Hello, Coukapecker!
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.