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using the law of sines to find a triangle with one obtuse angle if $\\a…

Question

using the law of sines to find a triangle with one obtuse angle if $\angle a = 48^\circ, a = 31, b = 34$. 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 \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Given \(A = 48^{\circ}\), \(a = 31\), and \(b = 34\). Substitute these values into the formula: \(\sin B=\frac{b\sin A}{a}\).

$$ \sin B=\frac{34\sin48^{\circ}}{31} $$
$$ \sin B=\frac{34\times0.7431}{31}\approx\frac{25.2654}{31}\approx0.815 $$

Step2: Find the acute and obtuse values of \(B\)

Using the inverse - sine function, the acute value \(B_1=\sin^{-1}(0.815)\approx54.6^{\circ}\). The obtuse value \(B_2 = 180^{\circ}-54.6^{\circ}=125.4^{\circ}\) (since \(\sin\theta=\sin(180 - \theta)\)).

Step3: Check the triangle - angle sum for the obtuse \(B\)

If \(B = 125.4^{\circ}\), then using the triangle - angle sum \(A + B+C=180^{\circ}\), we can find \(C\).
\(C=180^{\circ}-A - B=180^{\circ}-48^{\circ}-125.4^{\circ}=6.6^{\circ}\)

Step4: Use the Law of Sines again to find \(c\)

Using \(\frac{a}{\sin A}=\frac{c}{\sin C}\), we have \(c=\frac{a\sin C}{\sin A}\).

$$ c=\frac{31\sin6.6^{\circ}}{\sin48^{\circ}} $$
$$ c=\frac{31\times0.114}{0.7431}\approx\frac{3.534}{0.7431}\approx4.76 $$

Answer:

\(\angle B\) is \(125.4\) degrees, \(\angle C\) is \(6.6\) degrees, \(c = 4.76\)