686/VB03. problem / Math - Solved Math Problems

MR-686 / 6. problem

Calculate the area of a parallelogram whose heights are 3 cm and 2√3 cm, and the angle between them is 60°.

$A = a \cdot h_{a} = b \cdot h_{b}$

$\sin 60 ° = \frac{h_{b}}{a} = \frac{h_{a}}{b}$

$\frac{\sqrt{3}}{2} = \frac{2 \sqrt{3}}{a} = \frac{3}{b}$

$a \cdot \sqrt{3} = 2 \cdot 2 \cdot \sqrt{3}$

$A = a \cdot h_{a} = 4 \cdot 3$

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