Question

applications of the cosine law

1. two forest fires were reported on the same day in northern british columbia to the wildfire management branch in prince george. the first fire was reported in stewart (320 miles from prince george) and the second in dease lake (419 miles from prince george). an at-802 water bomber is dispatched from prince george to stewart then to dease lake. determine whether or not the water bomber can return safely to base on a single tank if the angle between the direct paths to stewart and dease lake is 22.3°. (note: the at-802 water bomber has a range of 800 miles on a full tank.)

Explanation:

Step1: Identify the triangle sides and angle

Let \( a = 419 \) (distance from Prince George to Dease Lake), \( b = 320 \) (distance from Prince George to Stewart), and \( C = 22.3^\circ \) (the included angle). We need to find the length of the side \( c \) (distance from Stewart to Dease Lake) using the Law of Cosines: \( c^2 = a^2 + b^2 - 2ab\cos(C) \).

Step2: Calculate \( c^2 \)

First, compute \( a^2 = 419^2 = 175561 \), \( b^2 = 320^2 = 102400 \), and \( 2ab\cos(C) = 2\times419\times320\times\cos(22.3^\circ) \). Calculate \( \cos(22.3^\circ) \approx 0.925 \). Then \( 2\times419\times320\times0.925 = 2\times419\times296 = 838\times296 = 248048 \). Now, \( c^2 = 175561 + 102400 - 248048 = 277961 - 248048 = 29913 \).

Step3: Find \( c \)

Take the square root of \( c^2 \): \( c = \sqrt{29913} \approx 172.95 \) miles.

Step4: Calculate total distance

The total distance the water bomber travels is \( 320 + 419 + 172.95 = 911.95 \) miles? Wait, no. Wait, the path is Prince George -> Stewart -> Dease Lake -> Prince George. Wait, no: Prince George to Stewart (320), Stewart to Dease Lake (c), Dease Lake to Prince George (419). So total distance is \( 320 + c + 419 \). Wait, no, I made a mistake earlier. Wait, the Law of Cosines: the triangle is Prince George (P), Stewart (S), Dease Lake (D). So sides: PS = 320, PD = 419, angle at P is \( 22.3^\circ \). So SD is the side we found as \( c \approx 172.95 \). Then the total distance is PS + SD + DP = 320 + 172.95 + 419 = 911.95? Wait, but the range is 800. Wait, no, maybe I messed up the path. Wait, the water bomber is dispatched from Prince George to Stewart (320), then to Dease Lake (SD), then return to Prince George (419). So total distance is 320 + SD + 419. Wait, but let's recalculate SD correctly.

Wait, Law of Cosines: \( c^2 = a^2 + b^2 - 2ab\cos(C) \), where \( a = 419 \), \( b = 320 \), \( C = 22.3^\circ \). So:

\( c^2 = 419^2 + 320^2 - 2\times419\times320\times\cos(22.3^\circ) \)

Calculate \( 419^2 = 175561 \), \( 320^2 = 102400 \), sum is 277961.

\( 2\times419\times320 = 267520 \)

\( \cos(22.3^\circ) \approx \cos(22^\circ18') \approx 0.9254 \) (more accurate: using calculator, \( \cos(22.3^\circ) \approx 0.92537 \))

So \( 267520\times0.92537 \approx 267520\times0.925 = 267520\times0.9 + 267520\times0.025 = 240768 + 6688 = 247456 \). More accurately, 267520*0.92537 ≈ 267520*0.92537 ≈ 247,500 (exact calculation: 267520 * 0.92537 = 267520 * 0.9 + 267520 * 0.02 + 267520 * 0.005 + 267520 * 0.00037 = 240768 + 5350.4 + 1337.6 + 98.9824 = 240768 + 5350.4 = 246118.4 + 1337.6 = 247456 + 98.9824 = 247554.9824)

So \( c^2 = 277961 - 247554.9824 = 30406.0176 \)

Then \( c = \sqrt{30406.0176} \approx 174.37 \) miles.

Now total distance: 320 (PG to S) + 174.37 (S to D) + 419 (D to PG) = 320 + 174.37 = 494.37 + 419 = 913.37 miles.

Wait, but the range is 800 miles. Wait, that can't be. Wait, maybe I misread the problem. Wait, the problem says "dispatched from Prince George to Stewart then to Dease Lake. Determine whether or not the water bomber can return safely to base on a single tank". So the path is PG -> S -> D -> PG. So total distance is PG-S + S-D + D-PG.

But let's check the Law of Cosines again. Wait, maybe the angle is between the two paths from PG, so triangle P-S-D, with PS=320, PD=419, angle at P is 22.3 degrees. So SD is the side opposite angle P? Wait, no: Law of Cosines is \( c^2 = a^2 + b^2 - 2ab\cos(C) \), where C is the angle between sides a and b. So if angle at P is 22.3 degrees, then sides adjacent to P are PS=320 and PD=419, so the side opposite P is SD, so \( SD^2 = PS^2 + PD^2 - 2\times PS\times PD\times\cos(\angle P) \). Yes, that's correct. So SD ≈ 174.37 miles. Then total distance is 320 + 174.37 + 419 = 913.37 miles, which is more than 800. Wait, but maybe I made a mistake in the angle? Wait, no, the problem says "the angle between the direct paths to Stewart and Dease Lake is 22.3°", so that's angle at PG, so that's correct.

Wait, but let's recalculate SD with more precise cosine. Let's use a calculator for \( \cos(22.3^\circ) \). 22.3 degrees: cos(22.3) ≈ cos(22 + 0.3) = cos(22°18') ≈ using calculator, cos(22.3) ≈ 0.92537.

So \( SD^2 = 320^2 + 419^2 - 2\times320\times419\times\cos(22.3^\circ) \)

320² = 102400, 419² = 175561, sum is 277961.

2×320×419 = 267520.

267520×0.92537 ≈ 267520×0.92537 ≈ let's compute 267520×0.92537:

267520 × 0.9 = 240768

267520 × 0.02 = 5350.4

267520 × 0.005 = 1337.6

267520 × 0.00037 = 267520 × 0.0003 + 267520 × 0.00007 = 80.256 + 18.7264 = 98.9824

Adding these: 240768 + 5350.4 = 246118.4 + 1337.6 = 247456 + 98.9824 = 247554.9824

So SD² = 277961 - 247554.9824 = 30406.0176

SD = √30406.0176 ≈ 174.37 miles.

Total distance: 320 + 174.37 + 419 = 913.37 miles. The range is 800 miles. So 913.37 > 800, so the water bomber cannot return safely.

Wait, but maybe I messed up the path. Wait, maybe the water bomber goes from PG to S (320), S to D (SD), then D to PG (419). So total distance is 320 + SD + 419. If SD is approximately 174, then total is 320 + 174 + 419 = 913, which is more than 800. So the answer is no, it cannot return safely.

Answer:

The water bomber cannot return safely to base on a single tank because the total distance traveled (approximately 913.37 miles) exceeds its range of 800 miles.