Find the diagonal $ d_1 $ of a rhombus if side $a = 12$ and angle $\beta = 30^o$.

$d_1 = 12 \sqrt{ 3 }$

STEP 1: find angle $ \alpha $

To find angle $ \alpha $ use formula:

$$ \alpha + \beta = 90^o $$

After substituting $ \beta = 30^o $ we have:

$$ \alpha + 30^o = 90^o $$ $$ \alpha = 90^o - 30^o $$ $$ \alpha = 60^o $$

STEP 2: find height $ h $

To find height $ h $ use formula:

$$ \sin \left( \alpha \right) = \dfrac{ h }{ a } $$

After substituting $ \alpha = 60^o $ and $ a = 12 $ we have:

$$ \sin( 60^o ) = \dfrac{ h }{ 12 } $$ $$ \frac{\sqrt{ 3 }}{ 2 } = \dfrac{ h }{ 12 } $$$$ h = \frac{\sqrt{ 3 }}{ 2 } \cdot 12 $$$$ h = 6 \sqrt{ 3 } $$

STEP 3: find diagonal $ d1 $

To find diagonal $ d1 $ use formula:

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

After substituting $ \alpha = 30^o $ and $ h = 6 \sqrt{ 3 } $ we have:

$$ \sin \left( \frac{ 60^o }{ 2 } \right) = \dfrac{ h }{ } $$ $$ \sin( 30^o ) = \dfrac{ 6 \sqrt{ 3 } }{ d_1 } $$ $$ \frac{ 1 }{ 2 } = \dfrac{ 6 \sqrt{ 3 } }{ d_1 } $$ $$ d_1 = \dfrac{ 6 \sqrt{ 3 } }{ \frac{ 1 }{ 2 } } $$ $$ d_1 = 12 \sqrt{ 3 } $$

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