In ∆leABC , i⨍ cotA2:cotB2:cotC2=3:7:9, then a:b:c=

Given, cot A2:cot B2:cot C2=3:7:9⇒s(s-a)3 ∆=s(s-b)7∆=s(s-c)9∆⇒s-a3=s-b7=s-c9 ⇒s-a=3k,s-b=7k,s-c=9k By adding all these three we get 3s-(a+b+c)=19k ⇒3s-2s=19k⇒s=19k ⇒a=16k,b=12k,c=10k ∴a:b:c=8:6:5

In ∆leABC , i⨍ cotA2:cotB2:cotC2=3:7:9, then a:b:c=

Given, cot A2:cot B2:cot C2=3:7:9⇒s(s-a)3 ∆=s(s-b)7∆=s(s-c)9∆⇒s-a3=s-b7=s-c9 ⇒s-a=3k,s-b=7k,s-c=9k By adding all these three we get 3s-(a+b+c)=19k ⇒3s-2s=19k⇒s=19k ⇒a=16k,b=12k,c=10k ∴a:b:c=8:6:5

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In $∆ l e {ABC}, i ⨍ cot \frac{A}{2} :cot \frac{B}{2} :cot \frac{C}{2} =3:7:9,$ then $a : b : c =$

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7:19:25

8:6:5

14:11:6

7:9:11

answer is D.

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Detailed Solution

Given, $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$

$\Rightarrow \frac{s (s - a)}{3_{} ∆} = \frac{s (s - b)}{7 ∆} = \frac{s (s - c)}{9 ∆} \Rightarrow \frac{s - a}{3} = \frac{s - b}{7} = \frac{s - c}{9} \Rightarrow s - a = 3 k, s - b = 7 k, s - c = 9 k B y a d d i n g a l l t h e s e t h r e e w e g e t 3 s - (a + b + c) = 19 k \Rightarrow 3 s - 2 s = 19 k \Rightarrow s = 19 k \Rightarrow a = 16 k, b = 12 k, c = 10 k ∴ a : b : c = 8 : 6 : 5$