Question

gustavo and aiden are walking due east through the forest when an angry grizzly bear appears behind them. gustavo runs away to the east for 300 m, and aiden runs away in a more southward line for 250 m. gustavo and aiden end up 262 m apart, as shown. how many degrees away from east did aiden turn before running? do not round during your calculations. round your final answer to the nearest degree.

Explanation:

Step1: Identify the triangle sides

We have a triangle with sides: \( a = 250 \) m (Bear to Aiden), \( b = 300 \) m (Bear to Gustavo), \( c = 262 \) m (Gustavo to Aiden). We need to find the angle at the Bear between the east direction (Bear - Gustavo) and the direction Bear - Aiden. Let's call this angle \( \theta \).

Step2: Apply the Law of Cosines

The Law of Cosines formula is \( c^2 = a^2 + b^2 - 2ab \cos\theta \). We can rearrange it to solve for \( \cos\theta \):
\[ \cos\theta = \frac{a^2 + b^2 - c^2}{2ab} \]
Substitute \( a = 250 \), \( b = 300 \), \( c = 262 \):
\[ \cos\theta = \frac{250^2 + 300^2 - 262^2}{2 \times 250 \times 300} \]
Calculate the numerator: \( 250^2 = 62500 \), \( 300^2 = 90000 \), \( 262^2 = 68644 \). So numerator is \( 62500 + 90000 - 68644 = 83856 \). Denominator is \( 2 \times 250 \times 300 = 150000 \). Then \( \cos\theta = \frac{83856}{150000} = 0.55904 \).

Step3: Find the angle \( \theta \)

Take the arccosine of \( 0.55904 \):
\[ \theta = \arccos(0.55904) \]
Using a calculator, \( \theta \approx 56.0^\circ \) (rounded to nearest degree). Wait, no, wait. Wait, actually, the angle we found is the angle between Bear - Gustavo (east) and Bear - Aiden. But let's check again. Wait, maybe I mixed up the sides. Wait, the triangle: Bear to Gustavo is 300 (east), Bear to Aiden is 250 (some angle south of east), and Gustavo to Aiden is 262. So the sides: Bear-Gustavo (b=300), Bear-Aiden (a=250), Gustavo-Aiden (c=262). So Law of Cosines: \( c^2 = a^2 + b^2 - 2ab \cos\theta \), where \( \theta \) is the angle at Bear between a and b. So solving for \( \theta \):

\( \cos\theta = (a^2 + b^2 - c^2)/(2ab) = (250² + 300² - 262²)/(2*250*300) \)

250²=62500, 300²=90000, 262²=68644. Sum of 250² and 300²: 62500+90000=152500. Subtract 262²: 152500-68644=83856. Then divide by (2*250*300)=150000. So 83856/150000=0.55904. Then arccos(0.55904) is approximately 56 degrees? Wait, no, arccos(0.559) is about 56 degrees? Wait, cos(56°) is about 0.559, yes. Wait, but let's check with calculator: arccos(0.55904) ≈ 56 degrees (since cos(56) ≈ 0.5592, which is close). So the angle Aiden turned from east is this angle, because the Bear's east direction is Bear-Gustavo, and Bear-Aiden is the direction Aiden ran from the Bear's initial position (which was behind them, so the initial direction was east, so Aiden turned south from east by angle \( \theta \). Wait, no, the Bear is behind them, so initially, both are walking east. So the Bear's position is west of them. Wait, maybe the diagram is: Bear is at a point, Gustavo runs east 300m, Aiden runs from the Bear's position (which is west of them) in a southward line 250m, and then Gustavo and Aiden are 262m apart. Wait, maybe the triangle is: Bear (B), Gustavo (G), Aiden (A). So BG is 300m east, BA is 250m (some angle south of east? Wait, no, the diagram shows Bear at the top left, Gustavo at top right (east), Aiden at bottom. So the angle at B (Bear) between BG (east) and BA (the line from Bear to Aiden) is the angle we need. So using Law of Cosines, we found that angle is approximately 56 degrees. Wait, but let's confirm. Alternatively, maybe using Law of Sines, but Law of Cosines is more direct here.

Wait, let's recalculate the numerator: 250² is 62500, 300² is 90000, sum is 152500. 262² is 262*262: 260²=67600, 2*260*2=1040, 2²=4, so (260+2)²=260²+2*260*2+2²=67600+1040+4=68644. So 152500-68644=83856. Denominator: 2*250*300=150000. 83856/150000=0.55904. Then arccos(0.55904) is equal to, let's use calculator: arccos(0.55904) ≈ 56 degrees (since cos(56°) ≈ 0.5592, which is very close). So the angle is approximately 56 degrees.

Answer:

56