Question

Answer 1

Xét ΔCHA vuông tại H có \(\tan HAC=\frac{CH}{HA}=\frac{CH}{HB+120}\)

=>\(HB+120=\frac{CH}{\tan25}\)

=>\(HB=\frac{CH}{\tan25}-120\)

Xét ΔCDH vuông tại H có \(\tan D=\frac{CH}{HD}\)

=>\(\frac{CH}{BH+270}=\tan20\)

=>\(BH+270=\frac{CH}{\tan20}\)

=>\(HB=\frac{CH}{\tan20}-270\)

=>\(\frac{CH}{\tan25}-120=\frac{CH}{\tan20}-270\)

=>\(\frac{CH}{\tan20}-\frac{CH}{\tan25}=-120+270=150\)

=>\(CH\left(\frac{1}{\tan20}-\frac{1}{\tan25}\right)=150\)

=>CH≃248,77(m)

=>Ngọn núi cao khoảng 248,77 mét