# Thread: Some Trig Wordproblem Guidance

1. ## Some Trig Wordproblem Guidance

Ok I am completely stuck on one and I would like a second opinion on the other two.
1. A major League baseball diamond square 90 feet on a side. The pitching rubber is located 60.5 feet from home plate on a line joining home plate and second base. A.) how far is it from the pitching rubber to first base? B.) If a pitcher faces home plate, through what angle does he need to turn to face first base?
I'm stuck because I only have two of the distances but no angles. I tried setting up a drawing but was still stuck I guess the angles of the square would be 90° and then each triangle would have 45° angles?

2. Coast Guard Station Zulu is located 120 miles due west of Station X-ray. A ship at sea sends an SOS call that is received by each station. The call to Station Zulu indicates that the bearing of the ship from Zulu is N40°E the call to Station X-Ray indicates that the bearing of the ship from X-ray is N30°W.
A.) How far is each station from the ship? B.) If a helicopter capable of flying 200 mph is dispatched from the nearest station to the ship, how long will it take to reach the ship?

I got 120/sin70°=a/sin50° a~97.8mi and 120sin60°/sin70°=110.6mi so the angles of the triangle would be 50°, 60° and 70°? and the sides would be 120mi 97.8mi and 110.6mi? Part b) would be 97.8(60)/200~29.34min or ~29min 20 seconds?

3.) At a county fair truck pull, two pickup trucks are attached to the back end of a monster truck. One of the pickups pulls with a force of 2000 pounds and the other pulls with a force of 3000 pounds with an angle of 45° between them. With how much force must the monster truck pull in order to remain unmoved? (It will be in the opposite direction of the resultant force and with the same magnitude.)
I got $\displaystyle |V3|^2= 3000^2+2000^2-2(2000)(3000)cos45°$ and that is ~2124 pounds?

Also on a side note, can anyone tell me a site where you can draw in the angles and make like graphs (sorry for not really knowing what I'm taking about)

2. Originally Posted by jasonk
Ok I am completely stuck on one and I would like a second opinion on the other two.
1. A major League baseball diamond square 90 feet on a side. The pitching rubber is located 60.5 feet from home plate on a line joining home plate and second base. A.) how far is it from the pitching rubber to first base? B.) If a pitcher faces home plate, through what angle does he need to turn to face first base?
I'm stuck because I only have two of the distances but no angles. I tried setting up a drawing but was still stuck I guess the angles of the square would be 90° and then each triangle would have 45° angles?

2. Coast Guard Station Zulu is located 120 miles due west of Station X-ray. A ship at sea sends an SOS call that is received by each station. The call to Station Zulu indicates that the bearing of the ship from Zulu is N40°E the call to Station X-Ray indicates that the bearing of the ship from X-ray is N30°W.
A.) How far is each station from the ship? B.) If a helicopter capable of flying 200 mph is dispatched from the nearest station to the ship, how long will it take to reach the ship?

I got 120/sin70°=a/sin50° a~97.8mi and 120sin60°/sin70°=110.6mi so the angles of the triangle would be 50°, 60° and 70°? and the sides would be 120mi 97.8mi and 110.6mi? Part b) would be 97.8(60)/200~29.34min or ~29min 20 seconds?

3.) At a county fair truck pull, two pickup trucks are attached to the back end of a monster truck. One of the pickups pulls with a force of 2000 pounds and the other pulls with a force of 3000 pounds with an angle of 45° between them. With how much force must the monster truck pull in order to remain unmoved? (It will be in the opposite direction of the resultant force and with the same magnitude.)
I got $\displaystyle |V3|^2= 3000^2+2000^2-2(2000)(3000)cos45°$ and that is ~2124 pounds?

Also on a side note, can anyone tell me a site where you can draw in the angles and make like graphs (sorry for not really knowing what I'm taking about)

Ok, for #1, first I have to say that I know nothing about baseball, but I assume that the pitching rubber is what the pitcher stands on and that it is of equal distance from each base, which would make it the center of the square.
If the pitching rubber is 60.5 feet from home base, then it would be 60.5 feet from each of the other bases as well. If a straight line runs from home base to second base, going through the pitching rubber, then from what we know about a square, this diagonal line would create two right, isoceles triangles. Drawing another line from 1st base to 3rd base would create 4 isoceles, right triangles. The pitcher would have to rotate 90 degrees to face 1st base

For #2, I think you may not be drawing you picture correctly (of course I may be misinterpreting the word "bearing") I drew mine with Zulu on the left bottom (it is 120 mi. due WEST of Xray) and then Xray on the right bottom. This is the bottom of the triangle. Then the bearings of the ship are given in terms of its location FROM each station, so the angle at Zulu would be 40° and the angle at Xray would be 30°, making the angle at the ship 110°
Using Law of Sines: $\displaystyle \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}$

$\displaystyle \frac{120}{sin(110)} = \frac{b}{sin(30)}$

etc.

3. Hello, jasonk!

1. A major League baseball diamond square 90 feet on a side.
The pitching rubber is located 60.5 feet from home plate.

(A) How far is it from the pitching rubber to first base?

(B) If a pitcher faces home plate, through what angle does he need to turn to face first base?
Did you make a sketch?
Code:
                *
*   *
*       *
*           *
*               *
*                   *
*                       *
*                           *
*                               o F
*             P       *     *
*           o  θ        *
*         |         *
*   60.5|       * 90
*     |     *
*   |45°*
* | *
o
H
Draw line segment $\displaystyle PF.$

The pitcher is at $\displaystyle P.$
Home plate is at $\displaystyle H\!:\;PH = 60.5$
First base is at $\displaystyle F\!:\;HF = 90$
Let $\displaystyle \angle FPH = \theta$
Note that $\displaystyle \angle PHF = 45^o.$

(A) Law of Cosines: .$\displaystyle PF^2 \:=\:60.5^2 + 90^2 - 2(60.5)(90)\cos45^o \:=\:4059.857183$

. . .Therefore: .$\displaystyle PF \:\approx\:63.7$ ft

(B) Law of Cosines: .$\displaystyle \cos\theta \:=\:\frac{60.5^2 + 63.7^2 - 90^2}{2(60.5)(63.7)} \:=\:-0.049568613$

. . .Therefore: .$\displaystyle \theta \;\approx\;92.8^o$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

And I completely agree with your reasoning and work on #2.

Good work!

4. can anyone help with #3

5. NVM I got #2.