Find the diagonal $ d_2 $ of a rhombus if side $a = 1$ and angle $\alpha = 60^o$.

$d_2 = 3.3461$

STEP 1: find angle $ \beta $

To find angle $ \beta $ use formula:

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

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

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

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 = 1 $ we have:

$$ \sin( 60^o ) = \dfrac{ h }{ 1 } $$ $$ \frac{\sqrt{ 3 }}{ 2 } = \dfrac{ h }{ 1 } $$$$ h = \frac{\sqrt{ 3 }}{ 2 } \cdot 1 $$$$ h = \frac{\sqrt{ 3 }}{ 2 } $$

STEP 3: find diagonal $ d2 $

To find diagonal $ d2 $ use formula:

$$ \sin \left( \frac{ \beta }{ 2 } \right) = \dfrac{ h }{ d_2 } $$

After substituting $ \beta = 15^o $ and $ h = \frac{\sqrt{ 3 }}{ 2 } $ we have:

$$ \sin \left( \frac{ 30^o }{ 2 } \right) = \dfrac{ h }{ } $$ $$ \sin( 15^o ) = \dfrac{ \frac{\sqrt{ 3 }}{ 2 } }{ d_2 } $$ $$ 0.2588 = \dfrac{ \frac{\sqrt{ 3 }}{ 2 } }{ d_2 } $$ $$ d_2 = \dfrac{ \frac{\sqrt{ 3 }}{ 2 } }{ 0.2588 } $$ $$ d_2 = 3.3461 $$

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