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,
[tex]\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}[/tex]
Taking last two ratio,
[tex]\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}[/tex]
Putting the value of BC , AC ,∠A,∠B
[tex]\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}[/tex]
[tex]\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}[/tex]
≈734.106 miles
Therefore they are 734.106 miles apart.
