Construct $∆$ LMN in base MN = 7 cm, $∠$ LMN = 40 $°$ and LM $-$ LN = 3cm.

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Solution

Steps of construction:

1. Draw MN = 7 cm.
2. Draw ray MX, such that $∠ NMX = 40 °$ .
3. With M as centre and radius 3 cm, draw an arc intersecting ray MX at point D.
4. Join D and N.
5. Draw the perpendicular bisector of seg DN, intersecting seg MX at point L.
6. Join L and N.
7. Hence, $△$ LMN is the required triangle.

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△ L M N

M N = 7 c m

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Construct $△$ LMN in which base MN = 7 cm, $∠$ LMN = $75^{\circ}$ and LM + LN = 9 cm.

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