In triangle ABC,if a cos^2 frac{C}{2} + c cos | vedclass

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

Question

(A) Given the equation: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}

Vedclass · Accepted Answer

$2b = a + c$

Vedclass · Answer

(A) Given the equation: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$: $a \left( \frac{s(s-c)}{ab} \right) + c \left( \frac{s(s-a)}{bc} \right) = \frac{3b}{2}$ $\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$ $\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$ Since $2s = a + b + c$,we have $2s - a - c = b$: $\frac{s}{b} (b) = \frac{3b}{2} \Rightarrow s = \frac{3b}{2}$ $\frac{a + b + c}{2} = \frac{3b}{2} \Rightarrow a + b + c = 3b$ $a + c = 2b$

In $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then

Similar Questions

In a triangle $ABC$,with usual notations,if $\tan \left(\frac{A}{2}\right) = \frac{5}{6}$ and $\tan \left(\frac{C}{2}\right) = \frac{2}{5}$,then:

In $\triangle ABC$,find the value of $\sqrt{\frac{r r_2}{r_3 r_1}}$.

With usual notations,if the lengths of the sides of a triangle are $7 \text{ cm}$,$4\sqrt{3} \text{ cm}$,and $\sqrt{13} \text{ cm}$,then the measure of the smallest angle is

In any triangle $ABC$, if $a: b: c = 2: 3: 4$, then $R: r =$

With usual notations, in $\triangle ABC$, if $\cos C = \frac{\sin A}{2 \sin B}$, then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module