\fAnswer : Consider the given diagram: A A B B x x C v C D D v vr At noon, ship reaches its closes approach so C 90 . Now in triangle ABC 90 45 180 180 135 45 Thus triangle ABC is an isosceles triangle (two angles are equal). Let the equal side AC BC x Hence using Pythagorean thorem in triangle ABC x 2 x 2 1.52 2 x 2 1.52 1.52 2 1.5 x 2 x2 Now in next 90 minutes or 1.5 hours distance traveled is 1.5 12 18 Hence the side length BD 18 1.5 19.06 2 Now using cosine rule in triangle ABD AD 1.52 19.062 2 1.5 19.06 cos 45 18.031 Using sine rule, sin sin 45 1.5 18.031 1.5sin 45 sin 18.031 1.5sin 45 sin 1 18.031 3.372 Since we know angular velocity v sin AD 12sin 3.372 18.031 0.039 rad/hr v , we have AD Second method: Using Sine rule, x sin 1.5 sin 135 dx d d cos sin 135 cos 135 sin dt dt dt 1.5 sin 2 135 dx d sin 135 cos cos 135 sin dt dt 1.5 sin 2 135 d 135 sin 12 dt 1.5 sin 2 135 2 d 8sin 135 dt sin135 When 131.628 (since 3.372 and one angle is 45 ), we have 2 d 8sin 135 131.628 dt sin135 0.039 rad/hr Answer : Consider the given diagram: A A B B x x C C D D v v vr At noon, ship reaches its closes approach so C 90 . Now in triangle ABC 90 45 180 180 135 45 Thus triangle ABC is an isosceles triangle (two angles are equal). Let the equal side AC BC x Hence using Pythagorean thorem in triangle ABC x 2 x 2 1.52 2 x 2 1.52 1.52 2 1.5 x 2 x2 Now in next 90 minutes or 1.5 hours distance traveled is 1.512 18 Hence the side length BD 18 1.5 19.06 2 Now using cosine rule in triangle ABD AD 1.52 19.062 2 1.519.06 cos 45 18.031 Using sine rule, sin sin 45 1.5 18.031 1.5sin 45 sin 18.031 1.5sin 45 sin 1 18.031 3.372 Since we know angular velocity v sin AD 12sin 3.372 18.031 0.039 rad/hr v , we have AD Second method: Using Sine rule, x sin 1.5 sin 135 dx d d cos sin 135 cos 135 sin dt dt dt 1.5 sin 2 135 dx d sin 135 cos cos 135 sin dt dt 1.5 sin 2 135 d sin 135 12 dt 1.5 sin 2 135 2 d 8sin 135 dt sin135 When 131.628 (since 3.372 and one angle is 45 ), we have 2 d 8sin 135 131.628 dt sin135 0.039 rad/hr