Task/26694747
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Дано: треугольник ABC ; A (3; -6) , B (5; 16) , C (12; 2) .
Найти : cos A, cos B, cos C
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решение :
a = BC =√ ( (12-5)² +(2-16)² )=√ (7² +(-14)² )=√ (7²+7²*2²) =7√5.
b = AC =√ ( (12-3)² +(2-(-6) )² )=√ (9² +8²)=√145.
c = AB =√ ( (5-3)² +(16-(-6) )² )=√ (2² +22²)=√ (2²+2²*11²) =2√122.
По теореме косинусов :
a² =b²+c² -2bc*cos∠A ⇒
cos∠A =(b²+c² -a²) /2bc=(145+488 -245) /2√ 145*2√122 = 97 /√(145*122)= 97 /√(145*122)= 97 /√17690 .
аналогично:
cosB =(a²+c² - b²) /2ac =(245 +488 -145) /(2*7√5*2√122)=588/4*7√5*√122=
=21/√ 610.
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cosC =(a²+b²- c²) / 2ab =(245 +145- 488) /2*7√5*√145 = -98 / 2*7*5√29 =
= - 7/5√29 = - 7√29 / 145 . * * * тупой угол * * *
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
можно через скалярное произведение векторов:
AC(12-3 ; 2 -(-6)) ,.т.е AC (9 ; 8) .
AB(5-3; 16-(-6) ) ,.т.е AB(2; 22 )
cos∠A =cos(AC ^AB) =AC *AB / |AC|* |AB| =(9*2+8*22) /√(9²+8²)*√(2²+22²) =
2(9+88) /√(9²+8²)*2√122=97/√145*√122 =97/√(145*122) =97/√(145*122)=
97/√17690. * * * ( 97√17690 ) / 17690 * * *
и т.д.