Answer:
35.7 m
Explanation:
Let
[tex]\mid A\mid=18.5 m[/tex]
[tex]\mid B\mid=41 m[/tex]
We have to find the distance between Joe's and Karl'e tent.
[tex]A_x=Acos\theta[/tex]
[tex]A_y=Asin\theta[/tex]
Substitute the values then we get
[tex]A_x=18.5cos23^{\circ}=17 m[/tex]
[tex]A_y=18.5sin 23^{\circ}=7.2 m[/tex]
[tex]B_x=41cos37.5^{\circ}=32.5 m[/tex]
[tex]B_y=41sin37.5^{\circ}=-24.96 m[/tex]
Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.
By triangle addition of vector
[tex]B=A+C[/tex]
[tex]C=B-A[/tex]
[tex]C_x=B_x-A_x=32.5-17=15.5 m[/tex]
[tex]C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m[/tex]
[tex]\mid C\mid=\sqrt{C^2_x+C^2_y}[/tex]
[tex]\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m[/tex]
Hence, the distance between Joe's and Karl's tent=35.7 m