Question

determine the remaining sides and angles of the triangle abc.
a = 114.22°, c = 22.47°, c = 140
b = 43.31°
a ≈ □
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{c}{\sin C}$. We know $A = 114.22^\circ$, $C=22.47^\circ$, and $c = 140$. We can solve for $a$ by rearranging the formula: $a=\frac{c\sin A}{\sin C}$.

Step2: Calculate the sines of the angles

First, find $\sin(114.22^\circ)$ and $\sin(22.47^\circ)$.
$\sin(114.22^\circ)=\sin(180^\circ - 65.78^\circ)=\sin(65.78^\circ)\approx0.9123$ (using the identity $\sin(180^\circ - x)=\sin x$)
$\sin(22.47^\circ)\approx0.3830$

Step3: Substitute the values into the formula for $a$

Substitute $c = 140$, $\sin A\approx0.9123$, and $\sin C\approx0.3830$ into $a=\frac{c\sin A}{\sin C}$:
$a=\frac{140\times0.9123}{0.3830}=\frac{127.722}{0.3830}\approx333.5$ (after rounding to the nearest tenth)

Answer:

$333.5$