# Thread: Circle Problem

1. ## Circle Problem

I am COMPLETELY stumped with this question. Can someone help me get started on a couple of steps? Hints? Thank you so much.

Streelights A, B, C, D, and E are placed 50 m apart on the main road. The light from a streetlight is effective up to a distance of 60m.

a. Determine the distance from A to the farthest point on the side road that is effectively illuminated.

b. Determine the length of the side road that is effectively illuminated by both streetlights C and D.

Thank you!

2. Since the lights only are good for 60 feet, you must find a point along the side road that is 60 feet from the line of lights.

x is the distance along the side road we must find.

$sin^{-1}(\frac{60}{x})=21$

$sin(21)=\frac{60}{x}$

$x=\frac{60}{sin(21)}$

$x=60csc(21)$

This is the distance along the side road that is 60 feet from the line of lights.

The point along the lights falls between D and E.

You can find that point by using trig or even Pythagoras.

3. Hello, nathan02079!

Streelights A, B, C, D, and E are placed 50 m apart on the main road.
The light from a streetlight is effective up to a distance of 60 m.

(a) Find the distance from A to the farthest point on the side road that is illuminated.

(b) Determine the length of the side road that is illuminated by both lights C and D.
Code:
                                    F     *
o
*      \
*             \
*                    \
*  21°                      \
o--------o--------o--------o-------o
A   50   B   50   C   50   D   50  E
From $E$, draw $EF$ perpendicular to the side road.

In right triangle $EFA\!:\;\;\sin21^o \:=\:\frac{EF}{200} \quad\Rightarrow\quad EF \:=\:200\sin21^o \:\approx\;72$ m.

Hence, streetlight $E$ does not effectively illuminate the side road.

Code:
                                    G     *
o
*    /
*         /
*              / 60
*  21°              /
o-------o-------o--------o
A   50  B   50  C   50   D

Draw $DG = 60$ so that $\angle D$ is obtuse.

Law of Sines: . $\frac{\sin G}{150} \:=\:\frac{\sin21^o}{60} \quad\Rightarrow\quad \sin G \:=\:0.895919874 \quad\Rightarrow\quad G \:\approx\:63.6^o$

. . Then: . $\angle D \:=\:180^o - 21^o - 63.6^o \:=\:95.4^o$

Law of Sines: . $\frac{AG}{\sin95.4^o} \:=\:\frac{60}{\sin21^o} \quad\Longrightarrow\quad AG \;\approx\;167.4\text{ m}$ .(a)

4. Soroban has shown how to do part (a). For (b), use similar techniques to find the distance XY in this picture.

[Confession. My only real reason for contributing to this thread is that I have been trying to learn a new system for drawing geometric figures in TeX, and this seemed like a good picture to try it out on.]

5. Originally Posted by nathan02079
...

b. Determine the length of the side road that is effectively illuminated by both streetlights C and D.

...
See attachment.

6. Originally Posted by earboth
Originally Posted by nathan02079
...

b. Determine the length of the side road that is effectively illuminated by both streetlights C and D.

...
See attachment.
This is a question of interpretation of English. I take "illuminated by both streetlights C and D" to refer to the intersection of the areas of illumination of C and D, namely the stretch of road illuminated by both those lights. If the union of these areas was intended, the wording would have been "illuminated by either of the streetlights C and D".