Question

maths question answer please​

Answer 1

Step-by-step explanation:

Here,

• (Y+20)° + (2y +30)° + (y+10)°=180°(L.P.P)
• Y + 2y + y +20+30+10 =180°
• 4y+60 =180°
• 4y =180 - 60°
• Y =120/4°
• Y = 30°

Now,

(Y+20)° = 30° +20° = 50°

2Y° + 30° = 2×30° + 30° = 90°

Y° + 10° = 30° + 10°=40°

Answer 2

Provided that:

• AOB is a straight line.

To find:

• Find ∠AOC, ∠BOD and ∠BOC

Solution:

~ According to the figure,

• AOC = (y + 20)°
• BOC = (2y+30)°
• BOD = (y+10)°

~ According to the question,

• AOB is a straight line

Knowledge required: If any angle is on a straight line then it's sum is always 180 degrees. Therefore,

→ AOC + BOC + BOD = 180° (∵ Linear Pair)

→ (y + 20)° + (2y+30)° + (y+10)° = 180°

→ y + 20 + 2y + 30 + y + 10 = 180°

→ y + 2y + y + 20 + 30 + 10 = 180°

→ 1y + 2y + 1y + 20 + 30 + 10 = 180°

→ 4y + 20 + 30 + 10 = 180°

→ 4y + 60 = 180°

→ 4y = 180° - 60°

→ 4y = 120°

→ y = 120/4

→ y = 30°

We get the value for y as 30 degree

~ Now let us substitute the value of y in the each of three angles

Let us put y in the angle ∠AOC first

→ y + 20°

→ 30° + 20°

→ 50° is the measure of ∠AOC

Now let us put y in the angle ∠BOD

→ y + 10°

→ 30° + 10°

→ 40° is the measure of ∠BOD

Now let us put y in the angle ∠BOC

→ 2y + 30°

→ 2(30) + 30

→ 60° + 30°

→ 90° is the measure of ∠BOC $\:$