# Thread: Question on 3-d vectors

1. ## Question on 3-d vectors

The CN Tower is 553 m tall. A person is south of the tower, in a boat on Lake Ontario at a position that makes an angle of elevation of 10 degrees to the top of the tower. A second person is sitting on a park bench, on a bearing of 060 degrees relative to the tower, observing the top of the tower at an angle of elevation of 9 degrees. Determine the displacement between the two people. Could somebody help me visualize the situation because I thought that the angle between the two people would be 150 degrees rather than 120 degrees.

2. Originally Posted by dalbir4444
The CN Tower is 553 m tall. A person is south of the tower, in a boat on Lake Ontario at a position that makes an angle of elevation of 10 degrees to the top of the tower. A second person is sitting on a park bench, on a bearing of 060 degrees relative to the tower, observing the top of the tower at an angle of elevation of 9 degrees. Determine the displacement between the two people. Could somebody help me visualize the situation because I thought that the angle between the two people would be 150 degrees rather than 120 degrees.
Use the tan-function to calculate the blue and darkgreen line from boat and from the bench respectively.

Afterwards use the Cosine rule to calculate the distance between the boat and the bench.

3. Hello, dalbir4444!

The CN Tower is 553 m tall.
A person is south of the tower at a position that makes
an angle of elevation of 10° to the top of the tower.
A second person is on a bearing of 060 degrees relative to the tower,
observing the top of the tower at an angle of elevation of 9°.

Determine the displacement between the two people.

Looking down at the ground, we have this diagram:
Code:
      N       o B
:     */
:60°* /
: *y /
T o   /
|  /
x | /
|/
A o

The tower is at T.
The first observer is at A; the second is at B.
Let $x = AT,\;y = BT.$
$\angle NTB = 60^o \quad\Rightarrow\quad \angle BTA = 120^o$

Law of Cosines: . $AB^2 \;=\;x^2 + y^2 - 2xy\cos120^o$ .[1]

The observer at A has this right triangle:
Code:
                  *
*  |
*     | 553
* 10°    |
* - - - - - *
x

We have: . $\tan10^o \:=\:\frac{553}{x} \quad\Rightarrow\quad x \:=\:\frac{553}{\tan10^o}$ .[2]

The observer at B has this right triangle:
Code:
      *
|  *
553 |     *
|     9° *
* - - - - - *
y

We have: . $\tan9^o \:=\:\frac{553}{y} \quad\Rightarrow\quad y \:=\:\frac{553}{\tan9^o}$ .[3]

Substitute [2] and [3] into [1]:

. . $AB^2 \;=\;\left(\frac{553}{\tan10^o}\right)^2 + \left(\frac{553}{\tan9^o}\right)^2 - 2\left(\frac{553}{\tan10^o}\right)\left(\frac{553} {\tan9^o}\right)\cos120^o$

Go for it!