Question

using the law of sines to solve the all possible triangles if $\angle a = 104^\circ, a = 35, b = 17$. 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 $c$. question help: \boxed{video}

Explanation:

Step1: Apply the Law of Sines to find \(\angle B\)

The Law of Sines states that \(\frac{\sin A}{a}=\frac{\sin B}{b}\).
Substituting the given values \(A = 104^{\circ}\), \(a = 35\), and \(b = 17\) into the formula:

$$ \sin B=\frac{b\sin A}{a}=\frac{17\sin104^{\circ}}{35} $$

$$ \sin B=\frac{17\times0.97}{35}\approx\frac{16.49}{35}\approx0.471 $$

$$ B=\sin^{- 1}(0.471)\approx28.1^{\circ} $$

Step2: Find \(\angle C\)

Since the sum of angles in a triangle is \(180^{\circ}\), we have \(C = 180^{\circ}-A - B\).
Substituting \(A = 104^{\circ}\) and \(B\approx28.1^{\circ}\):

$$ C=180^{\circ}-104^{\circ}-28.1^{\circ}=47.9^{\circ} $$

Step3: Find side \(c\) using the Law of Sines

Again, using the Law of Sines \(\frac{\sin A}{a}=\frac{\sin C}{c}\).

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

Substituting \(a = 35\), \(A = 104^{\circ}\), and \(C = 47.9^{\circ}\):

$$ c=\frac{35\times\sin47.9^{\circ}}{\sin104^{\circ}} $$

$$ c=\frac{35\times0.742}{0.97}\approx\frac{25.97}{0.97}\approx26.8 $$

Answer:

\(\angle B\) is \(28.1\) degrees, \(\angle C\) is \(47.9\) degrees, \(c = 26.8\)