Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠C = 60°. Then construct another triangle, whose sides are 3/5 times the corresponding sides of ∆ABC.

Given,

BC = 7 cm

∠B = 45°

∠C = 60°

As we know sum of all the angles of a triangle = 180°

So,

∠A + ∠B + ∠C = 180°

∠A + 45° + 60° = 180°

∠A + 105° = 180°

∠A = 180° - 105°

∠A = 75°

Steps for constructing the triangles;

1. To draw a ∆ABC;

(a) First, draw a line BC of 7 cm.

(b) Taking B as center draw an arc intersecting BC.

(c) Taking point O as a center with the same radius draw an arc intersecting the previous arc at P.

(d) Taking P as center draw an arc intersecting the previous one at point Q.

(e) Now taking P and Q as centers with radius more than half of PQ, draw an arc intersecting each other at R.

(f) By drawing the line through the point where arc intersects we get an angle OBR 90°.

(g) By taking the distance between points O and I as radius draw a first from the point O and then from the point I respectively.

(h) Where both the arc intersects each other take that point as J and draw a line through it joining at point B.

(i) It makes the angle JBO 45°.

(j) Now for the 60° angle, we have to take point C as a center and draw an arc with any radius intersecting BC at O’.

(k) By keeping the same radius draw an arc intersecting the previous arc at point P’.

(l) By drawing a line through it we get an angle of 60°.

2. Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

3. Locate 5 Points at BX as B₁, B₂, B₃, B_4, and B₅

4. Join B₃ at C’ and B₅ at C. We can see B₃C’ and B₅C are parallel.

5. From point C’ draw a line parallel to AC intersecting line segment at A’.

So, the required ∆ABC is formed.

Proof;

We have;

And A’C’ is parallel to AC

They will take the same angle with line BC

So,

∠A’C’B = ∠ACB

Now,

We can say that ∆A’BC’ ∼ ∆ABC by AA similarity.

∵ ∠B = ∠B [Common angle]

∠A’C’B = ∠ACB

Since corresponding sides of similar triangles are in the same ratio.

So,

Hence proved.