# trigonometry with triangle

• May 19th 2008, 10:35 AM
NIKITA
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.
• May 19th 2008, 12:45 PM
Soroban
Hello, NIKITA!

Did you make some sketches?

Quote:

The point $T$ is the top of a vertical lighthouse.
Boats are moored at points $A, B\text{ and }C.$
The foot $O$ of the lighthouse, and $A,B,C$ are all at sea level.

Boat $A$ is 85m from $O$; the angle of elevation of $T$ from $A$ is 24°.
Boats $A\text{ and }B$ are due east of lighthouse and distance $AB$ is 40m.
Boat $C$ is due south of boat $A$; the angle of elevation of $T$ from $C$ is 18°.

Caculate:

1) The distance $OT$ to the nearest meter.

Code:

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

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

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

Quote:

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

Quote:

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

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

Quote:

4) The distance $CT$
Code:

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

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

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

Quote:

5) The distance $OC$

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

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

Quote:

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

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

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

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

Quote:

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

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

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

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