In a Delta ABC,a = a_1 = 2,b = a_2,c = a_3 su | vedclass

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

Question

(D) Given $a_1 = 2$. For $p = 1$: $a_2 = \frac{5^1}{3^{2-1}} a_1 \left( 2^{2-1} - \frac{4(1)-2}{5^1} a_1 \right) = \frac{5}{3} \cdot 2 \left( 2 - \fra

Vedclass · Accepted Answer

$r_2 = 3r_1$

Vedclass · Answer

(D) Given $a_1 = 2$. For $p = 1$: $a_2 = \frac{5^1}{3^{2-1}} a_1 \left( 2^{2-1} - \frac{4(1)-2}{5^1} a_1 \right) = \frac{5}{3} \cdot 2 \left( 2 - \frac{2}{5} \cdot 2 \right) = \frac{10}{3} \left( 2 - \frac{4}{5} \right) = \frac{10}{3} \cdot \frac{6}{5} = 4$. So,$b = 4$. For $p = 2$: $a_3 = \frac{5^2}{3^{2-2}} a_2 \left( 2^{2-2} - \frac{4(2)-2}{5^2} a_2 \right) = \frac{25}{1} \cdot 4 \left( 1 - \frac{6}{25} \cdot 4 \right) = 100 \left( 1 - \frac{24}{25} \right) = 100 \cdot \frac{1}{25} = 4$. So,$c = 4$. The sides of the triangle are $a = 2$,$b = 4$,$c = 4$. Semi-perimeter $s = \frac{2+4+4}{2} = 5$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5(5-2)(5-4)(5-4)} = \sqrt{5 \cdot 3 \cdot 1 \cdot 1} = \sqrt{15}$. Exradii are given by $r_1 = \frac{\Delta}{s-a} = \frac{\sqrt{15}}{5-2} = \frac{\sqrt{15}}{3}$,$r_2 = \frac{\Delta}{s-b} = \frac{\sqrt{15}}{5-4} = \sqrt{15}$,$r_3 = \frac{\Delta}{s-c} = \frac{\sqrt{15}}{5-4} = \sqrt{15}$. Thus,$r_2 = r_3 = 3 \left( \frac{\sqrt{15}}{3} \right) = 3r_1$. Therefore,$r_2 = 3r_1$.

In a $\Delta ABC$,$a = a_1 = 2$,$b = a_2$,$c = a_3$ such that $a_{p+1} = \frac{5^p}{3^{2-p}} a_p \left( 2^{2-p} - \frac{4p-2}{5^p} a_p \right)$ where $p = 1, 2$,then:

Similar Questions

In $\Delta ABC,$ if $\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in $H.P.,$ then $a, b, c$ will be in

Observe the following statements:
$(I)$ In $\triangle ABC$,$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
Which of the following is correct?

If in a triangle $ABC$,$b \cos^2 \frac{A}{2} + a \cos^2 \frac{B}{2} = \frac{3}{2} c$,then $a, b, c$ are in:

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C = $

In a triangle $ABC$,with usual notations,$\frac{\cos B+\cos C}{b+c}+\frac{\cos A}{a}$ has the value

Vedclass Test Series

Exam Paper Generator

Online Exam Module