task/26859237
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|AB| =3 ;
|AD| =5 ;
α =∠(AB,AD)=180° -120° =60°. * * * α₁ =∠(BA, AD)=120°. * * *
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φ =∠(AB,AC) -?
Скалярное произведение двух векторов :
a*b = |a|*|b|*cos∠(a,b) * * *a*a = |a|* |a|*cos∠(a,a) =|a|²*cos0 =|a|² * * *
AC =AB + AD ;
AC² =(AB +AD)² = AB² +AD²+2AB*AD =|AB|² +|AD|² +2*|AB|*|AD|*cosα=
3²+5²+2*3*5*cos60°=49 =7². ⇒ |AC| =7.
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AB*AC =AB*(AB +AD) =AB*AB +AB*AD =|AB|²+|AB|*|AD|*cosα.
|AB|*|AC|*cos(∠(AB,AC) = |AB|*( |AB|+|AD|*cosα ) .
|AC|*cosφ = |AB|+|AD|*cosα .
7*cosφ =3+5*1/2 ⇒ cosφ =11 /14.
φ =arccos(11/14) .
ответ: arccos(11/14). * * * ≈ 38,2° * * *