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Ram Kushwah 6 years, 10 months ago
{tex}\begin{array}{l}if\;\tan\theta=\frac1{2\sqrt6}=0.2041\\then\;\;\theta\;=11.53^\circ\end{array}{/tex}
I have answered your question as per you write but I think there is
something wrong in your question please checek
Posted by Durga Appa 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
DE=9,AB=?
Now as area is different so triangle cannot be congruent so are similar
{tex}\begin{array}{l}Now\;\triangle ABC\;\sim\triangle DEF\\\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}\\\frac{64}{81}=\frac{AB^2}{9^2}\\AB^2=64\\AB=8\end{array}{/tex}
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Aashish Singh???? 6 years, 10 months ago
0Thank You