Questions: Solve the following oblique triangle for all of its missing sides angles. Round all answers to the nearest whole number. Angle A= Angle B= side c=

Solution Steps

• Angle \( C = 22^\circ \)
• Side \( b = 122 \)
• Side \( a = 192 \)

The Law of Sines states: \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] Using the given values: \[ \frac{192}{\sin A} = \frac{122}{\sin 22^\circ} \] Solving for \( \sin A \): \[ \sin A = \frac{192 \cdot \sin 22^\circ}{122} \] \[ \sin A \approx \frac{192 \cdot 0.3746}{122} \approx 0.589 \] \[ A \approx \sin^{-1}(0.589) \approx 36^\circ \]

Since the sum of angles in a triangle is \( 180^\circ \): \[ B = 180^\circ - A - C \] \[ B = 180^\circ - 36^\circ - 22^\circ = 122^\circ \]

Using the Law of Sines again: \[ \frac{c}{\sin C} = \frac{a}{\sin A} \] \[ c = \frac{a \cdot \sin C}{\sin A} \] \[ c = \frac{192 \cdot \sin 22^\circ}{\sin 36^\circ} \] \[ c \approx \frac{192 \cdot 0.3746}{0.588} \approx 122 \]

Final Answer