Find the diagonal $ d_2 $ of a rhombus if diagonal $d_1 = 12 \sqrt{ 3 }$ and height $h = 6 \sqrt{ 3 }$.

$d_2 = 6 \sqrt{ 3 }$

STEP 1: find angle $ \alpha $

To find angle $ \alpha $ use formula:

$$ \sin \left( \frac{ \alpha }{ 2 } \right) = \dfrac{ h }{ d_1 } $$

After substituting $ h = 6 \sqrt{ 3 } $ and $ d_1 = 12 \sqrt{ 3 } $ we have:

$$ \sin \left( \frac{ \alpha }{ 2 } \right) = \dfrac{ 6 \sqrt{ 3 } }{ 12 \sqrt{ 3 } } $$ $$ \sin \left( \frac{ \alpha }{ 2 } \right) = \frac{ 1 }{ 2 } $$ $$ \frac{ \alpha }{ 2 } = \arcsin\left( \frac{ 1 }{ 2 } \right) $$ $$ \frac{ \alpha }{ 2 } = 30^o $$$$ \alpha = 30^o \cdot 2 $$$$ \alpha = 60^o $$

STEP 2: find diagonal $ d2 $

To find diagonal $ d2 $ use formula:

$$ \sin \left( \frac{ \alpha }{ 2 } \right) = \dfrac{ d_2 }{ d_1 } $$

After substituting $ \alpha = 30 $ and $ d_1 = 12 \sqrt{ 3 } $ we have:

$$ \sin \left( \frac{ 60^o }{ 2 } \right) = \dfrac{ d_2 }{ 12 \sqrt{ 3 } } $$ $$ \sin( 30 ) = \dfrac{ d_2 }{ 12 \sqrt{ 3 } } $$ $$ \frac{ 1 }{ 2 } = \dfrac{ d_2 }{ 12 \sqrt{ 3 } } $$$$ d_2 = \frac{ 1 }{ 2 } \cdot 12 \sqrt{ 3 } $$$$ d_2 = 6 \sqrt{ 3 } $$

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