Construct a triangle $A B C$ in which side $B C = 54$ cm, angle $A B C = 60^{0}$ and length of perpendicular from vertex $A$ on side $B C = 30$ cm.

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Solution

Given : $B C = 54 c m, A C = 3 c m$ & $∠ B = 60 ° .$
$i)$ Draw a base $B C = 54 c m .$
$i i)$ Construct $∠ A B C = 60 °$ at $B .$
$i i i)$ Extend the ray from $∠ B .$ From the point ray $B X .$ set of $A D = 30 c m .$
$∴$ $A D ⊥ B C,$ we got $△ A B C .$

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