frac{n+2}{2n},\] from which $n=4.$ By substitution, the answer is $\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.$

~MRENTHUSIASM

Solution 2 (Law of Cosines Only)

This solution uses the same variable definitions as Solution 1 does. Moreover, we conclude that $\cos\theta=\frac{n+5}{2(n+2)}$ from the second paragraph of Solution 1.

We apply the Law of Cosines to solve for $\cos(2\theta):$ \[\cos(2\theta)=\frac{a^2+b^2-c^2}{2ab}=\frac{n-3}{2n}.\] By the Double-Angle Formula $\cos(2\theta)=2\cos^2\theta-1,$ we set up an equation for $n:$ \[\frac{n-3}{2n}=2\left(\frac{n+5}{2(n+2)}\right)^2-1,\] from which $n=-3,-\frac12,4.$ Recall that $n$ is a positive integer, so $n=4.$ By substitution, the answer is $\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.$

~MRENTHUSIASM

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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

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