In triangle ABC, ∠ ABC=120°, AB=3, and BC=4. If perpendiculars constructed to AB at A and to BC at C meet at D, then CD=

1992 AMC 12 Problem 25

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25.

In triangle

ABC,

\angle ABC=120^\circ,

AB=3,

and

BC=4.

If perpendiculars constructed to

\overline{AB}

A

and to

\overline{BC}

C

meet at

D,

then

CD=

3

\frac8{\sqrt3}

5

\frac{11}{2}

\frac{10}{\sqrt3}

Place

B=(0,0)

and

A=(3,0)

Use the

120^\circ

angle to locate

C,

then parametrize the line through

C

perpendicular to

BC

Set

B=(0,0)

and

A=(3,0).

Then

\begin{aligned} C&=4(\cos120^\circ,\sin120^\circ)\\ &=(-2,2\sqrt3). \end{aligned}

The perpendicular to

AB

A

x=3.

A vector perpendicular to

BC=(-2,2\sqrt3)

(\sqrt3,1),

so the other perpendicular is

C+t(\sqrt3,1).

Its

x

-coordinate is

3

when

t=\frac{5}{\sqrt3}.

Since

|(\sqrt3,1)|=2,

CD=2t=\frac{10}{\sqrt3}.

Thus the correct answer is E.