Question

pls solve these questions ​

Answer 1

Step-by-step explanation:

here is your answer

.........ok

Answer 2

Answer:

1. An angle on a straight line is 180°

⇒ 50° + 4x = 180°

4x = 180° - 50°

x = 130/4

x = 32.5

2. ∠POQ = x+y

Now, angle on a straight line is 180°

⇒ x+x+y+y = 180°

2x + 2y = 180°

2(x+y) = 180°

x+y = 180/2

x+y = 90°

⇒ ∠POQ = 90°

3. ∠COB = ∠AOD   { Vertically opposite angles }

⇒ ∠COB = ∠AOD = 5y

Also, angle on a straight line is 180°

⇒ ∠AOP + ∠QOD + ∠AOD = 180°

5y + 2y + 5y = 180°

12y = 180°

y = 180/12

y = 15°

4. ∠AOC = ∠BOD = x   { Vertically opposite angles}

and, ∠AOD + ∠BOC = y   { Vertically opposite angles}

Given, ∠AOC + ∠BOC + ∠BOD = 270°

⇒ x+y+x = 270°

2x+y = 270   -----1.

also, Angle on a straight line is 180°

⇒ ∠AOC + ∠AOD = 180°

x + y = 180  -----2.

Subtracting 2. from 1.

2x + y = 270

 x + y = 180

-    -      -      

x        = 90

Put value of x in 2.

90 + y =180

y = 180 - 90

y = 90

∴ ∠AOC = ∠BOD = x = 90°

and, ∠AOD + ∠BOC = y = 90°

5. Given: ∠AOE = ∠BOE = x

As we know: ∠FOE = ∠FOB + ∠BOE

                        180° = ∠FOB + x

                         180° - x = ∠FOB  -----1.

and, ∠FOE = ∠AOE + ∠FOA

        180° = x + ∠FOA

         180° - x = ∠FOA  -----2.

From 1. and 2.

∠FOA = ∠FOB

Step-by-step explanation:

Hope it helps you :)