Question

Construct an angle of 90° at the initial point of a given ray and justify the construction​

Answer 1

Step-by-step explanation:

first draw a line of ob

then put a diameter

then construct 60 then120

from120to60cutan arc above

then put a line straight to wards co

Answer 2

$\huge\star{\pink{\underline{\underline{\mathfrak{Explanation:}}}}}$

______________________________

$\bold{STEPS-OF-CONSTRICTION:}$

• Draw a ray OA

• Take O as a centre with any radius, draw an arc DCB is that cuts OA at B.

• With B as a centre with the same radius, mark a point C on the arc DCB.

• With C as a centre and the same radius, mark a point D on the arc DCB.

• Take C and D as centre, draw two arcs which intersect each other with the same radius at P.

• Finally, the ray OP is joined which makes an angle 90° with OP is formed

$\huge\star{\purple{\underline{\mathfrak{Justification:}}}}$

In order to prove this draw a dotted line from the point O to C and O to D and the angles formed are:

From the construction, it is observed that

OB= BC = OC

Therefore OBC is an equilateral triangle

So that, ∠BOC = 60°.

Similarly,

OD= DC = OC

Therefore DOC is an equilateral triangle

So that, ∠DOC = 60°.

From SSS triangle congruence rule

△OBC ≅ OCD

∠BOC= ∠DOC

Therefore, ∠COP = ½ ∠DOC = ½ (60°).

∠COP = 30°

To find the ∠POA = 90°:

∠POA =∠BOC+ ∠COP

∠POA = 60° +30°

∠POA =90°

$\large\star{\red{\mathrm{Hence, justified!}}}$