If in a triangle ABC,r_3 = r_1 + r_2 + r,then | vedclass

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

Question

(C) We know that the exradii and inradius are given by:$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$r_1 = 4R \sin \frac{A}{2} \cos \fra

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) We know that the exradii and inradius are given by:$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$$r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$Given $r_3 = r_1 + r_2 + r$,we have:$r_3 - r = r_1 + r_2$$4R \sin \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4R \cos \frac{C}{2} \sin \frac{A}{2} \cos \frac{B}{2} + 4R \cos \frac{C}{2} \cos \frac{A}{2} \sin \frac{B}{2}$Dividing by $4R$:$\sin \frac{C}{2} (\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2}) = \cos \frac{C}{2} (\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2})$Using the identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$ and $\sin(x+y) = \sin x \cos y + \cos x \sin y$:$\sin \frac{C}{2} \cos(\frac{A+B}{2}) = \cos \frac{C}{2} \sin(\frac{A+B}{2})$Since $A+B+C = \pi$,we have $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$:$\sin \frac{C}{2} \cos(\frac{\pi}{2} - \frac{C}{2}) = \cos \frac{C}{2} \sin(\frac{\pi}{2} - \frac{C}{2})$$\sin \frac{C}{2} \sin \frac{C}{2} = \cos \frac{C}{2} \cos \frac{C}{2}$$\sin^2 \frac{C}{2} = \cos^2 \frac{C}{2}$$	an^2 \frac{C}{2} = 1 \Rightarrow 	an \frac{C}{2} = 1$ (as $C$ is an angle of a triangle,$\frac{C}{2} > 0$)$\frac{C}{2} = 45^{\circ} \Rightarrow C = 90^{\circ}$Since $A+B+C = 180^{\circ}$,$A+B = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

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

Similar Questions

If in the $\Delta ABC$,$AB = 2BC$,then find the ratio $\tan \frac{B}{2} : \cot \left( \frac{C - A}{2} \right)$.

The area of an isosceles triangle is $9 \, cm^2$. If the equal sides are $6 \, cm$ in length,the angle between them is .... $^\circ$.

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

Two sides of a triangle are given by the roots of the equation $x^2-5x+6=0$ and the angle between the sides is $\frac{\pi}{3}$. Then,the perimeter of the triangle is

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module