a plane flies from A to B , a distance of 120 kilometres on a bearing of 035 degrees. he then changes direction and flies to a further distance of 200 kilometres in a bearing of i25 degrees. find the distance AC to the nearest kilometre

Answer 1

I will assume the 2nd direction is 125°

The angle between the two directions is 90°
so AC = √(120^2 + 200^2)
= 233.238 km

or

Suppose we did not see that 90° ,we could simply use vectors

AC = (120cos35,120sin35) + (200cos125,200sin125)
= (98.298.., 68.829..) + (-114.715.. , 163.830..)
= (--16.417.. , 232.659..)

whose magnitude = √( (-16.417)^2 + 232.959^2)
= appr 233.238 km