Steps of construction
(i) Draw a ray AB at the point A.
(ii) Taking A as centre and a convenient radius, draw an arc which intersect AB at C.
(iii) Taking C as center and with the same radius, draw an arc which intersect the previous arc at E.
(iv) Similarly, taking E as center and with the same radius, draw an arc which intersect at F.
(v) Taking E and F as centre, draw arcs with equal radius (more than half of EF), which intersect at H.
(vi) Draw a ray AG. ∠PAQ is the required angle of 90°.
Justification: Join AE, CE, EF, FG, and GE.
AC = CB = AE [∵ By Construction]
⇒ ΔACE is an equilateral Triangle.
⇒ ∠CAE = 60 …(1)
Similarly, ∠AEF = 60° …(2)
Hence, ∠CAE = ∠AEF [∵ from (1) and (2)]
∠CAE and ∠AEF alternate angles, therefore
FE ∥ AC
Here, FG = EG [∵ By Construction]
⇒ Point G lies on the perpendicular bisector of EF. ⇒ ∠GIE = 90°
Hence, ∠GAB = ∠GIE = 90° [∵ corresponding angles]