Question

Answer 1

Kẻ CH⊥AB

Áp dụng tslg:

\(\left\{{}\begin{matrix}tanA=\dfrac{CH}{AH}\\tanB=\dfrac{CH}{BH}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}CH=AH.tan51^0\\CH=BH.tan30^0\end{matrix}\right.\)

\(\Rightarrow\dfrac{AH}{tan30^0}=\dfrac{BH}{tan51^0}=\dfrac{AH+BH}{tan30^0+tan51^0}=\dfrac{AB}{tan51^0+tan30^0}\approx124\left(m\right)\)

\(\Rightarrow AH\approx71\left(m\right)\Rightarrow CH\approx88\left(m\right)\)