Answer:
[tex]let \: the \: angle \: be \: \alpha \\ \tan( \alpha ) = \frac{p}{b} \\ \tan(16.8) = \frac{2.7}{x} \\ x = \frac{2.7}{0.30 } \\ x = 9 \\ now \: by \: using \: pythagoras \: theorem \: \\ h = \sqrt{p {}^{2} } + b {}^{2} \\ h = \sqrt{ {2.7}^{2} } + {9 }^{2} \\ h = 9.396[/tex]
