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> | ||
− | == Solution 1 == | + | == 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> | ||
− | We apply the Law of Cosines to solve for <math>\cos\ | + | 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> | |
− | + | 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> | |
− | + | 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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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 | ||
− | == Solution 2 == | + | == 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. | |
− | We apply the Law of Cosines to solve for <math>\cos | + | 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> |
− | + | 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> | |
− | By the Double-Angle Formula <math>\cos(2\theta)=2\cos^2\theta-1,</math> we | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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== See Also == | == See Also == | ||
{{AHSME box|year=1982|num-b=22|num-a=24}} | {{AHSME box|year=1982|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:29, 15 September 2021
Contents
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
Solution 1 (Law of Sines and Law of Cosines)
In let and for some positive integer We are given that and we need
We apply the Law of Cosines to solve for We apply the Law of Sines to and By the Double-Angle Formula we simplify and rearrange to solve for We equate the expressions for from which By substitution, the answer is
~MRENTHUSIASM
Solution 2 (Law of Cosines Only)
This solution uses the same variable definitions as Solution 1 does. Moreover, we conclude that from the second paragraph of Solution 1.
We apply the Law of Cosines to solve for By the Double-Angle Formula we set up an equation for from which Recall that is a positive integer, so By substitution, the answer is
~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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