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 $\displaystyle x = AT,\;y = BT.$

$\displaystyle \angle NTB = 60^o \quad\Rightarrow\quad \angle BTA = 120^o$

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

The observer at A has this right triangle: Code:

*
* |
* | 553
* 10° |
* - - - - - *
x

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

The observer at B has this right triangle: Code:

*
| *
553 | *
| 9° *
* - - - - - *
y

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

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

. . $\displaystyle 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!*