Question

determining the number of possible triangles
in △mno, m = 20, n = 14, and m∠m = 51°. how many distinct triangles can be formed given these measurements?
there are no triangles possible.
there is only one distinct triangle possible, with m∠n ≈ 33°.
there are two distinct triangles possible, with m∠n ≈ 33° or m∠n ≈ 147°.
there is only one distinct triangle possible, with m∠n ≈ 147°.

Explanation:

Step1: Apply the Law of Sines

The Law of Sines states that $\frac{\sin M}{m}=\frac{\sin N}{n}$. We know $m = 20$, $n = 14$, and $m\angle M=51^{\circ}$. Plugging in the values: $\frac{\sin51^{\circ}}{20}=\frac{\sin N}{14}$.

Step2: Solve for $\sin N$

First, calculate $\sin51^{\circ}\approx0.7771$. Then, $\sin N=\frac{14\times\sin51^{\circ}}{20}=\frac{14\times0.7771}{20}\approx\frac{10.8794}{20}\approx0.54397$.

Step3: Find possible values of $N$

We know that if $\sin\theta = x$, then $\theta=\arcsin(x)$ or $\theta = 180^{\circ}-\arcsin(x)$. So, $N_1=\arcsin(0.54397)\approx33^{\circ}$, and $N_2 = 180^{\circ}- 33^{\circ}=147^{\circ}$. Now, check if $N_2$ is valid. The sum of angles in a triangle is $180^{\circ}$. If $N = 147^{\circ}$ and $M = 51^{\circ}$, then $M + N=51^{\circ}+147^{\circ}=198^{\circ}>180^{\circ}$, which is impossible. So only $N\approx33^{\circ}$ is valid, meaning only one triangle is possible.

Answer:

There is only one distinct triangle possible, with $m\angle N\approx33^{\circ}$.