Odpowiedź:
I AC I = y
I AB I = x
y - x = 12√2 ⇒ y = x + 12√2
I BC I = 20
Z tw. kosinusów mamy
( x + 12√2)² = x² + 20² -2*x*20*cos 135°
x² + 24√2 x + 288 = x² + 400 - 40 x*( - sin 45°)
24√2 x + 288 = 400 + 40*[tex]\frac{\sqrt{2} }{2}[/tex]* x
24√2* x + 288 = 400 + 20 √2 x
4√2 x = 112 / : 4√2
x = 14√2 i y = 26√2
===========
Pole ΔBC
P = 0,5*14√2*20*sin 135° = 140√2*cos 45° = 140√2*[tex]\frac{\sqrt{2}} {2}[/tex] = 140
P = [tex]\frac{a*b*c}{4 R}[/tex] a*b*c = 14√2*20*26√2 = 14 560
140 = [tex]\frac{14 560}{4 R}[/tex] ⇒ 14 560 = 560 R / : 560
R = 26
===========
II sposób: Tw. sinusów
I AC I = 26√2
2 R = [tex]\frac{26\sqrt{2} }{sin 135^o} = \frac{26\sqrt{2} }{0,5\sqrt{2} }[/tex] = 26*2 / : 2
R = 26
=========
Szczegółowe wyjaśnienie:
