Using the Law of Sines to solve the all possible triangles if $\angle A=116^{\circ}, a=29, b=15$. If no answer exists, enter DNE for all answers.
$\angle B$ is degrees
$\angle C$ is degrees
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c=
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Step 1 ：First, we need to understand the Law of Sines. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is the same for all three sides of the triangle. This can be written as: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

Step 2 ：Given that angle A = 116 degrees, side a = 29, and side b = 15, we can find angle B using the Law of Sines: \(\sin B = \frac{b \cdot \sin A}{a} = \frac{15 \cdot \sin(116)}{29}\)

Step 3 ：Calculating this gives us \(\sin B \approx 0.513\)

Step 4 ：Taking the arcsine of both sides to solve for B gives us \(B \approx 31.1\) degrees

Step 5 ：Since the sum of the angles in a triangle is 180 degrees, we can find angle C by subtracting the other two angles from 180: \(C = 180 - A - B = 180 - 116 - 31.1 = 32.9\) degrees

Step 6 ：Finally, we can find side c using the Law of Sines: \(c = \frac{a \cdot \sin C}{\sin A} = \frac{29 \cdot \sin(32.9)}{\sin(116)}\)

Step 7 ：Calculating this gives us \(c \approx 15.8\)

Step 8 ：So, the possible triangle is: \(\angle B\) is 31.1 degrees, \(\angle C\) is 32.9 degrees, c = 15.8

Step 9 ：However, there is another possible solution. Since \(\sin B = \sin(180 - B)\), there is another possible value for B, which is \(B' = 180 - 31.1 = 148.9\) degrees

Step 10 ：Then, \(C' = 180 - A - B' = 180 - 116 - 148.9 = -84.9\) degrees

Step 11 ：Since the angle in a triangle cannot be negative, there is no second solution

Step 12 ：So, the only solution is: \(\angle B\) is 31.1 degrees, \(\angle C\) is 32.9 degrees, \(\boxed{c = 15.8}\)