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0 votes
15.2k views
in Trigonometry by (53.3k points)

If r1, r2, r3 are the radii of the escribed circles of a triangle ABC and r is the radius of its in-circle then the root(s) of the equation x2 – r (r1r2 + r2r3 + r3r1)x + r1r2r3 – 1 = 0 is/are

(A)  1 

(B)  r1 + r2 + r3

(C)  r

(D)  r1r2r3 – 1

1 Answer

+1 vote
by (53.1k points)
selected by
 
Best answer

Correct option (A), (D)

We have 

Therefore, x2 – r(r1r2 + r2r3 + r3r1)x + r1r2r3 – 1 = 0 is satisfied by x = 1.

So, one root is x = 1 and other root is r1r2r3 – 1.

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