Question

If the angles \[A{\text{ }} < {\text{ }}B{\text{ }} < {\text{ }}C\] of a triangle \[ABC\] are in A.P. then
\[\left( 1 \right)\]${c^2} = {a^2} + {b^2} - a{b^2}$
\[\left( 2 \right)\]${c^2} = {a^2} + {b^2}$
\[\left( 3 \right)\]${b^2} = {a^2} + {c^2} - ac$
\[\left( 4 \right)\]${a^2} = {b^2} + {c^2} - bc$

Answer 1

Hint: We have to find the relation between the angles of a triangle . We solve the question using the concept of three terms of an A.P. and the cosine law of trigonometry . First we will be using the concept of A.P. of three terms and hence we can find an angle . Using the value of the angle and applying cosine law we can determine the relation between the angles of the given triangle .

Complete step-by-step solution:
Given : angles \[A{\text{ }},{\text{ }}B\]and \[C\]of triangle ABC are in A.P.
As the angles of the triangle are in A.P. . So let the angles of the triangle be such that as given :
$\left( {a - d} \right)$ , $a$ , $\left( {a + d} \right)$ which are the angles \[A{\text{ }},{\text{ }}B\]and \[C\]of triangle ABC respectively .
As we know that the sum of angles of a triangle is always \[{180^o}\] .
Now , using the value of sum of angles of the triangles , we can write the expression as :
\[(a - d) + a + (a + d) = {180^o}\]
\[3a = {180^o}\]
On further simplifying , we get the expression as :
\[a = {60^o}\]
So , we get the value of angle \[B\] as :
\[B = {60^o}\]
Now , we know that the formula of cosine law is given as :
$\cos B = \dfrac{{({A^2} + {C^2} - {B^2})}}{{2ac}}$
Using the formula , we can write the expression as :
\[\cos {60^o} = \dfrac{{{A^2} + {C^2} - {B^2}}}{{2AC}}\]
As , we know that the value of \[\cos {60^o}\] is given as :
\[\cos {60^o} = \dfrac{1}{2}\]
Putting the value in above expression , we can write the expression as :
$\dfrac{1}{2} = \dfrac{{({a^2} + {c^2} - {b^2})}}{{2ac}}$
On further solving , we get
$({a^2} + {c^2} - {b^2}) = ac$
On simplifying , we get
$({a^2} + {c^2} - ac) = {b^2}$
Thus , the relation between angles of the triangle ABC is $({a^2} + {c^2} - ac) = {b^2}$
Hence , the correct option is\[\left( 3 \right)\].

Note: We considered the angles such as taken above as it is given that the angles of the triangle are in A.P. Hence we considered the angles as given. Had it been the angles of a quadrilateral then we would have considered the angles in the same way such that they form an A.P.
The formula of triangle law of sine is given as :
\[\dfrac{A}{{\sin A}} = \dfrac{B}{{\sin B}} = \dfrac{C}{{\sin C}}\]
Where \[A\] , \[B\] , \[C\] are the sides of the triangle ABC .
The formula of triangle law of cosine is given as :
$\cos A = \dfrac{{({B^2} + {C^2} - {A^2})}}{{2BC}}$