1982 AHSME Problems/Problem 23

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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