In triangle JKL, we have JK = JL = 25 and KL = 20. Find the circumradius
Find the circumradius.
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Cosine law:
cos(α)=b2+c2−a22bc=2⋅252−2022⋅25⋅25=0,68α=acos(0,68)=47.156°α2=23.578°
cos(α2)=252rr=12.5cos(23.578°)=13.639
The circumradius is r=13.639.
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