1982 AMC 12 Problem 23

Question

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

\frac34

\frac7{10}

\frac23

\frac9{14}

none of these

Answer: A

Concepts:law of sines law of cosines trigonometric identity

Difficulty rating: 2150

Small hint:

Let the smallest and largest angles be

\theta

and

2\theta

Big hint:

Use the law of sines to compare the shortest and longest consecutive sides

Solution:

Let the consecutive sides be

n,

n+1,

n+2,

opposite angles

\theta,

\phi,

2\theta.

By the law of sines,

\frac{n+2}{n}=\frac{\sin2\theta}{\sin\theta}=2\cos\theta.

The law of cosines at the smallest angle also gives

\cos\theta=\frac{n^2+6n+5}{2(n+1)(n+2)}.

Equating these expressions yields

n^2-3n-4=0,

so

n=4.

Hence the sides are

4,

5,

6

and

\cos\theta=\frac{6}{2\cdot4}=\frac{3}{4}.

Therefore, the correct answer is A.