Question

triangle abc is a sketch of a triangular flower bed that has an area of 63.1 square feet.
to the nearest foot, what amount of fencing is needed to surround the perimeter of the flower bed?
40 feet 17 feet
33 feet 50 feet
trigonometric area formula: area = \\( \frac { 1 } { 2 } a b \sin ( c ) \\)

Explanation:

Step1: Find the length of side \( b \)

Given the formula for the area of a triangle \( A=\frac{1}{2}ab\sin(C) \). Here \( a = 13\), \( C=50^{\circ}\), and \( A = 63.1\).
Substitute the values into the formula: \(63.1=\frac{1}{2}\times13\times b\times\sin(50^{\circ})\).
First, calculate \(\sin(50^{\circ})\approx0.766\).
The equation becomes \(63.1=\frac{1}{2}\times13\times b\times0.766\).
Simplify the right - hand side: \(\frac{1}{2}\times13\times0.766 = 4.979\).
So, \(63.1 = 4.979b\).
Solve for \( b\): \(b=\frac{63.1}{4.979}\approx12.7\).

Step2: Calculate the perimeter of the triangle

The perimeter \(P\) of triangle \(ABC\) is \(P=a + b+ c\), where \(a = 13\), \(b\approx12.7\), and \(c = 10\).
\(P=13 + 12.7+10=35.7\approx36\) (This part seems wrong. Let's re - check using the Law of Cosines.
Using the Law of Cosines \(b^{2}=a^{2}+c^{2}-2ac\cos(B)\). Wait, no, we should use the area formula correctly.
Wait, another way:
We know \(A=\frac{1}{2}ac\sin(B)\). But we were given \(A = 63.1\), \(a = 13\), \(c = 10\), \(C = 50^{\circ}\).
Using \(A=\frac{1}{2}ab\sin(C)\), \(63.1=\frac{1}{2}\times13\times b\times\sin(50^{\circ})\), \(b=\frac{2\times63.1}{13\times\sin(50^{\circ})}\), \(\sin(50^{\circ})\approx0.766\), \(b=\frac{126.2}{13\times0.766}=\frac{126.2}{9.958}\approx12.7\)
Using the Law of Cosines \(b^{2}=a^{2}+c^{2}-2ac\cos(C)\)
\(b^{2}=13^{2}+10^{2}-2\times13\times10\times\cos(50^{\circ})\)
\(b^{2}=169 + 100-260\times0.6428\)
\(b^{2}=269-167.128\)
\(b^{2}=101.872\), \(b\approx10.1\) (There was a miscalculation before. Let's start over with the correct formula)
The area formula \(A=\frac{1}{2}ab\sin(C)\), given \(A = 63.1\), \(a = 13\), \(C = 50^{\circ}\)
\(63.1=\frac{1}{2}\times13\times c\times\sin(50^{\circ})\) (Wait, no, sides: let \(a = 13\), \(c = 10\), then \(A=\frac{1}{2}ac\sin(B)\). But we are given \(C = 50^{\circ}\). Wait, the formula \(A=\frac{1}{2}ab\sin(C)\) where \(a\) and \(b\) are two sides and \(C\) is the included angle.
Let \(a = 13\), \(b\) be the unknown side adjacent to \(C = 50^{\circ}\), \(c = 10\)
\(A=\frac{1}{2}ab\sin(C)\), \(63.1=\frac{1}{2}\times10\times b\times\sin(50^{\circ})\)
\(\sin(50^{\circ})\approx0.766\)
\(63.1 = 5\times0.766\times b\)
\(b=\frac{63.1}{5\times0.766}=\frac{63.1}{3.83}\approx16.5\)
Now, using the Law of Cosines \(a^{2}=b^{2}+c^{2}-2bc\cos(A)\) (no, we need perimeter. The sides are \(a = 13\), \(b\approx16.5\), \(c = 10\)
Perimeter \(P=13 + 16.5+10=39.5\approx40\)

Answer:

\(40\) feet