Question

using the law of sines to find a triangle with one obtuse angle if $\angle a = 45^\circ, a = 27, b = 29$. if no answer exists, enter dne for all answers. $\angle b$ is \boxed{} degrees $\angle c$ is \boxed{} degrees $c = \boxed{}$ assume $\angle a$ is opposite side $a$, $\angle b$ is opposite side $b$, and $\angle c$ is opposite side $c$.

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Given \(A = 45^{\circ}\), \(a = 27\), and \(b = 29\), we can find \(\sin B\) as follows:

$$ \sin B=\frac{b\sin A}{a}=\frac{29\sin45^{\circ}}{27} $$

Since \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.707\), then \(\sin B=\frac{29\times0.707}{27}\approx\frac{20.503}{27}\approx0.759\)
So, \(B=\sin^{- 1}(0.759)\approx49.4^{\circ}\) or \(B = 180^{\circ}-49.4^{\circ}=130.6^{\circ}\) (we want the obtuse - angle, so \(B = 130.6^{\circ}\))

Step2: Find angle \(C\)

Since the sum of angles in a triangle is \(A + B + C=180^{\circ}\), then \(C=180^{\circ}-A - B\)
Substitute \(A = 45^{\circ}\) and \(B = 130.6^{\circ}\)
\(C=180^{\circ}-45^{\circ}-130.6^{\circ}=4.4^{\circ}\)

Step3: Find side \(c\)

Using the Law of Sines \(\frac{c}{\sin C}=\frac{a}{\sin A}\)
\(c=\frac{a\sin C}{\sin A}\)
Since \(a = 27\), \(A = 45^{\circ}\), and \(C = 4.4^{\circ}\), \(\sin4.4^{\circ}\approx0.077\), \(\sin45^{\circ}\approx0.707\)
\(c=\frac{27\times0.077}{0.707}=\frac{2.079}{0.707}\approx2.94\)

Answer:

\(\angle B\) is \(130.6\) degrees, \(\angle C\) is \(4.4\) degrees, \(c\approx2.94\)