Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 miles how far apart are they?

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships  is  135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC =  402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40'      [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

Taking last two ratio,

Putting the value of BC , AC ,∠A,∠B

         ≈734.106 miles

Therefore they are 734.106 miles apart.

is there any photos or whatever?

493 would be the correct answer

step-by-step explanation: