[math]\sin\left(180^\circ-\angle BAC\right)=\sin\angle BAC=\sqrt{\frac1{ctg^2\angle BAC+1}}[/math]

[math]\sin\angle BAK=\sqrt{\frac1{\left(\frac{15}{7\sqrt{15}}\right)^2+1}}=\sqrt{\frac{7^2\cdot15}{15^2+7^2\cdot15}}=\sqrt{\frac{7^2}{15+49}}=\frac78=0,875[/math]