Algebra 2 question! Please answer!

Question

Starting at the same point in a forest, two hikers take different paths. The first hiker walks due north at a speed of 3 miles per hour. The second hiker walks 60 degrees west of north at a speed of 4 miles per hour.

1.How far apart are the hikers after 1 hour?

2.The two hikers carry walkie-talkies with a range of 8 miles. After how much time are the two hikers out of range of each other?

3.Supposed after two hours the first hiker stops and tells the second hiker to meet her. How long will it take the second hiker to meet the first hiker? In what direction should the second hiker walk?

Answer 1

I drew the triangles to indicate that the hikers started at point A, and they are at points B and C after 1 hour. Take a look at the diagram:

 

https://drive.google.com/open?id=0BwfqpqJa4_wiQnU0bldUR3U2cVU&authuser=0

 

1. You can find the distance between them (length a) by using the rule of cosines

a² = b² + c² -2bcCosA

Plug in the values of b, c and angle A

a² = 4*4 + 3*3 -2*4*3*Cos60

a² = 16+9-12

a² =13

a = √13 = 3.61 miles

 

2. Let's say after t hours, hikers are 8 miles apart.

In t hours, first hiker will be at a distance = 3*t miles from point A => c=3t

In t hours, second hiker will be at a distance = 4*t miles from point A => b=4t

And the distance between them, a = 8 miles

Again, we apply the rule of cosines:

a² = b² + c² -2bcCosA

64 = 16t² + 9t² -2(4t)(3t)Cos60

64 = 25t² - 12t²

64 = 13t²

t = 8/√13 = 2.22 hours

 

3. Take a look at the figure on right hand side in the link, I posted at the beginning.

Going by the same logic as described above.

In 2 hours, first hiker would be c=6 miles from point A

In 2 hours, second hiker would be b=8 miles from point A

We again apply the rule of cosines to find the distance between the hikers

a² = b² + c² -2bcCosA

a² = 64+36-2(8)(6)Cos60

a² = 100-48

a² = 52

a = √52 = 7.21 miles

Since second hiker walks at 4 miles/hour, (s)he will take 7.21/4 = 1.8 hours to meet the first hiker

 

For finding the direction in which the second hiker should walk, it's angle DCB east of north (see figure), let's say this angle is θ degrees

Notice, that angle DCB = angle ABC

We can again apply the rule of cosines again to find angle ABC (same as angle B)

b² = a²+c²-2acCosB

64 = 52 + 36-2(√52)(6)CosB

2(√52)(6)CosB = 24

CosB = 0.277

B = 73.9 degrees, east of north

Answer 2

I’ve updated my answer:

 

Either the second hiker is walking at 5 miles per hour (which is running in a forest), or the person who created the problem drew the triangle upside down.  You draw it !

 

To be 30,60,90 and have sides of 3,4,5, with one side horizontal (90 degrees west of north), the second hiker must travel at 5 mph  (the hypotenuse) (at 60 degrees west of north).

Cool!  This problem has what I call “magic numbers” (numbers that make the calculations very simpl;e so that you can concentrate on what you are learning).

North is “Up” (did you see the movie?)

60% west of north just creates a 60% angle.

Now, I’d like to suggest an error in the problem statement.

 

The second hiker might be walking at 5 miles per hour.  Check it out!

 

My quick answer was:

 

The other side of the triangle is the side between the two walkers.
 
In just one hour the first hiker has travelled 3 miles (3 miles per hour, right?) and the second hiker has travelled 4 miles (actually, that’s pretty fast walking in a forest).  So, how far apart are they at the end of the first hour?  Use the Pythagorean Theorem, because the line between them will create a right triangle (with a horizontal line)  (because that’s what 3 and 4 and 60degrees does (but other values don’t) – so, remember that 3,4,5 triangles have angles of 30, 60,90 degrees.

 

Reread the problem, be sure to answer each part. Note in part (3) that the first hiker STOPS, so the second hiker has to travel that entire distance.

Go for it!