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S =S(ABCD) -?
S =AB*BC*sin1+tq²α =1/cos²α ;
cosα =1/√(1+tq²α) * * * 0<α<90° * * *<br>cosα =1/√(1+15) =1/4 ⇒sinα =tqα*cosα =(√15 ) /4 .
Из ΔABC по теореме косинусов:
AC² =AB² +BC² -2AB*BC*cosα =2AB²(1-cosα)⇒AB² =AC²/2(1-cosα) =4²/2(1-1/4) =32/3 .
S = AB² *sinα =32/3* √15 ) /4 = (8√15)/3 .