Question

please someone answer this ques....


rmuskan473 is asking

Answer 1

$\huge\bf { \red A \green N \pink S\blue W \orange E \purple R \green{...}}$

In Fig 8.33

$\longrightarrow \sf\angle AOD + \angle DOC + \angle COB = 180 \degree \\ \\\longrightarrow \sf x + 10 + x + x + 20 = 180 \\ \\\longrightarrow \sf 3x + 30 = 180 \\ \\\longrightarrow \sf 3x = 180 - 30 \\ \\ \longrightarrow \sf3x = 150 \\ \\\longrightarrow \sf x = \frac{150}{3} \\ \\\longrightarrow \boxed{ \sf x = 50}$

_____________________________

In Fig 8.34

$\longrightarrow \sf\angle AOB + \angle BOC + \angle COD + \angle DOE + \angle EOA = 360 \\ \\\longrightarrow \sf x + x + x + x + x = 360 \\ \\\longrightarrow \sf 5x = 360 \\ \\\longrightarrow \sf x = \frac{360}{5} \\ \\\longrightarrow \boxed{ \sf x = 72}$

Answer 2

Hey there here is your correct answer!!

I think it's to be prooved that How All these five angles sums 360°.

Step-by-step explanation:

Given≈

• A point O and rays OA, OB, OC, OD, OE making angles around O.

To be prooved ≈

• ∠ AOB +∠ BOC +∠ COD +∠ DOE + ∠ EOA=360°

Construction ≈

• Draw a ray OF opposite to ray OA.

Proof≈

Since ray OB stands on line FA.

» ∠ AOB + ∠ BOF = 180°

• ( Linear pair)

» ∠ AOB + ∠ BOC + ∠ COF = 180°

• [∠ BOF =∠ BOC +∠ COF ]..................(I)

again ,ray OD stands on line FA.

» ∠ FOD +∠ DOA = 180°

• (linear pair)

» ∠ FOD + ∠ DOA + ∠ EOA = 180° ...........(II)

• (Linear pair)

On adding (I) and (II) , we get

∠ AOB + ∠ BOC +∠ COF +∠ FOD + ∠DOE+ ∠ EOA

= 180°+180°

=> ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA=360°

• [∠ COF +∠ FOD =∠ COD]

__Hence prooved__

Hence The theorem used is “The sum of all the angels around a point is equal to 360°”

• Hope this will help you out
• Thanks a lot
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