
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
164.7k+ views
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 $\left[ 0-2\pi \right]$.
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 $\left[ 0-2\pi \right]$.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
