Question

two paths ascend separate hills from the same point. one is 18 m long, the other 25 m. a 30 m bridge connects their endpoints. what is the angle between the paths at the base? base angle = ?° round your answer to the nearest hundredth.

Explanation:

Step1: Identify the triangle sides

We have a triangle with sides \( a = 18 \, \text{m} \), \( b = 25 \, \text{m} \), and \( c = 30 \, \text{m} \). We need to find the angle \( \theta \) between sides \( a \) and \( b \).

Step2: Apply the Law of Cosines

The Law of Cosines states that \( c^2 = a^2 + b^2 - 2ab \cos \theta \). Rearranging for \( \cos \theta \):

$$ \cos \theta = \frac{a^2 + b^2 - c^2}{2ab} $$

Substitute \( a = 18 \), \( b = 25 \), \( c = 30 \):

$$ \cos \theta = \frac{18^2 + 25^2 - 30^2}{2 \times 18 \times 25} $$

Calculate numerator: \( 324 + 625 - 900 = 49 \)
Denominator: \( 2 \times 18 \times 25 = 900 \)
So, \( \cos \theta = \frac{49}{900} \approx 0.0544 \)

Step3: Find the angle

Take the arccosine: \( \theta = \arccos(0.0544) \approx 87.00^\circ \) (rounded to nearest hundredth)

Answer:

\( 87.00 \)