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 = 10, b = 30, A = 150°

Accepted Answer

Identify the given elements: side $$a = 10$$, side $$b = 30$$, and angle $$A = 150^\circ. $$Note that angle $$A is $$opposite side $$a. $$Use the Law of Sines to find the possible value(s) of angle $$B. $$The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B}. $$Substitute the known values to get $$\frac{10}{\sin 150^\circ} = \frac{30}{\sin B}. $$Solve for $$\sin B by $$rearranging the equation: $$\sin B = \frac{30 	imes \sin 150^\circ}{10}. $$Calculate the right side to find the value of $$\sin B (do $$not compute the final value here). Determine the number of possible triangles by analyzing the value of $$\sin B$$: if $$\sin B \u003e 1$$, no triangle exists; if $$\sin B = 1$$, one right triangle exists; if $$0 \u003c \sin B \u003c 1$$, there are two possible angles for $$B ($$one acute and one obtuse), so two triangles may exist. For each possible angle $$B$$, find angle $$C$$ using the triangle angle sum: $$C = 180^\circ - A - B. $$Then use the Law of Sines again to find side $$c$$: $$\frac{c}{\sin C} = \frac{a}{\sin A}. $$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 Angle Sum Property