trigonometry with triangle

Hello, NIKITA!

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

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2) The distance $OB$

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

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3) The angle of elevation of $T$ from $B$ to the nearest degree.

In $\Delta TOB\!:\;\;\tan B \:=\:\frac{38}{125} \:=\:0.304$

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

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

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5) The distance $OC$

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

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

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

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