Questions: Consider a triangle ABC like the one below. Suppose that A=82°, B=39°, and b=71. (The figure is not drawn to scale.) Solve the triangle. Round your answers to the nearest tenth. If there is more than one solution, use the button labeled "or". C=°, a=, c=

Solution Steps

The sum of the angles in a triangle is \(180^{\circ}\). We are given \(A = 82^{\circ}\) and \(B = 39^{\circ}\). Therefore, \(C = 180^{\circ} - A - B = 180^{\circ} - 82^{\circ} - 39^{\circ} = 180^{\circ} - 121^{\circ} = 59^{\circ}\).

We can use the Law of Sines to find side \(a\). \(\frac{a}{\sin A} = \frac{b}{\sin B}\) \(\frac{a}{\sin 82^{\circ}} = \frac{71}{\sin 39^{\circ}}\) \(a = \frac{71 \sin 82^{\circ}}{\sin 39^{\circ}}\) \(a = \frac{71 \times 0.9903}{0.6293}\) \(a \approx \frac{70.3113}{0.6293}\) \(a \approx 111.7\)

We can use the Law of Sines to find side \(c\). \(\frac{c}{\sin C} = \frac{b}{\sin B}\) \(\frac{c}{\sin 59^{\circ}} = \frac{71}{\sin 39^{\circ}}\) \(c = \frac{71 \sin 59^{\circ}}{\sin 39^{\circ}}\) \(c = \frac{71 \times 0.8572}{0.6293}\) \(c \approx \frac{60.8612}{0.6293}\) \(c \approx 96.7\)

Final Answer