Question

A cruise ship travels 310 miles due east before turning 20 degrees north of east. it travels 150 miles its new course. how far is the cruise ship from its initial position

A)295 miles
B)274 miles
C)454miles
D)160 miles

Answer 1

d^2=x^2+y^2

d^2=(310+150cos20)^2+(150sin20)^2

d^2=205991.41373309

d=453.86mi

So C. to the nearest mile.

Answer 2

454 miles.

Explanation:

Refer the attached figure.

A cruise ship travels 310 miles due east i.e. AB = 310 miles

Now a cruise turns 20 degrees north of east .i.e.∠CBE = 20°

Using linear pair :Sum of angles = 180°

∠CBA+∠CBE=180°

∠CBA=180°-20° =160°

It travels 150 miles its new course i.e. BC= 150 miles.

Now we are supposed to find How far is the cruise ship from its initial position

Now using cosine rule:
`c = √(a^2+b^2-2abcos\theta)`

a =BC = 150 miles

b= Ab = 310 miles

c = AC

`\theta = 160^(\circ)`

Substitute the values.

`c = \sqrt{150^2+310^2-2* 150 * 310 cos160^(\circ)}`

`c = 453.8 \sim 454`

Thus the cruise ship is 454 miles from its initial position.