Question

radar tracking station O is 40 km away from a ship at point A. The ship travels east 45 km to point C at 30km/h. How long will the ship be on the radar with a 20 km radius?

Answer 1

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

`(a)/(\sin A) = (b)/(\sin B)`

Thus,

`(|OC|)/(\sin \angle OAC) = (|OA|)/(\sin \angle OCB) \\ \\ (20)/(\sin21) = (40)/(\sin \angle OCB) \\ \\ \sin \angle OCB= (40\sin21)/(20) =0.7167 \\ \\ \angle OCB=\sin^(-1)(0.7167)=45.79^o`

Segment BC is a chord which makes angle 45.79 with the radius of the circle,
Using Pythagoras theorem, we have

`\cos45.79= (\left( (1)/(2) |BC|\right))/(|OC|)= (\left( (1)/(2) |BC|\right))/(20) \\ \\ \left( (1)/(2) |BC|\right)=20\cos45.79=13.95 \\ \\ |BC|=2(13.95)=27.89km`

Recall that time of a body travelling a distance, d, at v km/h is given by

`t= (d)/(v)`

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

`t= (27.89)/(30) =0.93 \ hours\approx56 \ minutes`