16) $$ \mathrm{m} \angle \mathrm{B}=84^{\circ}, a=18, b=9 $$

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We are given a triangle with  
- $$ m\angle B = 84^\circ $$  
- $$ a = 18 $$ (side opposite angle $$A$$)  
- $$ b = 9 $$ (side opposite angle $$B$$)

Using the Law of Sines, we have

$$
\frac{a}{\sin A}=\frac{b}{\sin B}.
$$

Substitute the given values:

$$
\frac{18}{\sin A}=\frac{9}{\sin 84^\circ}.
$$

Solve for $$\sin A$$ by cross-multiplying:

$$
18\sin 84^\circ = 9\sin A.
$$

Divide both sides by 9:

$$
\sin A = 2\sin 84^\circ.
$$

Since $$\sin 84^\circ$$ is very close to 1 (in fact, $$\sin 84^\circ\approx 0.9945$$), we have

$$
\sin A \approx 2\times 0.9945 = 1.989.
$$

But the sine of an angle must satisfy

$$
0 \leq \sin A \leq 1.
$$

Because $$1.989>1$$, the result is impossible. Therefore, no angle $$A$$ exists for which the sine equals $$2\sin 84^\circ$$.

Thus, no triangle can be formed with the given measurements.
No triangle can be formed with the given measurements because the sine of angle $$ A $$ exceeds 1, which is impossible.

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