how to construct a ΔABC in which <B=45°, <C=60° and the perpendicular from the vertex A to the base BC is 5cm.

Step 1:—

Make a line PQ of any length.

Step 2:—

On line PQ, make an angle of 90° at any point D.

Step 3:—

At a distance of 5 cm from D, mark a point A.

Step 4:—

Make an angle of 45° at A and where it meet the line DQ, name that point as B.

Step 5:—

At point A, on other side, make an angle 30° and where it meets the line PD, name that point C.

The required triangle ABC is the triangle which satisfies the given condition

i.e. = 45°, LC = 60° and AD = 5 cm

we can justify that ∠B = 45° as —

In ∆ADB,

∠EAB + ∠AEB + ∠EBA = 180°

45° + 90° + ∠EBA = 180°

⇒ ∠EBA = 180° – 135°