Question

determine the remaining sides and angles of the triangle abc.
a = 140°40, c = 20°20, ab = 1
b = 19°0
bc ≈ □
(do not round until the final answer. then round to the nearest hundredth as needed.)

Explanation:

Step1: Convert angles to decimal degrees

First, convert \(A = 140^{\circ}40'\) to decimal degrees. \(40'=\frac{40}{60}\approx0.67^{\circ}\), so \(A = 140.67^{\circ}\). Convert \(C = 20^{\circ}20'\) to decimal degrees. \(20'=\frac{20}{60}\approx0.33^{\circ}\), so \(C = 20.33^{\circ}\). \(B = 19^{\circ}0'=19^{\circ}\) (already in decimal - degree form as \(0' = 0\)).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). We know \(c = AB = 1\), \(A = 140.67^{\circ}\), \(C=20.33^{\circ}\), and we want to find \(a = BC\).
From \(\frac{a}{\sin A}=\frac{c}{\sin C}\), we can solve for \(a\):

$$a=\frac{c\sin A}{\sin C}$$

Substitute \(c = 1\), \(\sin A=\sin(140.67^{\circ})\approx0.63\), \(\sin C=\sin(20.33^{\circ})\approx0.35\)

$$a=\frac{1\times\sin(140.67^{\circ})}{\sin(20.33^{\circ})}$$

$$a=\frac{\sin(140.67^{\circ})}{\sin(20.33^{\circ})}$$

Using a calculator, \(\sin(140.67^{\circ})=\sin(180 - 39.33^{\circ})=\sin(39.33^{\circ})\approx0.63\), \(\sin(20.33^{\circ})\approx0.35\)

$$a=\frac{0.63}{0.35}=1.8$$

Answer:

\(BC\approx1.80\)