Question

В треугольнике ABC угол С равен 90 cosB=3/5 ,АС =4 , СН высота найдите ВН

Answer 1

$\mathtt{cosB= \dfrac{BC}{AB};\ BC=3x,\ AB=5x}\\\mathtt{AB^2=AC^2+BC^2;\ AC=\sqrt{AB^2-BC^2}}\\\mathtt{\sqrt{25x^2-9x^2;}=AC;\ AC=4x;\ AC=4}\\\mathtt{4x=4}\\\mathtt{x=1}\\\boxed{\mathtt{BC=3,AB=5,AC=4}}\\\mathtt{\bigtriangleup BHC\sim\bigtriangleup ABC\to \dfrac{BH}{BC} = \dfrac{BC}{AB} }}\\\mathtt{BC^2=BH*AB;\ BH= \dfrac{BC^2}{AB}= \dfrac{9}{5}=\boxed{1,8}}\\\mathcal{OTVET}:\mathtt{1,8}$

Answer 2

Ответ фоткан........