1. ## trigonometry with triangle

The pt T is the top of a vertical lighthouse.Boats are moored at pts A,B andC.The foot O of the lighthouse A,B and C are all at sea level.The boat A is 85m from the ptO and the angle of elevation of T from A is 24degress .The boats A and B are due east of lighthouse and the distance AB is 40m.The boat atC is due south of the boat at A and the angle of elevation of T from C is 18 degrees.
Caculate
1)The distance OT in m to the nearest m,
2)Thje distance OB
3)The angle of elevation of the pt T from the boat at Bin degrees to the nearest degree.
Calculate also in m to the nearest metre
4)the distance CT
5)The distance OC
6)The distance AC
7)Calculate the bearing of the boat at B from the boat at C in degrees to the nearest degree.

2. Hello, NIKITA!

Did you make some sketches?

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