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In DeltaABC, a = 10, A = (2pi)/(3), and ...

In `DeltaABC, a = 10, A = (2pi)/(3)`, and circle through B and C passes through the incenter. Find the radius of this circle

Text Solution

Verified by Experts

The correct Answer is:
10
`angle BIC = pi - ((B + C)/(2)) = (pi)/(2) + (A)/(2)`

Using the sine rule,
Diameter of required circle `= (BC)/(sin angle BIC) = (a)/(cos.(A)/(2)) = 20`
`:.` Radius = 10
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