91Ó°ÊÓ

A cruise ship maintains an average speed of $15$knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of $600$nautical miles.

The captain maintains the $15$- knot speed for $10$hours, so the distance covered is $150$miles. Let -

Now, $a^2=b^2+c^2-2bc\cos A$

$=22500+360000-2×150×600\cos 20°$

localid="1646922354456" $a≈461.9$

We have to find angle B -

localid="1646921671452" $\frac{\sin A}{a}=\frac{\sin B}{b}$ use Law of Sine

localid="1646921685369" $\frac{\sin 20°}{461.9}≈\frac{\sin B}{150}$

localid="1646921702873" ${\sin }^{-1}\frac{150\sin 20°}{461.9}≈B$

$6.4°≈B$

We know, $20°+6.4°+C=180°$

$C=153.6°$

The angle the captain should turn to head directly to Barbados is -

$180°-153.6°=26.4°$

A cruise ship maintains an average speed of $15$knots in going from San Juan, Puerto Rico, to Barbados, West Indies, a distance of $600$nautical miles.

We have,

The distance after turning the ship to reach Barbados is$461.9$miles and $15$- knot speed is maintained. So,

Timerole="math" localid="1646922797759" $≈\frac{461.9}{15}≈30.8$hours