Question

in $\triangle stu$, $t = 100$ cm, $m\angle u = 159^\circ$ and $m\angle s = 18^\circ$. find the length of $s$, to the nearest 10th of a centimeter.

Explanation:

Step1: Find angle T

Sum of angles in triangle: $m\angle T = 180^\circ - 159^\circ - 18^\circ = 3^\circ$

Step2: Apply Law of Sines

Relate sides and angles: $\frac{s}{\sin(m\angle S)} = \frac{t}{\sin(m\angle T)}$

Step3: Solve for s

Rearrange and substitute values:
$s = \frac{t \cdot \sin(m\angle S)}{\sin(m\angle T)} = \frac{100 \cdot \sin(18^\circ)}{\sin(3^\circ)}$
Calculate trig values: $\sin(18^\circ) \approx 0.3090$, $\sin(3^\circ) \approx 0.0523$
$s \approx \frac{100 \cdot 0.3090}{0.0523} \approx 590.8$

Answer:

$590.8$ centimeters