Question

A boat leaves Port A and travels due east. Port B is 15 mi due north of Port A, and Port C is 24 mi due south of Port A. The boat is 25 mi from Port C. How far is the boat from Port B? Round to the nearest tenth of a mile. ____mi

Answer 1

With this problem, we have two right triangles meaning we will end up using the Pythagorean Theorem. For the sake of the problem, let's say the boat is at point D. The two triangles we have are ΔACD and ΔABD. Let's first go through and define the lines.

Let's start with ΔACD:
Port A is 24 miles from Port C so AC = 24mi. We will represent this as line a
Port C is 25 miles from Boat D so CD (the hypotenuse) = 25mi. We will represent this as line c
We do not know the distance between Port A and Boat D (AD also known as line b) so let's calculate that since we will need it at the end.

We can calculate line b's length using the theorem: a² + b² = c²
a = 24mi
b = ?
c = 25mi

So

(24)² + b² = (25)²
576 + b² = 625
Subtract 576 from both sides to isolate b²

576 + b² - 576 = 625 - 576
b² = 49
Now take the square root of both sides

√b² = √49
b = 7 miles.
Now we know that the distance between Port A and Boat D is 7 miles.

Now let's move on to ΔABD
Port B is 15 miles from Port A so AB = 15mi. We will represent this as line a
Boat D is 7 miles from Port A so AD = 7mi. This will still be line b
We do not know the distance between Port B and Boat D (BD also known as line c).

Time to use the Pythagorean Theorem once more!
a² + b² = c²
15² + 7² = c²
225 + 49 = c²
274 = c²
Take the square root of both sides

√274 = √c²
c ≈ 16.6mi when we round to the nearest tenth

So the distance between Port B and the boat is 16.6 miles