Question

given the triangle below, find the angle a and length of side x
note: picture is not drawn to scale, but you can assume an angle that appears acute is acute and angle that appears obtuse is obtuse.
(the triangle has sides 16, 12, and x, with a 23° angle opposite side 12? or adjacent? the diagram shows a triangle with side 16, side 12, angle 23° at the base, and side x at the base, angle a at the bases other end.)
a = degrees
x =

Explanation:

Step1: Use the Law of Sines to find angle opposite to side 12

The Law of Sines states that \(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\). Let the angle opposite to side 12 be \(B\). Then \(\frac{\sin23^{\circ}}{12}=\frac{\sin B}{16}\). Cross - multiply: \(16\sin23^{\circ}=12\sin B\). So, \(\sin B=\frac{16\sin23^{\circ}}{12}\). Calculate \(\sin23^{\circ}\approx0.3907\), then \(\sin B=\frac{16\times0.3907}{12}\approx0.5209\). \(B = \sin^{-1}(0.5209)\approx31.4^{\circ}\) or \(B = 180^{\circ}- 31.4^{\circ}=148.6^{\circ}\). But if \(B = 148.6^{\circ}\), then \(A=180^{\circ}-23^{\circ}-148.6^{\circ}=-11.6^{\circ}\) (not possible). So \(B = 31.4^{\circ}\).

Step2: Calculate angle \(A\)

Using the angle - sum property of a triangle (\(A + B+23^{\circ}=180^{\circ}\)), \(A=180^{\circ}-23^{\circ}-31.4^{\circ}=125.6^{\circ}\)

Step3: Use the Law of Sines to find \(x\)

\(\frac{\sin23^{\circ}}{12}=\frac{\sin125.6^{\circ}}{x}\). Cross - multiply: \(x\sin23^{\circ}=12\sin125.6^{\circ}\). \(\sin125.6^{\circ}=\sin(180^{\circ}-54.4^{\circ})=\sin54.4^{\circ}\approx0.813\), \(\sin23^{\circ}\approx0.3907\). Then \(x=\frac{12\times0.813}{0.3907}\approx25\)

Answer:

\(A = 125.6\) degrees, \(x = 25\)