Explore a diverse range of topics and get answers from knowledgeable individuals on IDNLearn.com. Join our community to receive prompt and reliable responses to your questions from knowledgeable professionals.

A rescue helicopter flew from its home base for 35 kilometers on a course of 330 degrees to pick up an accident victim. It then flew 25 kilometers on the course of 90 degrees to the hospital. what distance and on what course will the helicopter fly to return directly to its home base?

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)








View image Ollieboyne
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.