Question

In a triangle ABC, if a=3, b=5 and $\cos C = \dfrac{1}{5}$, how do you find the exact value of $c$ ?

Answer 1

Since we have been given the sides of the triangle and the value of one of the angles is given it is to be understood that the standard cosine formula should be used. You can make variations in the sides and angles of the formula when different angles of the triangles are given.

Complete step by step answer:
Since we have been given two sides and the value of $\cos C = \dfrac{1}{5}$and told to find the value of the remaining side, we can use the standard cosine rule. Standard cosine rule is given by ${a^2} = {b^2} + {c^2} - 2cb\cos A$. Since we have been told to find the value of $\cos C = \dfrac{1}{5}$, we will have to make adjustments in the above standard formula in respect to the sides a, b, c and the angles of the triangle.It will become,
${c^2} = {a^2} + {b^2} - 2ab\cos C$
Now we have been given all the values, we will substitute in the above equation.
We get
${c^2} = {3^2} + {5^2} - \left( {2 \times 3 \times 5 \times \dfrac{1}{5}} \right)$
On further solving we get

\Rightarrow {c^2} = 28$$ Taking square root on both sides, we get $$ \therefore c = 2\sqrt 7 $$ **Hence we have found the exact value of $c$ which is $$2\sqrt 7 $$.** **Note:** This rule can be applied when any of the information related to sides or angles is missing to find these missing parts. The cosine rule tries to relate all the sides of the triangle to one angle of the same triangle. It is true that the Pythagorus theorem which is true for right angled triangles is generalized by the law of cosines.