In a Delta ABC,the expression left( frac{a^2} | vedclass

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

Question

(B) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Thus,the expression becomes $(a(2R) + b(2R) + c(2R)) \si

Vedclass · Accepted Answer

$\Delta$

Vedclass · Answer

(B) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Thus,the expression becomes $(a(2R) + b(2R) + c(2R)) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.$= 2R(a + b + c) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.Since $a+b+c = 2s$,we have $2R(2s) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4Rs \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.We know that $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{4R}$,where $r$ is the inradius.Substituting this,we get $4Rs \left( \frac{r}{4R} \right) = rs$.Since $\Delta = rs$,the expression simplifies to $\Delta$.

In a $\Delta ABC$,the expression $\left( \frac{a^2}{\sin A} + \frac{b^2}{\sin B} + \frac{c^2}{\sin C} \right) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ simplifies to,where $\Delta$ is the area of the triangle.

Similar Questions

In $\triangle ABC$,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

Let $ABC$ and $ABC^{\prime}$ be two non-congruent triangles with sides $AB=4$,$AC=AC^{\prime}=2\sqrt{2}$ and $\angle B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

In a $\Delta ABC$,if the sides are $a = 3, b = 5$,and $c = 4$,then $\sin \frac{B}{2} + \cos \frac{B}{2}$ is equal to

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

In a $\Delta ABC,$ if $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13},$ then $\cos C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module