Question

1. determine the lengths of the three unknown sides of the tract of land shown in figure below. (figure shows quadrilateral (abcd) with (ad = 500) ft, (angle a = 125^circ), (angle adb = 25^circ), (angle cdb = 55^circ), (angle c = 50^circ), and dashed line (bd) dividing (abcd) into (\triangle abd) and (\triangle bcd))

Explanation:

Step1: Analyze Triangle ABD

In triangle \(ABD\), we know \(\angle A = 125^\circ\), \(\angle ADB = 25^\circ\), so \(\angle ABD=180^\circ - 125^\circ - 25^\circ = 30^\circ\). Using the Law of Sines: \(\frac{AB}{\sin 25^\circ}=\frac{BD}{\sin 125^\circ}=\frac{AD}{sin 30^\circ}\). Given \(AD = 500\) ft. So \(\frac{AB}{\sin 25^\circ}=\frac{500}{\sin 30^\circ}\), \(AB=\frac{500\sin 25^\circ}{\sin 30^\circ}\approx\frac{500\times0.4226}{0.5}\approx422.6\) ft. And \(\frac{BD}{\sin 125^\circ}=\frac{500}{\sin 30^\circ}\), \(BD=\frac{500\sin 125^\circ}{\sin 30^\circ}\approx\frac{500\times0.8192}{0.5}\approx819.2\) ft.

Step2: Analyze Triangle BCD

In triangle \(BCD\), \(\angle C = 50^\circ\), \(\angle CDB = 55^\circ\), so \(\angle CBD=180^\circ - 50^\circ - 55^\circ = 75^\circ\). Using the Law of Sines: \(\frac{BC}{\sin 55^\circ}=\frac{CD}{\sin 75^\circ}=\frac{BD}{\sin 50^\circ}\). We know \(BD\approx819.2\) ft. So \(\frac{BC}{\sin 55^\circ}=\frac{819.2}{\sin 50^\circ}\), \(BC=\frac{819.2\sin 55^\circ}{\sin 50^\circ}\approx\frac{819.2\times0.8192}{0.7660}\approx877.6\) ft. And \(\frac{CD}{\sin 75^\circ}=\frac{819.2}{\sin 50^\circ}\), \(CD=\frac{819.2\sin 75^\circ}{\sin 50^\circ}\approx\frac{819.2\times0.9659}{0.7660}\approx1030.4\) ft.

Answer:

\(AB\approx423\) ft, \(BC\approx878\) ft, \(CD\approx1030\) ft (or more precise values based on calculation, here rounded for simplicity)