# Trigonometry problem

• Nov 23rd 2012, 07:41 AM
aritech
Trigonometry 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.
• Nov 23rd 2012, 08:09 AM
chiro
Re: 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.
• Nov 23rd 2012, 09:30 AM
aritech
Re: 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.
• Nov 23rd 2012, 11:49 AM
aritech
Re: Trigonometry problem
• Nov 23rd 2012, 11:56 AM
skeeter
Re: Trigonometry problem
http://mathhelpforum.com/attachments...-problem-1.jpg

first sighting ...

distance from the airport, $d_1 = \frac{8000}{\tan(40)} \approx 9534 \, m$

second sighting ...

distance from the airport, $d_2 = \frac{5500}{\tan(35)} \approx 7855 \, m$

distance traveled parallel to the earth's surface, $r = \sqrt{9534^2 + 7855^2 - 2(9534)(7855)\cos(70)} \approx 10068 \, m$

descent angle $\theta = \arctan\left(\frac{8000-5500}{10068}\right) \approx 13.945^\circ = 13^\circ \, 57'$

speed $\approx 231 m/s = 829.9 \, km/hr$

next time, DON'T BUMP.
• Nov 23rd 2012, 01:07 PM
aritech
Re: Trigonometry problem
• Nov 24th 2012, 01:51 AM
aritech
Re: 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?

• Nov 24th 2012, 06:02 AM
skeeter
Re: 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.