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We use the sum of the angles of triangle being $180^{\circ}$ to write $$90^{\circ} + 20^{\circ} + A = 180^{\circ}.$$ Solving for $A$ yields $$A = 180^{\circ} - 20^{\circ} - 90^{\circ} = 70^{\circ}.$$

$$\sin\left( 20^{\circ} \right) = \dfrac{5}{c}.$$ Therefore $$c = \dfrac{5}{\sin\left(20^{\circ}\right)}.$$

One could use the Pythagorean theorem for this, but it is easier to simply notice that $$\tan\left(20^{\circ} \right) = \dfrac{5}{a},$$ so $$a = \dfrac{5}{\tan\left(20^{\circ}\right)}.$$

Using the law of sines, $$\dfrac{\sin\left(20^{\circ}\right)}{4} = \dfrac{\sin(C)}{6}.$$ Therefore $$\sin(C) = \dfrac{6\sin\left( 20^{\circ} \right)}{4}.$$ To isolate $C$ take $\sin^{-1}$ of both sides (recall that $\sin^{-1}(\sin(C))=C$) to get $$C = \sin^{-1} \left( \dfrac{6\sin \left( 20^{\circ} \right)}{4} \right) = 30.87^{\circ}.$$

Use the fact that the sum of angles in a triangle is $180^{\circ}$ to see $$20^{\circ} + 30.87^{\circ} + A = 180^{\circ}.$$ Therefore $$A = 180^{\circ} - 20^{\circ} - 30.87^{\circ} = 129.13^{\circ}.$$

By the law of sines, $$\dfrac{\sin\left(129.13^{\circ} \right)}{a}= \dfrac{\sin \left( 20^{\circ} \right)}{4},$$ so $$a = \dfrac{4\sin \left( 129.13^{\circ} \right)}{\sin \left(20^{\circ} \right)}.$$

Use the fact that the sum of angles in a triangle is $180^{\circ}$ to see $$20^{\circ} + 149.13^{\circ} + A = 180^{\circ}.$$ Therefore $$A = 180^{\circ} - 20^{\circ} - 149.13^{\circ}=10.87^{\circ}.$$

By the law of sines, $$\dfrac{\sin \left( 10.87^{\circ} \right)}{a} = \dfrac{\sin(20^{\circ})}{4},$$ so $$a = \dfrac{4 \sin \left( 10.87^{\circ} \right)}{\sin(20^{\circ})}.$$

By the law of sines we see $$\dfrac{\sin(B)}{7} = \dfrac{\sin(70^{\circ})}{3}.$$ Therefore $$\sin(B) = \dfrac{7\sin(70^{\circ})}{3} = 2.19.$$ But we may not take $\sin^{-1}$ of $2.19$ (it is too big) and so this triangle does not exist.

We have no choice other than to use the law of cosines to solve for the angles.

Using the law of cosines, we see $$9^2 = 6^2 + 4^2 - 2(6)(4)\cos(A).$$ This means $$81 = 36 + 16 - 48\cos(A).$$ Isolating $\cos(A)$ gives us $$\cos(A) = \dfrac{-29}{48}.$$ Therefore take $\cos^{-1}$ to isolate $A$: $$A = \cos^{-1} \left( -\dfrac{29}{48} \right) = 127.2^{\circ}.$$

Using the law of cosines, we see $$6^2 = 9^2 + 4^2 - 2(9)(4)\cos(B).$$ Simplifying yields $$36 = 81 + 16 - 72\cos(B).$$ Isolating $\cos(B)$ yields $$\cos(B) = \dfrac{-61}{-72}.$$ Therefore take $\cos^{-1}$ to get $$B = \cos^{-1} \left( \dfrac{-61}{-72} \right) =32.09^{\circ}.$$

We will use the fact that the sum of the angles in a triangle is $180^{\circ}$. This yields $$127.2^{\circ} + 32.09^{\circ} + C = 180^{\circ},$$ so $$C = 180^{\circ}-127.2^{\circ} - 32.09^{\circ} = 20.71^{\circ}.$$

Using the law of cosines, $$c^2 = 6^2+4^2-2(6)(4)\cos \left( 60^{\circ} \right),$$ so $$c= \sqrt{52-48\cos \left( 60^{\circ} \right)} = 5.29.$$ There are two missing angles. We will use the law of cosines to find one of the missing angles and then use the sum of the angles equalling $180^{\circ}$ to find the third.

Using the law of cosines, $$4^2 = 6^2 + 5.29^2 - 2(6)(5.29)\cos \left( B \right),$$ yielding $$\dfrac{-47.98}{-63.48} = \cos(B),$$ so take $\cos^{-1}$ of each side to get $$B = \cos^{-1} \left( \dfrac{-47.98}{-63.48} \right)=40.9^{\circ}.$$