Originally Posted by
Nina7
1-Tas is 80kts,the wind correction angle is +19 and headwind component is 20kts.What is the ground speed ?
2-True course is 055,wind 310/25.variation 6W and aircraft is flying with a true airspeed of 113 kts.What is the magnetic course ?
3-What is the difference between the latitude of the point A and the point B, which are located on following parallels of latitude:
A:15dgr 54` 30"N
B:10dgr 33` 30"S
Thank you,Nina
I'll give you some direction and you can try it from there. I'm assuming you have some understanding of how to use a navigation calculator like an E6B.
1- You know the TAS value in the direction of the heading you're holding to maintain course (heading is 19 degrees more than the desired course, i.e. You're crabbing 19 degrees right). In order to answer the question, you need the value of your ground speed component in the direction of your course. You already know the wind component (headwind component) in the opposite direction along that course.
2- You need to know if the wind information is true or magnetic. If it's from Atis, tower or some other source while airborne, it's magnetic. Bottom line is make sure your course and wind information are both the same (either magnetic or true) to start out. Calculate your course using the winds given along with your true airspeed. If you did it with true course and winds, then you need to apply the variation to your answer (in this case, since you add West variation (subtract East) add 6 degrees of variation to get magnetic heading). If your winds were already magnetic, you could have applied the variation correction to your true course of 055 and gotten 061 magnetic course. Calculate the course that way and your answer is already your magnetic course.
3- If I understand the question correctly, they're just asking how far apart the two points are in latitude. In other words how many degrees in north/south distance separate them. North latitudes are measured as degrees above the equator and South latitudes are measured in degrees below the equator. So, if you can determine how far away point A is from a known reference line.....like the equator

and do the same for point B, you can add those two degree amounts to get the total answer.