
If the angles $[A,B,C]$ of a triangle are in A.P. and the sides $[a,b,c]$ opposite to these angles are in G. P. then$ [{a^2},{b^2},{c^2}]$ are in
A. A.P.
B. H.P.
C. G.P.
D. None of these
Answer
161.1k+ views
Hint: The hint for the problem is to use the law of sines. Given that $[A,B,C]$ are in A.P., \[\;a2{\rm{ }} = {\rm{ }}b2{\rm{ }} + {\rm{ }}c2\] and so G. P. also contains these values$[\left( {A - B = G,{\rm{ }}B - C = H} \right)].$
According to the sine rule, if a, b, and c are a triangle's side lengths and a, b, and c are its angles, with an opposite a, etc., then$[asinA = bsinB = csinC].$
When two angles and one side of a triangle are known, a process known as triangulation can be used to calculate the other sides using the law of sines. When two sides and one of the non-enclosed angles are known, it can also be used.
When we are provided either a) two angles and one side or b) two sides and an excluded angle, the sine rule is used. When we have either a) three sides or b) two sides and the included angle, we apply the cosine rule.
Complete step by step solution: As the angle $[A,B,C]$are in A.P.,
So, $[2B = A + C]$
$ = > 3B = A + B + C$
$ = > 3B = 180^\circ $
Then, the value of B becomes,
$ = > B = 60^\circ $
$ = > \cos B = \dfrac{1}{2}$
$ = > \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}} = \dfrac{1}{2}$
$ = > {c^2} + {a^2} - {b^2} = ac$
So, the a,b,c are in G.P
$ = > {c^2} + {a^2} - {b^2} = {b^2}$
$ = > {c^2} + {a^2} = 2{b^2}$
Hence, the$ {a^2},{b^2},{c^2}$are in A.P.
So, Option ‘A’ is correct
Note: The notes for the problem are as follows:
If the angles$ [A,B,C] $of a triangle are in A.P. and the sides$ [a,b,c] $opposite to these angles are in G. P. then$ [{a^2},{b^2},{c^2}]$ are in A.
The notes for the problem are as follows:
If the angles $[A,B,C]$ of a triangle are in A.P. and the sides$ [a,b,c]$ opposite to these angles are in G. P. then$ [a{x^2} + b{y^2} + cz = g\left( {a - x} \right)]$, where $[g] $is an inverse function of$ [P]$ such that $[z = - g\left( a \right)].$
In this case, $[z = - 9/12]$ because $[9/12 \times \left( { - 1} \right) = - 9]$ which leaves on each side (since $[3]$ is not divisible by $[6]).$
According to the sine rule, if a, b, and c are a triangle's side lengths and a, b, and c are its angles, with an opposite a, etc., then$[asinA = bsinB = csinC].$
When two angles and one side of a triangle are known, a process known as triangulation can be used to calculate the other sides using the law of sines. When two sides and one of the non-enclosed angles are known, it can also be used.
When we are provided either a) two angles and one side or b) two sides and an excluded angle, the sine rule is used. When we have either a) three sides or b) two sides and the included angle, we apply the cosine rule.
Complete step by step solution: As the angle $[A,B,C]$are in A.P.,
So, $[2B = A + C]$
$ = > 3B = A + B + C$
$ = > 3B = 180^\circ $
Then, the value of B becomes,
$ = > B = 60^\circ $
$ = > \cos B = \dfrac{1}{2}$
$ = > \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}} = \dfrac{1}{2}$
$ = > {c^2} + {a^2} - {b^2} = ac$
So, the a,b,c are in G.P
$ = > {c^2} + {a^2} - {b^2} = {b^2}$
$ = > {c^2} + {a^2} = 2{b^2}$
Hence, the$ {a^2},{b^2},{c^2}$are in A.P.
So, Option ‘A’ is correct
Note: The notes for the problem are as follows:
If the angles$ [A,B,C] $of a triangle are in A.P. and the sides$ [a,b,c] $opposite to these angles are in G. P. then$ [{a^2},{b^2},{c^2}]$ are in A.
The notes for the problem are as follows:
If the angles $[A,B,C]$ of a triangle are in A.P. and the sides$ [a,b,c]$ opposite to these angles are in G. P. then$ [a{x^2} + b{y^2} + cz = g\left( {a - x} \right)]$, where $[g] $is an inverse function of$ [P]$ such that $[z = - g\left( a \right)].$
In this case, $[z = - 9/12]$ because $[9/12 \times \left( { - 1} \right) = - 9]$ which leaves on each side (since $[3]$ is not divisible by $[6]).$
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

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

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
