Question

Barbados A cruise ship maintains a speed of 5 knots (nautical miles per hour) sailing from San Juan to Barbados, a distance of 600 nautical miles. To avoid a tropical storm, the captain heads out of San Juan at a direction of 14 off a direct heading to Barbados. The captain maintains the 5-knot speed for 5 hours, after which time the path to Barbados becomes clear of storms (a) Through what angle should the captain turn to head directly to Barbados? (b) Once the turn is made, how long will it be before the ship reaches Barbados if the same 5 knot speed is maintained? San Juan (a) The captain should head through an angle of (Do not round until the final answer. Then found to one decimal place as needed) (b) The time it will take for the ship to reach Barbados is hours. (Do not round until the final answer. Then round to one decimal place as needed.) Enter your answer in each of the answer boxes.

Answer 1

Answers

(a)The captain should head through an angle of x=165.4°.

(b)the time ship will take to reach Barbados is 115.2 hours.

Given:-

Speed of ship=5 knots(nautical miles per km)

Distance between Saun Juan and Barbados

= 600 nautical miles

To avoid a tropical storm captain move for 5 hr at 14° to the Saun Juan.

Solution-

Let us suppose the position of Saun Juan and Barbados are denoted by A and C respectively.

Captain heads the ship at point B after which storm is clear.

Distance travelled by ship in 5hr

=speed×time

=5×5=25 nautical miles

So, Distance AB=25 nautical miles

This forms a ∆ABC.Let the opposite sides of Angles A,B and C are a, b and c respectively.Let Captain has to head at an angle of x° at point B.

chBarbodox) 600=6 A csaun Juan

(As shown in figure below.)

In ∆ABC, we have

<A=14° ,b=600 nautical miles,c=25 nautical miles

Now, using the formula of cosine we will find the side a of the ∆ABC.

We know cosine formula is

$a=\sqrt{b^2+c^2-2(b)(c)\cos( A)}$

On putting the values,we get

a = V(600)2 + (25)2 – 2(600) (25) cos(14)

$a=\sqrt{360000+625-(30000)(0.970295726)}$

$a=\sqrt{360600-29,108.8718}$

$a=\sqrt{331,516.128}$

$a=575.774372$

So, BC=a= 575.774372 nautical miles

(a)Let us first find angle B to find angle (because <x=180°- <B)

Now, let us use sine formula to find angle B of the triangle ABC.

We know that sine formula is

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

On putting the available values,we get

$\frac{575.774372}{sin(14^{\circ})}=\frac{600}{sinB}=\frac{c}{sinC}$

$\frac{575.774372}{(0.241921896)}=\frac{600}{sinB}=\frac{c}{sinC}$

$2,380.00107=\frac{600}{sinB}=\frac{c}{sinC}$

This gives

$2,380.00107=\frac{600}{sinB}$

On cross multiplication,we get

$2,380.00107sinB=600$

$sinB=\frac{600}{2,380.00107}$

$sinB=0.252100727$

This gives B =14.6018573°

Now,

< x =180° –<B

<x=180°-14.6018573°

<x=165.398143°

<x ≈ 165.4°

Hence, the captain should head through an angle of x=165.4°.

(b)

Time taken from point B to reach point C(Barbados)

= distance BC(a)÷speed of ship

(Since speed of ship=5 knots and

a=575.774372 nautical miles.)

=575.774372/5

=115.154874

≈ 115.2 hours

Hence, the time ship will take to reach Barbados is 115.2 hours.