How do you solve the triangles when the sides of $a=6,b=10$ and $\angle A={{31}^{\circ }}{{10}^{'}}$. Find the other sides and measurement of angles.

Hint: We first use the theorem for properties of triangles \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\]. We also use the relation of angles that sum of all the angles of a triangle is ${{180}^{\circ }}$. We put the values in the relation to get the remaining angles and the sides.

Complete step by step solution:
We have been given the values of two sides’ length and measurement of one angle.
We are going to use the relation for properties of triangles between angles and sides of a general triangle.
The relation gives that for $\Delta ABC$, we have \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\] where $a,b,c$ are the lengths of the sides and $A,B,C$ are corresponding opposite angles of the sides $a,b,c$ respectively.
We have been given that $a=6,b=10$ and $\angle A={{31}^{\circ }}{{10}^{'}}$.

We put these values in the equation \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\] to get \[\dfrac{6}{\sin \left( {{31}^{\circ }}{{10}^{'}} \right)}=\dfrac{10}{\sin B}=\dfrac{c}{\sin C}\]
We get \[\sin \left( {{31}^{\circ }}{{10}^{'}} \right)=0.5175\].
We get the first linear equation of \[\dfrac{6}{0.5175}=\dfrac{10}{\sin B}\].
Solving the equation, we get \[\sin B=\dfrac{10\times 0.5175}{6}=0.8626\].
Taking trigonometric inverse, we get $\angle B={{59.695}^{\circ }}$.
We now find the third angle using the relation of the angles of a triangle where we get the sum of all the angles as ${{180}^{\circ }}$.
So, $\angle A+\angle B+\angle C={{180}^{\circ }}$ which gives $\angle C={{180}^{\circ }}-\angle A-\angle B$.
We put the values to get \[\angle C={{180}^{\circ }}-{{31.17}^{\circ }}-{{59.695}^{\circ }}={{89.228}^{\circ }}\].
We take sin ratio on this angle to get \[\sin \left( {{89.228}^{\circ }} \right)=0.999\]
Now we use the second relation of \[\dfrac{6}{\sin \left( {{31}^{\circ }}{{10}^{'}} \right)}=\dfrac{c}{\sin C}\] which gives \[\dfrac{6}{0.5175}=\dfrac{c}{0.999}\].
Solving we get \[c=\dfrac{6\times 0.999}{0.5175}=11.593\].
Therefore, the other angles are $\angle B={{59.695}^{\circ }},\angle C={{89.228}^{\circ }}$ and the third side is \[c=11.593\].

Note: The individual ratios in the relation \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\] is equal to the $2R$ where $R$ is the circum-radius of the triangle. So, the relation is \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\]. If We know two sides and the angle between them, use the cosine rule and plug in the values for the remaining sides and the angles.