Hi all,

I was doing the problem attached, but cannot understand what is meant by E 70 S. Can someone please help me out. the Answer should be 13.57.

I also attached the sketch diagram.

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- November 23rd 2012, 08:41 AMaritechTrigonometry problem
Hi all,

I was doing the problem attached, but cannot understand what is meant by E 70 S. Can someone please help me out. the Answer should be 13.57.

I also attached the sketch diagram. - November 23rd 2012, 09:09 AMchiroRe: Trigonometry problem
Hey aritech.

For this problem E 70 S refers to a bearing which I think is east 70 south, or in translation a bearing of 70 degrees from either the south or east axis in either clock-wise or counter-clockwise direction.

However, the diagram doesn't indicate that either of the above are correct. - November 23rd 2012, 10:30 AMaritechRe: Trigonometry problem
Hi chiro, the diagram attached is sketched be me, so it can be incorrect. I searched on the net for bearings and i found the picture attached. The E 70 S is most probably referring to the angle between 90 and 60. But still, i don't know which is the angle of descent which will result to 13.57.

- November 23rd 2012, 12:49 PMaritechRe: Trigonometry problem
Any help please?

- November 23rd 2012, 12:56 PMskeeterRe: Trigonometry problem
http://mathhelpforum.com/attachments...-problem-1.jpg

first sighting ...

distance from the airport,

second sighting ...

distance from the airport,

distance traveled parallel to the earth's surface,

descent angle

speed

next time,**DON'T BUMP**. - November 23rd 2012, 02:07 PMaritechRe: Trigonometry problem
Thanks fro the reply

- November 24th 2012, 02:51 AMaritechRe: Trigonometry problem
Hi again, I cannot understand the working 100%, so I sketched a new diagram. Is the angle of descent the one marked with the green ?? and where in the diagram is E 70 S please?

Thanks in advance. - November 24th 2012, 07:02 AMskeeterRe: Trigonometry problem
the descent angle is relative to the horizontal. it is formed by the change in altitude, 8000 - 5500 = 2500 m, over the distance traveled parallel to the earth's surface, 10068 m.