How do you solve the triangle ABC given a = 5, b = 8, c = 12?

Hint: To solve a triangle means to find the values of all three sides and the three angles of that very triangle. Here, we are given the values of all three sides and we have to find the value of the three angles. For this, we will use ‘The Law of Cosines’. It is given by ${c^2} = {a^2} + {b^2} - 2ab\cos C$, where $a$, $b$ and $c$ are the sides of the triangle and $C$ is the angle between sides $a$ and $b$.


Complete step by step solution:
We have ${c^2} = {a^2} + {b^2} - 2ab\cos C$
It is given that $a = 5$, $b = 8$ and $c = 12$. On substituting the values of $a$, $b$ and $c$ in the cosine law given above, we get
$ \Rightarrow {12^2} = {5^2} + {8^2} - 2 \times 5 \times 8 \times \cos C$
$
   \Rightarrow 144 = 25 + 64 - 80\cos C \\
   \Rightarrow 144 = 89 - 80\cos C \\
 $
Now rearranging the like expression on one side, we get
$
   \Rightarrow 80\cos C = 89 - 144 \\
   \Rightarrow 80\cos C = - 55 \\
 $
On dividing both sides by $80$, we get
$
   \Rightarrow \dfrac{{80\cos C}}{{80}} = \dfrac{{ - 55}}{{80}} \\
   \Rightarrow \cos C = - 0.6875 \\
 $
Now. To find angle $C$, we have to find the inverse cosine of $ - 0.6875$
$
   \Rightarrow C = {\cos ^{ - 1}}( - 0.6875) \\
   \Rightarrow C = {133.43^o} \\
 $
So we have angle $C = {133.43^o}$.
Similarly for angle $B$, we will use the cosine law as
${b^2} = {a^2} + {c^2} - 2ac\cos B$
On substituting the known values of $a$, $b$ and $c$ in the cosine law given above, we get
$ \Rightarrow {8^2} = {5^2} + {12^2} - 2 \times 5 \times 12 \times \cos B$
$
   \Rightarrow 64 = 25 + 144 - 120\cos B \\
   \Rightarrow 64 = 169 - 120\cos B \\
 $
Now rearranging the like expression on one side, we get
$
   \Rightarrow 120\cos B = 169 - 64 \\
   \Rightarrow 120\cos B = 105 \\
 $
On dividing both sides by $120$, we get
$
   \Rightarrow \dfrac{{120\cos B}}{{120}} = \dfrac{{105}}{{120}} \\
   \Rightarrow \cos B = 0.875 \\
 $
Now. To find the angle $B$, we have to find the inverse cosine of $0.875$
$
   \Rightarrow B = {\cos ^{ - 1}}(0.875) \\
   \Rightarrow B = {28.95^o} \\
 $
So we have angle $B = {28.95^o}$.
Now. We have angle $B = {28.95^o}$ and angle $C = {133.43^o}$
As we know the sum of all the angles of a triangle is ${180^o}$.
So in $\Delta ABC$, we have
$ \Rightarrow A + B + C = {180^o}$
On substituting the values of $B$ and $C$, we get
$
   \Rightarrow A + {28.95^o} + {133.43^o} = {180^o} \\
   \Rightarrow A + {162.38^o} = {180^o} \\
   \Rightarrow A = {17.62^o} \\
 $

Hence, we have the values of all the three angles as $A = {17.62^O}$, $B = {28.95^o}$ and $C = {133.43^o}$.

Note: The Law of Cosines can also be used when the value of two sides of a triangle and the angle between the two sides is given. Also, it is evident that the law of cosine is not only restricted to right-angle triangles, it can be used for all types of triangles.