Question

Construct a $△ABC$ in which $BC=7$ cm, $\angle B={75}^0$ and $AB+BC=13$ cm.

Answer 1

Steps of construction,

i) Draw BC=7cm

ii) Construct ∠YBC = 45°

iii) From Ray BY, cut off line segment BD=AB+AC=13cm

iv) Join CD

v) Draw the perpendicular bisector of CD meeting BY at A.

vi) Join AC to obtain the required triangle ABC.

Justification,

Since A lies on the perpendicular bisector of CD.

Therefore AC=AD

Now, BD = 13cm

=> BA+AD = 13cm

=> BA+AC = 13cm

Hence, ABC is the required triangle.