Let A, B and C be the angles of a plane trian | vedclass

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

Question

(A) In a triangle $ABC$,the sum of angles is $A + B + C = \pi$.Therefore,$\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.Taking tangent on both sides,$\

Vedclass · Accepted Answer

$7/9$

Vedclass · Answer

(A) In a triangle $ABC$,the sum of angles is $A + B + C = \pi$.Therefore,$\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.Taking tangent on both sides,$	an \left( \frac{A+B}{2} ight) = 	an \left( \frac{\pi}{2} - \frac{C}{2} ight) = \cot \frac{C}{2}$.Using the formula $	an \left( \frac{A}{2} + \frac{B}{2} ight) = \frac{	an \frac{A}{2} + 	an \frac{B}{2}}{1 - 	an \frac{A}{2} 	an \frac{B}{2}}$,we get:$\cot \frac{C}{2} = \frac{\frac{1}{3} + \frac{2}{3}}{1 - (\frac{1}{3})(\frac{2}{3})} = \frac{1}{1 - \frac{2}{9}} = \frac{1}{\frac{7}{9}} = \frac{9}{7}$.Since $\cot \frac{C}{2} = \frac{9}{7}$,then $	an \frac{C}{2} = \frac{7}{9}$.

Let $A, B$ and $C$ be the angles of a plane triangle and $\tan \frac{A}{2} = \frac{1}{3}, \tan \frac{B}{2} = \frac{2}{3}$. Then $\tan \frac{C}{2}$ is equal to

Similar Questions

In $\triangle ABC$, with usual notation, if $a = 13, b = 14, c = 15$, then the sum of the values of $\sin(\frac{A}{2})$ and $\sin A$ is....

If in a $\triangle ABC$,$r_3 = r_1 + r_2 + r$,then $\angle A + \angle B$ is equal to (in $^{\circ}$)

In a $\Delta ABC$,evaluate $\frac{b \sin(C - A)}{c^2 - a^2} + \frac{c \sin(A - B)}{a^2 - b^2}$.

In $\Delta ABC,$ $\text{cosec } A(\sin B \cos C + \cos B \sin C) = $

With the usual notations, if the lengths of the sides of a triangle are $3 \text{ units}$, $5 \text{ units}$, and $7 \text{ units}$, then the largest angle of the triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module