Sagot :
I've drawn a triangle to represent the journey (see attached.
The angle from Home to Victim to Hospital is:
a = [tex]90 - (360 - 330) = 90- 30 = 60^o[/tex]
Using the cos rule, we can solve for the length x:
[tex]a^2 = b^2 +c^2 - 2bccosA \\x^2 = 35^2+25^2-2(35)(25)cos60 \\x^2 = 1850 - 875 \\x^2 = 975 \\x = \sqrt{975} \\x = 31.2249...[/tex]
Using the cos rule, we can solve for angle b:
[tex]a^2 = b^2 +c^2 - 2bccosA \\35^2 = 31.22...^2+25^2-2(31.22...)(25)cosb \\1225 = 975 + 625 - 1561.249...cosb \\-375 = -1561.249...cosb \\b = cos^{-1}( \frac{-375}{-1561.249...}) \\b = 76.1...[/tex]
So the bearing the helicopter has to travel is:
270 - 76.1 = 193.9 degrees (1 dp)
And the distance it travels is 31.2 km (1 dp)
The angle from Home to Victim to Hospital is:
a = [tex]90 - (360 - 330) = 90- 30 = 60^o[/tex]
Using the cos rule, we can solve for the length x:
[tex]a^2 = b^2 +c^2 - 2bccosA \\x^2 = 35^2+25^2-2(35)(25)cos60 \\x^2 = 1850 - 875 \\x^2 = 975 \\x = \sqrt{975} \\x = 31.2249...[/tex]
Using the cos rule, we can solve for angle b:
[tex]a^2 = b^2 +c^2 - 2bccosA \\35^2 = 31.22...^2+25^2-2(31.22...)(25)cosb \\1225 = 975 + 625 - 1561.249...cosb \\-375 = -1561.249...cosb \\b = cos^{-1}( \frac{-375}{-1561.249...}) \\b = 76.1...[/tex]
So the bearing the helicopter has to travel is:
270 - 76.1 = 193.9 degrees (1 dp)
And the distance it travels is 31.2 km (1 dp)