If in triangle ABC,with usual notations,a cdo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the identity $\cos^2 	heta = \frac{1 + \cos(2\

Vedclass · Accepted Answer

$a, b, c$ are in $A$.$P$.

Vedclass · Answer

(C) Given the equation: $a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the identity $\cos^2 	heta = \frac{1 + \cos(2	heta)}{2}$,we get: $a \left( \frac{1 + \cos C}{2} ight) + c \left( \frac{1 + \cos A}{2} ight) = \frac{3b}{2}$ Multiplying by $2$: $a(1 + \cos C) + c(1 + \cos A) = 3b$ $a + a \cos C + c + c \cos A = 3b$ Using the projection formula $b = a \cos C + c \cos A$: $a + c + b = 3b$ $a + c = 2b$ This condition implies that $a, b, c$ are in $A$.$P$.

If in $\triangle ABC$,with usual notations,$a \cdot \cos^2 \frac{C}{2} + c \cdot \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

Similar Questions

If in a triangle $ABC$,$\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6}$,then the value of $\cos A + \cos B + \cos C$ is equal to

In a $\triangle ABC$,if $a=4, b=5$ and $c=7$,then $\sin \left(\frac{A}{2}\right) = $

If,in a $\triangle ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then $a, b, c$ are such that :

In a triangle $ABC$,if $r_1 : r_2 = 3 : 4$ and $r_2 : r_3 = 2 : 3$,then $a : b : c =$

In a $\triangle ABC$,it is known that $AB=AC$. Suppose $D$ is the mid-point of $AC$ and $BD=BC=2$. Then,the area of the $\triangle ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module