Question

4 | P a g e

Q5:- In the given figure, AOB is a straight line and the

ray OC stands on it.

∠AOC = (5x + 30)° ∠BOC = (3x − 50)°,

ℎ x. , ∠AOC ∠BOC​

Answer 1

Step-by-step explanation:

According to question:-AOC+COB=180°

:-(5x+30)°+(3x-50)°=180°

:-8x+30°-50°=180°

:-8x-20°=180°

:-8x=180°+20°

:-8x=200°

200

:-x=-------

8

:-x=25°

5x+30°=5*25+30°=125+30°=155°

3x-50°=3*25-50°=75-50°=25°

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Answer 2

CORRECT QUESTION:

♦ Line AOB is a straight line

Ray OC stands on it . If ∠AOC = (5x + 30)° & ∠BOC = (3x − 50)°. Then find ∠AOC & ∠BOC

ANSWER:

Given

• ∠AOC = (5x + 30)°
• ∠BOC = (3x - 50)°

TO FIND:

• ∠AOC
• ∠BOC

Line AOB = 180° (.°. straight line)

∠AOC + ∠BOC = 180°

5x + 30° + 3x - 50° = 180°

8x - 20° = 180°

8x = 200°

x = 200/8

x = 25°

Now,

• ∠AOC = 5x + 30°

Substitute the value of x

• ∠AOC = 5(25°) + 30° = 125° + 30° = 155°
• ∠BOC = 3x - 50° = 3(25°) - 50 = 75° - 50° = 25°

Hence, ∠AOC = 155° & ∠BOC = 25°

• Note: I've told that straight line = 180°
• Because , If we divide it into 2 equal parts they are perpendicular to each other.
• perpendicular = 90°
• 2(perpendicular) = 2(90°) = 180°
• Hence straight line = 180°