In a triangle ABC,with usual notations,if tan | vedclass

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

Question

(C) Given $	an \left(\frac{A}{2}ight) = \frac{5}{6}$ and $	an \left(\frac{C}{2}ight) = \frac{2}{5}$.Using the formula $	an \left(\frac{B}{2}i

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $	an \left(\frac{A}{2}ight) = \frac{5}{6}$ and $	an \left(\frac{C}{2}ight) = \frac{2}{5}$.Using the formula $	an \left(\frac{B}{2}ight) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$,we know that $	an \left(\frac{A}{2}ight) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \left(\frac{C}{2}ight) = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$.Let $x = \sqrt{s-a}, y = \sqrt{s-b}, z = \sqrt{s-c}$. Then $	an \left(\frac{A}{2}ight) = \frac{y}{x} \cdot \frac{z}{\sqrt{s}} = \frac{5}{6}$ and $	an \left(\frac{C}{2}ight) = \frac{x}{z} \cdot \frac{y}{\sqrt{s}} = \frac{2}{5}$.Multiplying these,$	an \left(\frac{A}{2}ight) 	an \left(\frac{C}{2}ight) = \frac{y^2}{s} = \frac{5}{6} 	imes \frac{2}{5} = \frac{1}{3}$.Since $s = \frac{a+b+c}{2}$,we have $\frac{s-b}{s} = \frac{1}{3} \implies 3(s-b) = s \implies 3s - 3b = s \implies 2s = 3b$.Substituting $2s = a+b+c$,we get $a+b+c = 3b$,which simplifies to $a+c = 2b$.This condition implies that $a, b, c$ are in $A$.$P$.

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:

Similar Questions

In a triangle $ABC$,$a = 5, b = 7$ and $\sin A = \frac{3}{4}$,how many such triangles are possible?

In $\triangle ABC$,if $(r_2-r_1)(r_3-r_1)=2 r_2 r_3$,then $2(r+R)=$

If $A$ is the area and $2s$ is the sum of the $3$ sides of a triangle,then :

In a $\triangle ABC$,if $a=5, b=6, c=7$,then the length of the median drawn from $B$ is

If the sides of a triangle are in the ratio $\sqrt{3} : \sqrt{5} : \sqrt{8+\sqrt{15}}$,then the largest angle in that triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module