Question

1. a triangular piece of land is bounded by 135 ft of fencing on one side, 145 ft of stone wall on another side, and 245 ft of road frontage on the third side. what are the interior angles formed by the boundary lines?

Explanation:

We have a triangle with sides \( a = 135 \) ft, \( b = 145 \) ft, and \( c = 245 \) ft. We can use the Law of Cosines to find the angles. The Law of Cosines states that for a triangle with sides \( a \), \( b \), \( c \) opposite angles \( A \), \( B \), \( C \) respectively, \( c^{2}=a^{2}+b^{2}-2ab\cos(C) \), \( a^{2}=b^{2}+c^{2}-2bc\cos(A) \), and \( b^{2}=a^{2}+c^{2}-2ac\cos(B) \)

Step 1: Find angle \( C \) (opposite side \( c = 245 \))

We use the formula \( \cos(C)=\frac{a^{2}+b^{2}-c^{2}}{2ab} \)
Substitute \( a = 135 \), \( b = 145 \), \( c = 245 \)
\( a^{2}=135^{2} = 18225\), \( b^{2}=145^{2}=21025\), \( c^{2}=245^{2} = 60025\)
\( \cos(C)=\frac{18225 + 21025-60025}{2\times135\times145}=\frac{39250 - 60025}{38850}=\frac{- 20775}{38850}\approx - 0.5347\)
\( C=\arccos(- 0.5347)\approx122.3^{\circ}\)

Step 2: Find angle \( A \) (opposite side \( a = 135 \))

We use the formula \( \cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc} \)
Substitute \( a = 135 \), \( b = 145 \), \( c = 245 \)
\( \cos(A)=\frac{21025+60025 - 18225}{2\times145\times245}=\frac{62825}{70550}\approx0.8905\)
\( A=\arccos(0.8905)\approx27.1^{\circ}\)

Step 3: Find angle \( B \) (opposite side \( b = 145 \))

We know that the sum of angles in a triangle is \( 180^{\circ} \), so \( B=180^{\circ}-A - C\)
\( B = 180-(27.1 + 122.3)=180 - 149.4 = 30.6^{\circ}\)

Answer:

The interior angles are approximately \( 27.1^{\circ} \), \( 30.6^{\circ} \) and \( 122.3^{\circ} \)