Question

1. triangular garden plot

a triangular garden has sides of 18 m and 25 m with an included angle of 42°.
a) use the law of cosines to find the length of the third side.
b) find the area of the garden.

Explanation:

Part (a)

Step1: Recall Law of Cosines

For a triangle with sides \(a\), \(b\), included angle \(C\), the third side \(c\) is given by \(c^{2}=a^{2}+b^{2}-2ab\cos C\). Let \(a = 18\), \(b = 25\), \(C=42^{\circ}\).

Step2: Substitute values into formula

\(c^{2}=18^{2}+25^{2}-2\times18\times25\times\cos(42^{\circ})\)
Calculate \(18^{2}=324\), \(25^{2}=625\), \(2\times18\times25 = 900\), \(\cos(42^{\circ})\approx0.7431\)
\(c^{2}=324 + 625-900\times0.7431\)
\(c^{2}=949 - 668.79\)
\(c^{2}=280.21\)

Step3: Solve for \(c\)

\(c=\sqrt{280.21}\approx16.74\) m

Part (b)

Step1: Recall area formula for triangle

The area \(A\) of a triangle with two sides \(a\), \(b\) and included angle \(C\) is \(A=\frac{1}{2}ab\sin C\).

Step2: Substitute values

\(a = 18\), \(b = 25\), \(C = 42^{\circ}\), \(\sin(42^{\circ})\approx0.6691\)
\(A=\frac{1}{2}\times18\times25\times0.6691\)
\(A = 9\times25\times0.6691\)
\(A=225\times0.6691\approx150.55\) \(m^{2}\)

Part (a) Answer:

The length of the third side is approximately \(\boldsymbol{16.74}\) meters.

Part (b) Answer:

The area of the garden is approximately \(\boldsymbol{150.55}\) square meters.

Answer:

Step1: Recall area formula for triangle

The area \(A\) of a triangle with two sides \(a\), \(b\) and included angle \(C\) is \(A=\frac{1}{2}ab\sin C\).

Step2: Substitute values

\(a = 18\), \(b = 25\), \(C = 42^{\circ}\), \(\sin(42^{\circ})\approx0.6691\)
\(A=\frac{1}{2}\times18\times25\times0.6691\)
\(A = 9\times25\times0.6691\)
\(A=225\times0.6691\approx150.55\) \(m^{2}\)

Part (a) Answer:

The length of the third side is approximately \(\boldsymbol{16.74}\) meters.

Part (b) Answer:

The area of the garden is approximately \(\boldsymbol{150.55}\) square meters.