in $\\triangle mno$, $m = 8.5$ inches, $m\\angle n = 52^\\circ$ and $m\\angle o = 23^\\circ$. find the…

in $\\triangle mno$, $m = 8.5$ inches, $m\\angle n = 52^\\circ$ and $m\\angle o = 23^\\circ$. find the length of $o$, to the nearest 10th of an inch.\n\nanswer attempt 1 out of 2

in $\\triangle mno$, $m = 8.5$ inches, $m\\angle n = 52^\\circ$ and $m\\angle o = 23^\\circ$. find the length of $o$, to the nearest 10th of an inch.\n\nanswer attempt 1 out of 2

Answer

Explanation:

Step1: Find $\angle M$

The sum of angles in a triangle is $180^\circ$. $\angle M = 180^\circ - 52^\circ - 23^\circ = 105^\circ$

Step2: Apply Law of Sines

Relate sides and opposite angles: $\frac{o}{\sin\angle O} = \frac{m}{\sin\angle M}$

Step3: Solve for $o$

Rearrange to isolate $o$ and substitute values. $o = \frac{m \cdot \sin\angle O}{\sin\angle M} = \frac{8.5 \cdot \sin(23^\circ)}{\sin(105^\circ)}$ Calculate $\sin(23^\circ) \approx 0.3907$, $\sin(105^\circ) \approx 0.9659$ $o \approx \frac{8.5 \cdot 0.3907}{0.9659} \approx \frac{3.32095}{0.9659} \approx 3.4$

Answer:

3.4 inches