Question

In Exercises 17–32, two sides and an angle (SSA) of a triangle are given. Determine whether the given measurements produce one triangle, two triangles, or no triangle at all. Solve each triangle that results. Round to the nearest tenth and the nearest degree for sides and angles, respectively.
a = 30, b = 40, A = 20°

Accepted Answer

Identify the given elements: side $$a = 30$$, side $$b = 40$$, and angle $$A = 20^\circ. $$Since we have two sides and a non-included angle (SSA), this is the ambiguous case in triangle solving. Use the Law of Sines to find the possible value(s) of angle $$B. $$The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B}. $$Rearranged to find $$\sin B$$, we get $$\sin B = \frac{b \sin A}{a}. $$Calculate $$\sin B$$ using the values given: $$\sin B = \frac{40 	imes \sin 20^\circ}{30}. $$Then determine if $$\sin B is $$less than, equal to, or greater than 1 to decide the number of possible triangles: - If $$\sin B \u003e 1$$, no triangle exists. - If $$\sin B = 1$$, exactly one right triangle exists. - If $$\sin B \u003c 1$$, there may be one or two triangles depending on the angle values. If $$\sin B \u003c 1$$, find angle $$B by $$taking $$B = \sin$$^{-1}$$(\sin B). $$Then find the second possible angle $$B' = 180^\circ - B to $$check if a second triangle is possible (since sine is positive in both the first and second quadrants). For each valid triangle, find the third angle $$C$$ using the angle sum property: $$C = 180^\circ - A - B (or$$ $$C = 180^\circ - A - B'$$ for the second triangle). Then use the Law of Sines again to find side $$c$$: $$c = \frac{a \sin C}{\sin A}. $$Round all sides to the nearest tenth and angles to the nearest degree.

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Key Concepts

Law of Sines

Ambiguous Case of SSA Triangles

Triangle Solution and Rounding