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In triangle JKL, we have JK = JL = 25 and KL = 20. Find the circumradius

 Aug 25, 2021
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Find the circumradius.

 

Hello Guest!

 

Cosine law:

\(cos(\alpha)=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{2\cdot 25^2-20^2}{2\cdot 25\cdot 25}=0,68\\ \alpha=acos(0,68)=47.156°\\ \color{blue}\frac{\alpha}{2}=23.578°\)

\(cos(\frac{\alpha}{2})=\dfrac{\frac{25}{2}}{r}\\ r=\dfrac{12.5}{cos(23.578°)}=13.639\)

\(The\ circumradius\ is\ \color{blue}r=13.639\).

laugh  !

 Aug 25, 2021

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