Hello, NIKITA!

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The point $\displaystyle T$ is the top of a vertical lighthouse.

Boats are moored at points $\displaystyle A, B\text{ and }C.$

The foot $\displaystyle O$ of the lighthouse, and $\displaystyle A,B,C$ are all at sea level.

Boat $\displaystyle A$ is 85m from $\displaystyle O$; the angle of elevation of $\displaystyle T$ from $\displaystyle A$ is 24°.

Boats $\displaystyle A\text{ and }B$ are due east of lighthouse and distance $\displaystyle AB$ is 40m.

Boat $\displaystyle C$ is due south of boat $\displaystyle A$; the angle of elevation of $\displaystyle T$ from $\displaystyle C$ is 18°.

Caculate:

1) The distance $\displaystyle OT$ to the nearest meter. Code:

T *
| * *
| * *
| * *
| * *
| 24° * *
* - - - - - * - - - - - *
O 85 A 40 B

In right triangle $\displaystyle TOA\!:\;\;\tan24^o \:=\:\frac{OT}{85} \quad\Rightarrow\quad OT \:=\:85\tan24^o$

Therefore: .$\displaystyle OT \:=\:37.844... \:\approx\: \boxed{38\text{ m}}$

2) The distance $\displaystyle OB$ $\displaystyle OB \;=\;OA + AB \;=\;85 + 40 \;=\;\boxed{125\text{ m}}$

3) The angle of elevation of $\displaystyle T$ from $\displaystyle B$ to the nearest degree. In $\displaystyle \Delta TOB\!:\;\;\tan B \:=\:\frac{38}{125} \:=\:0.304$

Therefore: .$\displaystyle B \:=\:\tan^{-1}(0.304) \:=\:16.909... \:\approx\:\boxed{17^o}$

4) The distance $\displaystyle CT$ Code:

T *
| *
| *
38 | *
| *
| 18° *
* - - - - - - - - - - - *
O C

In $\displaystyle \Delta TOC\!:\;\;\sin18^o \:=\:\frac{38}{CT}\quad\Rightarrow\quad CT \:=\:\frac{38}{\sin18^o}$

Therefore: .$\displaystyle CT \:=\:122.970... \:\approx\:\boxed{123}\text{ m}$

5) The distance $\displaystyle OC$

In $\displaystyle \Delta TOC\!:\;\;\tan18^o \:=\:\frac{38}{OC} \quad\Rightarrow\quad OC \:=\:\frac{38}{\tan18^o}$

Therefore: .$\displaystyle OC \:=\:116.951... \:\approx\:\boxed{117\text{ m}}$

6) The distance $\displaystyle AC$ Looking straight down at the ground, we have: Code:

85
O * - - - - - - - - * A
* |
* |
117 * |
* |
* |
* C

From Pythagorus: .$\displaystyle AC^2 + 85^2 \:=\:117^2$

Therefore: .$\displaystyle AC \:=\:\sqrt{117^2-85^2} \:=\:80.399... \:\approx\:\boxed{80\text{ m}}$

7) Calculate the bearing of $\displaystyle B$ from $\displaystyle C$, nearest degree. Again, looking straight down at the ground . . .

Code:

40
A * - - - - - - - - * B
| *
| *
80 | *
| θ *
| *
*
C

We have: .$\displaystyle \tan\theta \:=\:\frac{40}{80}\quad\Rightarrow\quad \theta \:=\:\tan^{-1}(0.5)$

Therefore: .$\displaystyle \theta \:=\:26.565... \:\approx\:\boxed{27^o}$