Two ships leave the same port in different directions, forming a 120° angle between them. One ship travels 70 mi. and the other 52 mi. before they each drop their anchors. What is the distance between the ships to the nearest mile?

Question

(1 point)
Responses

87 mi.
87 mi.

98 mi.
98 mi.

106 mi.
106 mi.

122 mi.
122 mi.

Answer 1

To find the distance between the two ships, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides $a$, $b$, and $c$, and the angle $C$ opposite side $c$:

$$c^2=a^2+b^2-2ab\cdot \cos (C)$$

In this case, let:

• $a=70$ miles (distance traveled by the first ship)
• $b=52$ miles (distance traveled by the second ship)
• $C={120}^{\circ }$ (angle between the two ships)

First, we need to compute $\cos ({120}^{\circ })$: $$\cos ({120}^{\circ })=-\frac{1}{2}$$

Now, plug the values into the Law of Cosines formula: $$c^2={70}^2+{52}^2-2\cdot 70\cdot 52\cdot (-\frac{1}{2})$$

Calculate each term: $${70}^2=4900$$ $${52}^2=2704$$ $$-2\cdot 70\cdot 52\cdot (-\frac{1}{2})=70\cdot 52=3640$$

Now substitute back into the equation: $$c^2=4900+2704+3640$$ $$c^2=4900+2704=7604$$ $$c^2=7604+3640=11244$$ $$c=\sqrt{11244}\approx 106$$

Thus, the distance between the two ships is approximately $106$ miles.

The answer is: 106 mi.