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Circumradius of an isosceles DeltaABC w...

Circumradius of an isosceles `DeltaABC` with `angleA=angleB` is 4 times its in radius, then cosA is root of the equation :

A

`x^(2)-x-8=0`

B

`8x^(2)-8x+1=0`

C

`x^(2)-x-4=0`

D

`4x^(2)-4x+1=0`

Text Solution

Verified by Experts

The correct Answer is:
B
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