Question

# In an isosceles triangle, the length of equal sides is b and the base angle α is less than π4. Then which of the following is/are true ?

A
the circumradius of the triangle is b2cosec α
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B
the inradius of the triangle is bsin2α2(1+cosα)
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C
distance between circumcenter and incenter is bcos(3α/2)2sin(α/2)cos(α/2)
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D
distance between circumcenter and incenter is bcos(3α/2)2sinαcos(α/2)
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Solution

## The correct options are A the circumradius of the triangle is b2cosec α B the inradius of the triangle is bsin2α2(1+cosα) D distance between circumcenter and incenter is ∣∣∣bcos(3α/2)2sinαcos(α/2)∣∣∣ Using sine rule in △ABC, bsinα=2R⇒R=b2cosec α Let r be the distance of incenter (I) from the side BC. r=Δs=12b2sin(π−2α)12(b+b+2bcosα) =bsin2α2(1+cosα) Distance of circumcenter (O) from the side BC is, |RcosA|=|Rcos(π−2α)|=Rcos2α Hence, distance between circumcenter and incenter is given by OI=|r+Rcos2α| =∣∣∣bsin2α2(1+cosα)+b cosec αcos2α2∣∣∣ =b∣∣∣sin2α4cos2(α/2)+cos2α4sin(α/2)cos(α/2)∣∣∣ =b∣∣∣cos2αcos(α/2)+sin2αsin(α/2)4sin(α/2)cos2(α/2)∣∣∣ =∣∣∣bcos(3α/2)2sinαcos(α/2)∣∣∣

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