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

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

Question

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

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

(A) Given: $a^2 \sin^2 \frac{C}{2} + c^2 \sin^2 \frac{A}{2} = \frac{b^2}{2}$ Using the half-angle formulas $\sin^2 \frac{C}{2} = \frac{(s-a)(s-b)}{ab}$ and $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$: $a^2 \left( \frac{(s-a)(s-b)}{ab} \right) + c^2 \left( \frac{(s-b)(s-c)}{bc} \right) = \frac{b^2}{2}$ $\Rightarrow \frac{a(s-a)(s-b)}{b} + \frac{c(s-b)(s-c)}{b} = \frac{b^2}{2}$ $\Rightarrow \frac{s-b}{b} [a(s-a) + c(s-c)] = \frac{b^2}{2}$ $\Rightarrow (s-b) [as - a^2 + cs - c^2] = \frac{b^3}{2}$ Since $2s = a+b+c$,we have $s-b = \frac{a+c-b}{2}$. Substituting and simplifying leads to $a+c = 2b$. Thus,$\frac{a+c}{b} = \frac{2}{1}$,so $a+c : b = 2 : 1$.

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

Similar Questions

If the lengths of the sides of a triangle are $7 \, cm$,$4\sqrt{3} \, cm$,and $\sqrt{13} \, cm$,then the smallest angle is .....$^o$.

Which of the following sets of data does not uniquely determine an acute-angled $\Delta ABC$ ($R$ = circum-radius)?

If two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^{\circ}$. If the third side is $3$,then the remaining fourth side is :-

In $\triangle ABC$, with usual notations, if $a = 4$, $b = 5$, and $c = 6$, then $\cos A$ and $\cos C$ are calculated using the Law of Cosines. Find the value of $\cos C$ and $\cos A$ to determine the relationship between $\angle C$ and $\angle A$.

In $\triangle ABC$,$\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) \tan \frac{C}{2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module