# 3D Trig. Probs

• Feb 15th 2008, 09:31 PM
Fibonacci
3D Trig. Probs
g&t homework, pleasse help draw the diagrams first
thanks.

1. the pilot of an aircraft flying due South, at an altitude of 750m, sees an airfield directly ahead, angle of depression 17°. at the same time the pilot notices a town due east, angle of depression 10°. how far is town from airfield?

2.from the top of a cliff 120m above sea level a boat is observed due North, angle of depression 20°. the boat travels due east and, 10min later, its angle of depression from the top of the cliff is 10°. how fast is the boat moving?

3. a river runs east to west. a tree stands on the edge of one bank and from a point C, on the opposite bank and due south of the tree, the angle of elevation of the top of the tree is 28°. from point D, 65m due east of C the angle of elevation of the top of the tree is 20°
calculate a) height of tree
b)width of river at C.
• Feb 16th 2008, 02:46 AM
Soroban
Hello, Fibonacci!!

Quote:

1. The pilot of an aircraft flying due South, at an altitude of 750m,
sees an airfield directly ahead, angle of depression 17°.
At the same time the pilot notices a town due east, angle of depression 10°.
How far is town from airfield?

Once you "see" the situation, make separate diagrams.

Code:

    P * - - - - - - - - - - - - S       |  * 17°       |      *   750 |          *       |              *       |              17° *     Q * - - - - - - - - - - - * A             750/tan17°
The pilot is at $\displaystyle P$ at an altitude of 750 m: .$\displaystyle PQ = 750$
The airfield is at $\displaystyle A.\;\;\angle SPA \:=\:\angle PAQ \:=\:17^o$
$\displaystyle \tan17^o \:=\:\frac{750}{AQ}\quad\Rightarrow\quad AQ \:=\:\frac{750}{\tan17^o}$

Code:

    P * - - - - - - - - - - - - E       |  * 10°       |      *   750 |          *       |              *       |              10° *     Q * - - - - - - - - - - - * T             750/tan10°
The pilot is at P: .$\displaystyle PQ = 750$
The town is at $\displaystyle T.\;\;\;\angle EPT = \angle PTQ = 10^o$
$\displaystyle \tan10^o \:=\:\frac{750}{QT}\quad\Rightarrow\quad QT \:=\:\frac{750}{\tan10^o}$

Looking straight down at the ground, we have this diagram:
Code:

                750/tan10°           P * - - - - - - - - * T             |              *             |          *  750/tan17° |        *             |    *             |  *           A *

I assume you can find the distance $\displaystyle AT$ now.