Question

a dog walks 320 ft east, then turns 81° north of west and walks 180 ft. how far is the dog from its start?
? ft
180 ft
81°
320 ft
round to the nearest hundredth.

Explanation:

Step1: Identify the triangle sides and angle

We have a triangle with two sides: \( a = 320 \) ft (east), \( b = 180 \) ft (north of west), and the included angle \( C = 180^\circ - 81^\circ = 99^\circ \) (since the angle between east and north of west is supplementary to \( 81^\circ \)).

Step2: Apply the Law of Cosines

The Law of Cosines states that for a triangle with sides \( a \), \( b \), and \( c \), and included angle \( C \), \( c^2 = a^2 + b^2 - 2ab \cos(C) \).
Substitute \( a = 320 \), \( b = 180 \), and \( C = 99^\circ \):

$$ c^2 = 320^2 + 180^2 - 2 \times 320 \times 180 \times \cos(99^\circ) $$

Calculate each term:

• \( 320^2 = 102400 \)
• \( 180^2 = 32400 \)
• \( 2 \times 320 \times 180 = 115200 \)
• \( \cos(99^\circ) \approx -0.1564 \) (using calculator)

Substitute \( \cos(99^\circ) \):

$$ c^2 = 102400 + 32400 - 115200 \times (-0.1564) $$

$$ c^2 = 134800 + 115200 \times 0.1564 $$

$$ c^2 = 134800 + 17997.28 $$

$$ c^2 = 152797.28 $$

Step3: Solve for \( c \)

Take the square root of \( c^2 \):

$$ c = \sqrt{152797.28} \approx 390.89 $$

Answer:

\( 390.89 \)