A plane leaves the airport in galisteo and flies 170 km at 68 degrees east of north; then it changes direction to fly 230 km at 36 degrees south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?

Accepted Solution

Answer:4.68Β° south east 317.36 km Step-by-step explanation:We can find the angle between the two distances (vectors) because according to the diagram, we can draw two right triangles between them.The complement of the 36 degree angle is 54 (90-36=54), and the complement of the 68 angle is 22, (90-68=22) the sum of 22 and 54 is 76. So the angle between the two distances is 76.Then we apply the cosine law[tex]b^{2} =a^{2} +c^{2} -2*a*c*cosB\\ \\b^{2} =230^{2} +170^{2} -2*230*170*cos(76)\\\\b=\sqrt{230^{2} +170^{2} -2*230*170*cos(76)} \\\\b=317.36 km[/tex]then we apply the sin law[tex]\frac{sin(C)}{c} =\frac{sin(B)}{b} \\\\sin(C)=c*\frac{sin(B)}{b}\\\\sin(C)=170*\frac{sin(76)}{317.36}\\\\sin(C)=0.52\\\\arcsin(0.52)=C=31.32\\\\[/tex]and because in any triangle, the sum of the inside angles is equal to 180 [tex]180=76+31.32+68+y\\\\y=180-76-31.32-68\\\\y=4.68^{o}[/tex]180= 76+C+(68+Y)y=180-76-C-68So the emergency plane has to travel 317.36 km, 4.68Β° southeast.