Question

10,33 find x further find angle BOC ,Angel COD, angle AOD​

Answer 1

Solution:-

Given,

Angle BOC =x+20°

Angel COD =x°

Angle AOD=x+10°

To find-:

Actual measurements of Angle BOC ,Angle COD and Angle AOD

As we know,

Angle AOB=Angle BOC +Angle COD +Angle AOD... (i)

Also,

Angle AOB=180° (by linear pair axiom) ...(ii)

Therefore:-

From (i) and (ii),

Angle BOC +Angle COD +Angle AOD=180°

Putting the algebraic values,

(x+20°) +(x) +(x+10°) =180°

3x+30°=180°

3x=180°-30°=150°

x=50°

Now, x+20°=70°

And x+10°=60°

Hence, measurements of Angle BOC, Angle COD and Angle AOD are 70°, 50° and 60° respectively.

Answer 2

Given:-

• AOB is a Line segment
• ∠AOD=(x+10)°
• ∠DOC=x°
• ∠BOC=(x+20)°

To Find:-

• The Value of x.
• ∠AOD
• ∠DOC
• ∠BOC

Solution:-

We Know that AOB is a Line

$\therefore \: ∠AOB = 180\degree$

Also,

$∠AOB = ∠AOD + ∠DOC + ∠BOC$

Because of Angles are lying on a Same Line

Now, Putting the values

$∠AOB = (x + 20) + x + (x + 10)$

$= > 180 = 3x + 30$

$= > 3x = 180 - 30$

$= > 3x = 150$

$= > x = \frac{150}{3}$

$= > x = 50\degree$

Now Substituting the Value of x

$\sf \: ∠AOD=(x+10)° = 50 + 10 = 60\degree$

$\sf∠DOC=x° = 50\degree$

$\sf∠BOC=(x+20)° = 50 + 20 = 70\degree$