Question

consider a triangle abc like the one below. suppose that a = 54°, b = 98°, and c = 58. (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\.

Explanation:

Step1: Find angle C

The sum of angles in a triangle is 180°. So $C=180^{\circ}-A - B$. Substituting $A = 54^{\circ}$ and $B=98^{\circ}$, we get $C=180^{\circ}-54^{\circ}-98^{\circ}=28^{\circ}$.

Step2: Use the Law of Sines to find side a

The Law of Sines states that $\frac{a}{\sin A}=\frac{c}{\sin C}$. We know $c = 58$, $A = 54^{\circ}$, and $C=28^{\circ}$. So $a=\frac{c\sin A}{\sin C}=\frac{58\times\sin54^{\circ}}{\sin28^{\circ}}$. Using a calculator, $\sin54^{\circ}\approx0.809$, $\sin28^{\circ}\approx0.469$, then $a=\frac{58\times0.809}{0.469}\approx100.1$.

Step3: Use the Law of Sines to find side b

Again, by the Law of Sines $\frac{b}{\sin B}=\frac{c}{\sin C}$. We know $B = 98^{\circ}$, $c = 58$, and $C=28^{\circ}$. So $b=\frac{c\sin B}{\sin C}=\frac{58\times\sin98^{\circ}}{\sin28^{\circ}}$. Since $\sin98^{\circ}\approx0.990$, then $b=\frac{58\times0.990}{0.469}\approx122.6$.

Answer:

$C = 28^{\circ}, a\approx100.1, b\approx122.6$