Expert: Josh - 10/3/2005

Question
Hello, I received a question on my test and I did not know how to approach the question. My teacher would like us to re-do the questions that we got wrong. Here's the one that I need help in:

City A is 30 miles directly north of City C, and City B is 40 miles directly east of City C. Cities A and B are connected by a straight road. Find the length of the shortest path from City C to the road that connects A and B.

Hint: The shortest path, segment CD, is the length of the perpendicular from C to segment AB. First, find AB, AD, and BD.



I would really appreciate if you will be able to take out the time to help me. Thank you in advance!

Answer
Hi,

No problem. Let's draw a diagram so that we are both on the same page.

A
|
|
|
|------D
|
|
|
C---------------B

Figure is not to scale. CD is suppose to be perpendicular to AB. So, you can label angle ^ADC and ^BDC as 90 degrees.

Also, since town A is north of C, town B is east of C, the angle ^ACB is also 90 degrees (again, mark this in the diagram).

Now, using Pythagoras theorem, which states that the hypotenuse (the longest side directly opposite to the right angle, |AB|) is the square root of the sum of the squared distance of the two adjacent sides, viz., |AC|^2+|CB|^2, we have

|AB|= sqrt(|AC|^2+|CB|^2) ...[#1]

N.B. |AC| represents the length of AC, |AC|^2 =|AC|*|AC| etc.

Distance |AB|= sqrt(3^2+4^2) = sqrt(9+16)= sqrt(25) = 5 miles

Part (ii) Finding |CD|,|AD| and |BD|

OBSERVATION: Triangle ACD and BCD are similar to ABC.
If we label angle ^ABC as "y", then, angle ^BAC must be "90-y", since the sum of internal angles in a triangle must add up to 180. [Check that ^ACB+^ABC+^BAC=180 degrees]

So, it follows that ^BCD must be "90-y" degrees. How do we know this? As we've said, triangle BCD is similar to triangle ABC. Since, we already know that ^BDC=90, and ^CBD=y, and ^CBD+^BDC+^DCB=180,

^DCB=180-(^CBD+^BDC)=180-(y+90)=90-y.

Now, to make things really clear, I want you to draw triangles ABC and BCD as two separate diagrams. It's important that you trace these triangles onto another piece of paper from the original, such that triangle BCD has the right proportions in relation to triangle ABC.

Label the following angles,
In triangle ABC: ^ACB=90, ^CBA=y, ^BAC=90-y
In triangle BCD: ^BCD=90, ^CBD=y, ^BCD=90-y.

Question: why do we refer to ABC and BCD as similar triangles?
Answer: they satisfy the following property.

The ratio subtended by two corresponding sides are the same.
e.g., |AC|/|AB| = |CD|/|CB| = sin(y)
Here is another one,
     |BC|/|AB| = |BD|/|CB| = cos(y) = cos(90-y) ...[#2]

From [#2], we have |BD|=|CB|*cos(y), but
cos(y) is just the ratio |BC|/|AB|, both these lengths are known.

Thus, |BD|=|CB|*(|BC|/|AB|)=40*40/50=32

Using Pythagoras,
|CD|=sqrt(|BC|^2-|BD|^2)
   =sqrt(40*40-32*32)
   =24

Cheers.