1963 AMC 12 Problem 34

Question

In triangle

ABC,

side

a=\sqrt3,

side

b=\sqrt3,

and side

c\gt3.

Let

x

be the largest number such that the magnitude, in degrees, of the angle opposite side

c

exceeds

x.

Then

x

equals:

150

120

105

90

60

Use the law of cosines for the angle

C

opposite

c

Compare

c\gt3

with the limiting case

c=3

The law of cosines gives

\cos C=\frac{3+3-c^2}{2\sqrt3\sqrt3}=\frac{6-c^2}{6}.

Since

c\gt3,

\cos C\lt-\frac{1}{2},

so

C\gt120^\circ.

Values can approach

120^\circ

as

c

approaches

3

from above, so the largest guaranteed bound is

120.

Therefore, the correct answer is B.

Problem 34 in other years