Question

consider a triangle ( abc ) like the one below. suppose that ( b = 41 ), ( a = 65 ), and ( b = 27^{circ} ). (the figure is not drawn to scale.) solve the triangl
carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.
if no such triangle exists, enter
o solution.\ if there is more than one solution, use the button labeled \or\.

( a=square^{circ}, c=square^{circ}, c=square )

Explanation:

Step1: Use the Law of Sines to find angle \(A\)

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
Substituting the given values \(a = 65\), \(b = 41\), and \(B=27^{\circ}\), we get \(\sin A=\frac{a\sin B}{b}=\frac{65\sin27^{\circ}}{41}\).
Calculating \(\sin27^{\circ}\approx0.4540\), then \(\sin A=\frac{65\times0.4540}{41}\approx\frac{29.51}{41}\approx0.7198\).
So \(A=\sin^{- 1}(0.7198)\approx46.0^{\circ}\) or \(A = 180^{\circ}-46.0^{\circ}=134.0^{\circ}\).
But if \(A = 134^{\circ}\), then \(A + B=134^{\circ}+27^{\circ}=161^{\circ}\), and \(C=19^{\circ}\). Using the Law of Sines \(\frac{c}{\sin C}=\frac{b}{\sin B}\), \(c=\frac{b\sin C}{\sin B}\). \(\sin19^{\circ}\approx0.3256\), \(c=\frac{41\times0.3256}{0.4540}\approx29.5\). But checking the triangle inequality: \(a = 65\), \(b = 41\), \(c\approx29.5\), \(b + c=41+29.5 = 70.5>65\), \(a + c=65 + 29.5=94.5>41\), \(a + b=65+41 = 106>29.5\). Also, using the Law of Sines for the first case of \(A\):

Step2: Find angle \(C\) when \(A = 46.0^{\circ}\)

Since the sum of angles in a triangle is \(180^{\circ}\), \(C=180^{\circ}-(A + B)\). Substituting \(A = 46.0^{\circ}\) and \(B = 27^{\circ}\), we get \(C=107.0^{\circ}\).

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

By the Law of Sines \(\frac{c}{\sin C}=\frac{b}{\sin B}\). We know \(b = 41\), \(B = 27^{\circ}\), \(C = 107^{\circ}\), \(\sin107^{\circ}\approx0.9563\), \(\sin27^{\circ}\approx0.4540\). Then \(c=\frac{41\times\sin107^{\circ}}{\sin27^{\circ}}=\frac{41\times0.9563}{0.4540}\approx86.0\).

Answer:

\(A = 46.0^{\circ},C = 107.0^{\circ},c = 86.0\)