# Trig application question -- Help

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• Feb 4th 2010, 04:21 PM
Coukapecker
Trig application question -- Help
From a position 110km northwest of a coast guard station, an oil tanker makes radio
contact with the coast guard. The tanker is traveling due south at 25km/h. The radar
unit at the coast guard station has a range of 90km. For what length of time can the
coast guard expect the tanker to be visible on the radar screen, to the nearest tenth
of an hour?
Answer(Suppose to be, anyway): 3.6h

Can someone please explain how to do this problem correctly? i worked it out before but i keep getting 4.5h as the answer. I cant think of any alternate ways of tackling it and I keep getting 4.5h, how do you solve this problem?

any help greatly appricated,

~Coukapecker~
• Feb 4th 2010, 05:34 PM
Soroban
Hello, Coukapecker!

Quote:

From a position 110 km northwest of a coast guard station,
an oil tanker makes radio contact with the coast guard.
The tanker is traveling due south at 25 km/hr.
The radar unit at the coast guard station has a range of 90 km.
For what length of time can the coast guard expect the tanker
to be visible on the radar screen, to the nearest tenth of an hour?
Answer: 3.6 hrs

Code:

    T *       : *     _ :  *  110  55√2 :    *       :      *       :    45° *     B * - - -_- - * S       :  55√2       :
The tanker is at $\displaystyle T$, sailing toward $\displaystyle B$ at 25 km/hr.
The station is at $\displaystyle S.$
Since $\displaystyle \angle S = 45^o$, then: .$\displaystyle TB\,=\,BS\,=\,\frac{110}{\sqrt{2}} \:=\:55\sqrt{2}$

Code:

    T *       |       |       |     A *       |  *       |    *  90       |        *       |          *       |            θ *     B * - - - - -_- - - * S               55/2
Some time later, the tanker has moved to point $\displaystyle A,$
. . where $\displaystyle AS \,=\,90$ km.

Let $\displaystyle \theta = \angle ASB$

Then: .$\displaystyle \cos\theta \:=\:\frac{55\sqrt{2}}{90} \:=\:0.864241621 \quad\Rightarrow\quad \theta \:\approx\:30.2^o$

And: .$\displaystyle \sin30.2^o \,=\,\frac{AB}{90}\quad\Rightarrow\quad AB \:=\:90\sin30.2^o \:\approx\;45.27$ km.

The tanker will be in range from 45.27 km north of $\displaystyle B$ to 45.27 km south of $\displaystyle B.$

That is. It will be in range for 90.54 km.

At 25 km/hr, it will be in range for: .$\displaystyle \frac{90.54}{25} \:\approx\:3.6$ hours.

• Feb 4th 2010, 05:46 PM
Archie Meade
Quote:

Originally Posted by Coukapecker
From a position 110km northwest of a coast guard station, an oil tanker makes radio
contact with the coast guard. The tanker is traveling due south at 25km/h. The radar
unit at the coast guard station has a range of 90km. For what length of time can the
coast guard expect the tanker to be visible on the radar screen, to the nearest tenth
of an hour?
Answer(Suppose to be, anyway): 3.6h

Can someone please explain how to do this problem correctly? i worked it out before but i keep getting 4.5h as the answer. I cant think of any alternate ways of tackling it and I keep getting 4.5h, how do you solve this problem?

any help greatly appricated,

~Coukapecker~

Using Pythagoras' theorem, since the tanker is 45 degrees above and to the left of the line of latitude of the station...

$\displaystyle x^2+x^2=110^2$

$\displaystyle 2x^2=110^2$

$\displaystyle x=\frac{110}{\sqrt{2}}$

$\displaystyle x^2+y^2=90^2$

$\displaystyle y=\sqrt{90^2-\frac{110^2}{2}}=45.28$

$\displaystyle 2y='90km\ range\ of\ station'=90.56km$

The tanker travels at 25km/hr, so the time interval the tanker remains within range is

$\displaystyle \frac{90.56}{25}=3.62\ hours$