Question

Construct angle ABC = ${45}^{\circ }$ in which BC = 5 cm and AB = 4.6 cm.

Answer 1

Steps of Construction :

1. Draw a line segment BC = 5 cm

2. Taking B as centre, draw an arc of any suitable radius, which cuts BC at the point D.

3. With D as centre and the same radius, as taken in step 2, draw an arc which cuts the previous arc at point E.

4. With E as centre and the same radius, draw one more arc which cuts the first arc at point F.

5. With E and F as centres and radii equal to more than half the distance between E at F, draw arc which cut each other at point P.

6. Join BP to meet EF at L and produce to point O. Then $\angle$OBC = ${90}^{\circ }$

7. Draw BA, the bisector of angle OBC. [With D, L as centres and suitable radius draw two arc meeting each other at Q produced it to R]

$\Rightarrow \angle ABC={45}^{\circ }$

$[\therefore \text{BA is bisector of}\angle OBC\therefore \angle ABC=\frac{{90}^{\circ }}{2}{45}^{\circ }]$