Question

consider a triangle \\(abc\\) like the one below. suppose that \\(b = 129^\circ\\), \\(a = 33\\), and \\(c = 72\\). (the figure is not drawn to scale.)
solve the triangle.

carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.
if there is more than one solution, use the button labeled \or\.

\\(a = \box^\circ\\), \\(c = \box^\circ\\), \\(b = \box\\)

Explanation:

Apply the Law of Cosines to find side b

Using the Solving Triangles knowledge point

$$ b^2 = a^2 + c^2 - 2ac \cos(B) $$

$$ b^2 = 33^2 + 72^2 - 2(33)(72) \cos(129^\circ) $$

$$ b^2 = 1089 + 5184 - 4752 \cos(129^\circ) $$

$$ b^2 \approx 6273 - 4752 (-0.629320) \approx 6273 + 2990.5286 \approx 9263.5286 $$

$$ b \approx \sqrt{9263.5286} \approx 96.2472 \approx 96.2 $$

Apply the Law of Sines to find angle A

Using the Law of Sines knowledge point

$$ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} $$

$$ \sin(A) = \frac{a \sin(B)}{b} \approx \frac{33 \sin(129^\circ)}{96.2472} $$

$$ \sin(A) \approx \frac{33 (0.777146)}{96.2472} \approx \frac{25.6458}{96.2472} \approx 0.266458 $$

$$ A = \arcsin(0.266458) \approx 15.452^\circ \approx 15.5^\circ $$

Apply the Triangle Angle Sum Theorem to find angle C

Using the Triangle Angle Sum knowledge point

$$ C = 180^\circ - B - A $$

$$ C \approx 180^\circ - 129^\circ - 15.452^\circ = 35.548^\circ \approx 35.5^\circ $$

Answer:

Consider a triangle \(ABC\) like the one below. Suppose that \(B = 129^\circ\), \(a = 33\), and \(c = 72\). Solve the triangle.

\(A =\) <blank>\(15.5\)</blank>\(^\circ\), \(C =\) <blank>\(35.5\)</blank>\(^\circ\), \(b =\) <blank>\(96.2\)</blank>