Kẻ đường cao CH của tam giác ABC
\(\left\{{}\begin{matrix}tanB=\dfrac{CH}{BH}\Rightarrow BH=\dfrac{CH}{tan62^0}\\tanA=\dfrac{CH}{AH}\Rightarrow AH=\dfrac{CH}{tan87^0}\end{matrix}\right.\)
\(\Rightarrow BH+AH=\dfrac{CH}{tan62^0}+\dfrac{CH.}{tan87^0}=500\Rightarrow CH\approx856\left(m\right)\)
\(\left\{{}\begin{matrix}sinB=\dfrac{CH}{BC}\Rightarrow BC=\dfrac{CH}{sin62^0}\approx969,5\left(m\right)\\sinA=\dfrac{CH}{AC}\Rightarrow AC=\dfrac{CH}{sin87^0}\approx857,2\left(m\right)\end{matrix}\right.\)
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