(A) We have the formulas for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$ Given that $r_1

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(A) We have the formulas for the exradii of a triangle: $r_1 = \frac{\Delta}{s-a}, r_2 = \frac{\Delta}{s-b}, r_3 = \frac{\Delta}{s-c}$ Given that $r_1 $\frac{\Delta}{s-a} Since $\Delta$ is the area of the triangle and is positive, taking the reciprocal reverses the inequality signs: $s-a > s-b > s-c$ Subtracting $s$ from all parts: $-a > -b > -c$ Multiplying by $-1$ reverses the inequality signs again: $a < b < c$

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

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If in a $\triangle ABC$, $r_1 < r_2 < r_3$, then:

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AP EAMCET 2003 All AP EAMCET

A
$a < b < c$
B
$a > b > c$
C
$b < a < c$
D
$a < c < b$

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Trigonometrical Equations

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Relation between sides and angles, Solutions of triangles

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MediumMHT CET 2024

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In $\Delta ABC$,if $\angle C = 90^\circ$,$\angle A = 30^\circ$,and $c = 20$,then the values of $a$ and $b$ are:

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In a $\Delta ABC$,if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=2$,then its area is

EasyKCET 2013

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If in a $\triangle ABC$, $r_1 < r_2 < r_3$, then:

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