Finding distances to the frisbee

[kingsala]

Any help would help people:)

George and his two friends Mark and Kevin are playing frisbee on the beach. Mark is standing at a point M, 22 metres due south of Kevin, who is standing at point K. George is standing at a point G, 48◦ west of north of Mark and 25 metres from point M. (You may assume that all distances are flat and in a straight line.)

(a) Draw a diagram showing the points M, K and G, marking in the angle and the lengths that you are given.

(b) Find the distance between George and Kevin, that is, the length of the side GK.

(c) When George throws the frisbee, the wind catches it so that it lands on the ground between Mark and Kevin (on the line MK) due east of George (G). Add a line to your diagram that shows the shortest distance from George’s position G to where the frisbee lands at point F (on the line MK). What angle does GF make with MK?

(d) Find the distance that George would have to walk to retrieve the frisbee, i.e. the length of side GF.

(e) As Kevin is nearer, he decides to retrieve the frisbee instead. Find the distance that Kevin must walk to get it.

 

[soroban]

Hello, kingsala!

I'll get you started . . .

George, Mark and Kevin are playing Frisbee on the beach.
Mark (M) is standing 22 metres due south of Kevin (K).

George (G) is standing at a point 48^o west of north of M
and 25 metres from M.
(Assume that all distances are flat and in a straight line.)

(a) Draw a diagram showing the points M, K and G,
\(\displaystyle \quad\)labeling the given angle and lengths.

                  o K
              *   *
          *       *
    G o           *
        *         * 22
          *       *
         25 *     *
              *48[SUP]o[/SUP]*
                * *
                  o
                  M

(b) Find the distance between George and Kevin,
that is, the length of the side GK.

Law of Cosines:

\(\displaystyle \;\;\;GK^2 \;=\;25^2 + 22^2 - 2(25)(22)\cos48^o\)

(c) When George throws the Frisbee, the wind catches it so
that it lands on the ground between M and K due east of G.
Draw a line that shows the shortest distance
from G to where the Frisbee lands at point F.
What angle does GF make with MK?

                  K
                  o
              *   *
          *       *
    G o - - - - - o F
        *         *
          *       *
         25 *     *
              *48[SUP]o[/SUP]*
                * *
                  o
                  M

(d) Find the distance that George would have to walk
to retrieve the Frisbee, i.e. the length of side GF.

\(\displaystyle \sin48^o \;=\;\frac{GF}{25}\)

(e) As Kevin is nearer, he decides to retrieve the Frisbee
instead.\(\displaystyle \;\) Find the distance that Kevin must walk to get it.

Can you solve this one on your own?

 

[Poppypod]

Hi! I'm new here and would like to practice my trigonometry skills. I've stumbled across this question and I'm stuck on the last part. I've calculated GK to be 19m and GF to be 19m as well (which I'm not sure is right). I've also calculated angle GKF to be 74 degrees but I'm not sure if any of this is correct. Any help you could give would be greatly appreciated!

Thanks!

 

[MarkFL]

Hi! I'm new here and would like to practice my trigonometry skills. I've stumbled across this question and I'm stuck on the last part. I've calculated GK to be 19m and GF to be 19m as well (which I'm not sure is right). I've also calculated angle GKF to be 74 degrees but I'm not sure if any of this is correct. Any help you could give would be greatly appreciated!

Thanks!

According to what soroban posted, we find:

\(\displaystyle \overline{GK}=\sqrt{1109-1100\cos\left(48^{\circ}\right)}\approx19.312\)

\(\displaystyle \overline{GF}=25\sin\left(48^{\circ}\right)\approx18.579\)

To find $\angle GKF$, we can use:

\(\displaystyle \sin(\angle GKF)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\overline{GF}}{\overline{GK}}=\frac{25\sin\left(48^{\circ}\right)}{\sqrt{1109-1100\cos\left(48^{\circ}\right)}}\)

Hence:

\(\displaystyle \angle GKF=\arcsin\left(\frac{25\sin\left(48^{\circ}\right)}{\sqrt{1109-1100\cos\left(48^{\circ}\right)}}\right)\approx74.159^{\circ}\)