Question

Two ships leave a port at the same time.
Ship A sails 12 knots on a bearing of 035°
Ship B sails 16 knots on a bearing of 270°
Calculate the distance between the ships after 2 hours

(1 knot = 1 nautical mile per hour)

Answer 1

Answer:

49.8 nautical miles

Step-by-step explanation:

Recall that speed = distance/time

Time = 2hours

Speed = 12knots and 16 knots respectively

D = speed×time

D1 = 12×2 = 24

D2 = 16×2 = 32

Using the 'cosine rule' we have:

a^2 = b^2+c^2-2bc cos Θ

Where a =?

b =24

c = 32

Θ = 125°

a² = 24² + 32² - 2(24)(32)cos125°

a^2 = 576+1024 - 1536cos125°

a² = 1600 - 1536(-0.57357)

a² = 1600+881.0134

a² = 2481.0134

Then, a² = 2481.013406

a =√2481.013406

Hence, a = 49.8 nautical miles

Answer 2

In this exercise we must use the knowledge about triangles to calculate the distance that a ship will travel, in this way we find that:

49.8 nautical miles

First, remember the formula for distance, which is:

$Speed = distance/time$

And knowing that the data reported in the exercise are:

• Time = 2hours

• Speed = 12 knots and 16 knots respectively

So putting the values ​​informed in the distance formula, we have:

$D = speed*time\\D_1 = 12*2 = 24\\D_2 = 16*2 = 32$

Using the 'cosine rule' we have:

$a^2 = b^2+c^2-2bc cos \theta$

Find the a, will have:

$b =24 \ \ \ c = 32 \ \ \ \theta = 125\\a^2 = 24^2 + 32^2 - 2(24)(32)cos125\\a^2 = 576+1024 - 1536cos125\\a^2 = 1600 - 1536(-0.57357)\\a^2 = 1600+881.0134\\a^2 = 2481.0134\\a=49.8$

See more about triangles at brainly.com/question/25813512