# Applied trig problem

• Dec 16th 2009, 07:06 AM
clayson
Applied trig problem

An aircraft is sighted due west from a radar station at an elevation of 50 degrees at a height of 8000m and later at an elevation of 39 degrees and a height of 6000m on a bearing of 160 degrees. if it is descending uniformly find the angle of descent and the speed of the aircraft in km/h given that the time between the two observations is 1 minute.

i never knew whether to put this in the pre calculus section or trig section i hope i chose the right one.

thanks.
• Dec 19th 2009, 05:02 AM
Quote:

Originally Posted by clayson

An aircraft is sighted due west from a radar station at an elevation of 50 degrees at a height of 8000m and later at an elevation of 39 degrees and a height of 6000m on a bearing of 160 degrees. if it is descending uniformly find the angle of descent and the speed of the aircraft in km/h given that the time between the two observations is 1 minute.

i never knew whether to put this in the pre calculus section or trig section i hope i chose the right one.

thanks.

HI

I have attached a diagram , a terrible one .. sorry bout that .

The positions of the aircraft are A and E for the first and second observations respectively . By finding AE , then you can find its speed .

Note that $\angle COD is 110^o$ .

$\tan 50 =\frac{8000}{CO}\Rightarrow CO=\frac{8000}{\tan 50}$

Similarly , $DO=\frac{6000}{\tan 39}$

Now , use the cosine rule ,

$CD^2=(\frac{8000}{\tan 50})^2+(\frac{6000}{\tan 39})^2-2(\frac{8000}{\tan 50})(\frac{6000}{\tan 39})\cos 110$

$CD\approx 11575m$

Then $CD//BE$so $CD=BE=11575$

$AB=8000-6000=2000$

Phythagoras theorem , $AE\approx 11747m$

Using $s=vt$ , $11747=v(60)$ , v=196m/s

Convert this to km/h

You have got most of the information here , surely you can get the angle of descent .