How do you find the largest and smallest angle in a triangle with sides $3,6$ and $7?$

Hint: In this question, we are going to find the largest and smallest angle in a triangle for the given sides.
In this we have the values of the three sides of the triangle namely $3,6$ and $7$
First, we have to use the law of cosine to find one of the angles (angle $a$).
Next we have to find another side (angle $c$) by again using the law of cosine.
Finally, we can find angle $b$ by using angles of a triangle add to ${180^ \circ }$
By solving the triangle we can get the required result.

Formula used: The law of cosine is defined by
${a^2} = {b^2} + {c^2} - 2bc \times \cos A$

Complete step-by-step solution:
In this question, we are going to find the largest and the smallest angle of a triangle.
In this given the values of the three sides of a triangle namely $3,6$ and $7$
Here $a = 3,\,b = 6,\,c = 7$
First we are going to find the smallest angle (angle $a$).
Applying law of cosine we get,
$ \Rightarrow {\left( 3 \right)^2} = {\left( 6 \right)^2} + {\left( 7 \right)^2} - 2\left( 6 \right)\left( 7 \right) \times \cos A$
On simplify the term and we get,
$ \Rightarrow 9 = 36 + 49 - 2 \times 42 \times \cos A$
On adding and multiply the term and we get
$ \Rightarrow 9 = 85 - 84\cos A$
On rewriting we get,
$ \Rightarrow 84\cos A = 85 - 9$
Let us subtract the term and we get
$ \Rightarrow 84\cos A = 76$
Let us divide the term and we get
$ \Rightarrow \cos A = \dfrac{{76}}{{84}}$
On dividing the term and we get
$ \Rightarrow \cos A = \dfrac{{19}}{{21}}$
Then we get,
$ \Rightarrow \cos A = 0.90476$
On rewriting we get,
$ \Rightarrow A = {\cos ^{ - 1}}(0.90476)$
Then we get,
$A = 25.84^\circ$
Next we have to find the largest angle $c$ by again using the law of cosine.
$ \Rightarrow {c^2} = {a^2} + {b^2} - 2ab \times \cos C$
On putting the values and we get
$ \Rightarrow {\left( 7 \right)^2} = {\left( 3 \right)^2} + {\left( 6 \right)^2} - 2\left( 3 \right)\left( 6 \right) \times \cos C$
On squaring the term and we get
$ \Rightarrow 49 = 9 + 36 - 36 \times \cos C$
On adding the term and we get,
$ \Rightarrow 49 = 45 - 36\cos C$
On rewriting we get,
$ \Rightarrow 36\cos C = 45 - 49$
Let us subtract the term and we get
$ \Rightarrow 36\cos C = - 4$
On dividing the term and we get
$ \Rightarrow \cos C = \dfrac{{ - 4}}{{36}}$
On cancel, we get
$ \Rightarrow \cos C = \dfrac{{ - 1}}{9}$
Then we get,
$ \Rightarrow \cos C = - 0.11$
$ \Rightarrow C = {\cos ^{ - 1}}( - 0.11)$
$ \Rightarrow C = 96.38$Degree
Now we are going to find the angle $b$,
Angle $B = 180 - A - C$
Let us putting we get
\[ \Rightarrow B = 180 - 12.53 - 96.38\]
On adding the term and we get
\[ \Rightarrow B = 180 - 108.91\]
Let us subtract the term and we get
\[ \Rightarrow B = 71.09\]
The angle of $B$ is $71.09^\circ$
Now we have completely solved the triangle, that is we have found all its angles.

Here the largest angle is $96.38$ degree and the smallest angle is $25.84$ degree

Note: The angle opposite the smallest side of a triangle has the smallest measure. Likewise, the angle opposite the largest side has the largest measure. So, if given three side lengths, in order to put the angles in order from smallest to largest.