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Construct a `Delta ABC` in which `angleB=45^(@), angleC=60^(@)` and the perpendicular from the vertex A to base BC is 4.5 cm.

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Steps of Construction
(i) Draw any line XY.
(ii) Take any point D on XY n=and draw `DE bot XY`.
(iii) Cut off `DA=4.5` cm along DE.
(iv) Through A draw `LM||XY`
(v) Construct `angleLAB=45^(@)` and `anglwMAC=60^(@)`, meeting XY at B and C respectively. Then, `DeltaABC` is the required triangle.
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