From the given figure, segment OA = 40km, segment AC = 45km and segment OC = 20km
Angle OAC = 21 degrees.
Now, angle OBC = angle OCB because they are the angle formed at the intersection of the radius and the chord of a circle.
Using, sine rule, we find angle OCB.
Recall that sine rule states that
[tex] \frac{a}{\sin A} = \frac{b}{\sin B} [/tex]
Thus,
[tex] \frac{|OC|}{\sin \angle OAC} = \frac{|OA|}{\sin \angle OCB} \\ \\ \frac{20}{\sin21} = \frac{40}{\sin \angle OCB} \\ \\ \sin \angle OCB= \frac{40\sin21}{20} =0.7167 \\ \\ \angle OCB=\sin^{-1}(0.7167)=45.79^o[/tex]
Segment BC is a chord which makes angle 45.79 with the radius of the circle,
Using Pythagoras theorem, we have
[tex]\cos45.79= \frac{\left( \frac{1}{2} |BC|\right)}{|OC|}= \frac{\left( \frac{1}{2} |BC|\right)}{20} \\ \\ \left( \frac{1}{2} |BC|\right)=20\cos45.79=13.95 \\ \\ |BC|=2(13.95)=27.89km[/tex]
Recall that time of a body travelling a distance, d, at v km/h is given by
[tex]t= \frac{d}{v} [/tex]
Given that the ship travels at 30km/h, the time the ship will take on the radar covelling a distance of 27.89km is given by
[tex]t= \frac{27.89}{30} =0.93 \ hours\approx56 \ minutes[/tex]
