Question

help me fast.................​

Answer 1

Answer:

idk hehe

Step-by-step explanation:

Answer 2

Answer:

The image as for behind the mirror as the object is in front of the mirror.

Step-by-step explanation:

Given : An object OA placed at a point A, LM be a plane mirror, D be an observer and OB is the image.

To prove :The image is as far behind the mirror as the object is in front of the mirror i.e.,

OB=OA

Proof : ∴CN⊥ and AB⊥LM

⇒ AB∣∣CN

∠A=∠i [alternate interior angles]...(i)

∠B=∠r [corresponding angles]...(ii)

Also  ∠i=∠r [∵incident angle = reflected angle]...(iii)

From Eqs. (i), (ii) and (iii), ∠A=∠B

In ΔCOB and ΔCOA, ∠B=∠A [Proved above]

∠1=∠2 [each90∘]

and CO=CO[common side]

∴ ΔCOB≅ΔOAC [by AAS congruence rule]

⇒ OB=OA [by CPCT]

Alternate Method

InΔOBC and ΔOAC, ∠1=∠2 [each 90∘]

Also, ∠i=∠r [∵ incident angle =redlected angle]...(i)

On multiplying both sides of Eq. (i) by - 1 and than adding 90∘ both sides, we get

90∘−∠i=90∘−∠r

⇒ ∠ACO=∠BCO

and OC=OC[Commonside]

∴ ΔOBC≅ΔOAC [by ASA conguence rule]

⇒ OB=OA [by CPCT]

Hence, the image as for behind the mirror as the object is in front of the mirror.