Question

If $a=2,b=3,c=5$ in $\vartriangle ABC$, then $C=$
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{3}$
C. $\dfrac{\pi }{2}$
D. None of these.

Answer 1

Hint: We will use cosine rule or Law of cosine ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C$ by substituting all the given length of the sides to find the value the angle $C$.
The value of the cosine angle at \[\pi \] that is $\cos \pi $ is $\cos \pi =-1$

Formula Used: ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C$

Complete step by step solution: We are given the length of the sides $a=2,b=3,c=5$ of the triangle $\vartriangle ABC$ and we have to determine the value of the angle $C$.
We will substitute the given length of the sides of the triangle in the cosine law.
$\begin{align}
  & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C \\
 & {{5}^{2}}={{2}^{2}}+{{3}^{2}}-2(2\times 3)\cos C \\
 & 25=4+9-12\cos C \\
 & 12=-12\cos C \\
 & 1=-\cos C \\
 & \cos C=-1 \\
 & \cos C=\cos \pi \\
 & C=\pi
\end{align}$

The angle $C$ of the triangle $\vartriangle ABC$ having sides $a=2,b=3,c=5$ is $C=\pi $. Hence the correct option is (D).

Note: There are three laws of cosines with different sides and angles, so we must analyze which cosine rule should be taken according to the question. Here in this question we have to find the value of angle $C$, so we will choose the cosine law with two sides having angle $C$ hence ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos (C)$.
The Law of cosines states the relationship between length of the sides of a triangle with respect to the cosine of its angle. That is ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos (C)$ where $a,b,c$ are the sides of the triangle and $C$ is the angle opposite to side $c$.
The law of cosine can only be used in some triangles in contrast to the Law of sines which is true for all the triangles.
And to solve this kind of question we must be familiar with all the values of cosine and all the other trigonometric functions at every angle that is from $\left[ 0-2\pi \right]$.