In triangle ABC with the usual notations,if l | vedclass

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

Question

(C) We know that in $	riangle ABC$,$	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.Gi

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) We know that in $	riangle ABC$,$	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.Given $\left(	an \frac{A}{2}ight)\left(	an \frac{B}{2}ight)=\frac{3}{4}$.Substituting the formulas,we get $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} 	imes \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \frac{3}{4}$.Simplifying the expression,$\sqrt{\frac{(s-c)^2}{s^2}} = \frac{3}{4}$,which gives $\frac{s-c}{s} = \frac{3}{4}$.Substituting $s = \frac{a+b+c}{2}$,we have $\frac{\frac{a+b+c}{2} - c}{\frac{a+b+c}{2}} = \frac{3}{4}$.This simplifies to $\frac{a+b-c}{a+b+c} = \frac{3}{4}$.Cross-multiplying gives $4(a+b) - 4c = 3(a+b) + 3c$.Therefore,$a+b = 7c$.

In $\triangle ABC$ with the usual notations,if $\left(\tan \frac{A}{2}\right)\left(\tan \frac{B}{2}\right)=\frac{3}{4}$,then $a+b=\ldots$ (in $c$)

Similar Questions

Points $D$ and $E$ are taken on the side $BC$ of a triangle $ABC$ such that $BD = DE = EC$. If $\angle BAD = x$,$\angle DAE = y$,and $\angle EAC = z$,then the value of $\frac{\sin(x + y)\sin(y + z)}{\sin x \sin z}$ is:

If the sides of a triangle are $p, q$ and $\sqrt{p^2 + pq + q^2}$,then the largest angle is

If the sides of a triangle are $6, 10$ and $14$,then the triangle is

The perimeter of a $\Delta ABC$ is $6$ times the arithmetic mean of the sines of its angles. If the side $a$ is $1$,then the angle $A$ is

In $\Delta PQR$,if $\sin P, \sin Q$ and $\sin R$ are in $A$.$P$.,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module