Question

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?

Answer 1

Answer:

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,

$\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Taking last two ratio,

$\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}$

Putting the value of BC , AC ,∠A,∠B

$\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}$

$\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}$

         ≈734.106 miles

Therefore they are 734.106 miles apart.