Question

in the adjoining figure AOB is a straight line. Then find the measure of a (I) angle AOC (II) angle BOC if 2a-3b = 10 and a - b = 20 ​

Answer 1

Step-by-step explanation:

given 2a -3b = 10

2a = 10 + 3b

a = 10 + 3b /2........ (1)

given a - b = 20

-b = 20 - a

b = -20 + a .......(2)

substituting (2) in (1)

a = 10 + 3(-20 + a) / 2

2a = 10 + -60 + 3a

2a -3a = -50

-a = -50

therefore a = 50

a- b = 20

50 - b = 20

-b = 20 - 50

b = 30

angle AOC = 2a + b

= 2 × 50 + 30

= 130 DEGREES.

angle BOC

Angle AOC + Angle BOC = 180 Degree (linear pair)

130 degree + angle BOC = 180 DEGREE

Therefore angle BOC = 180 - 130

= 60 degrees.

Answer 2

Answer:

(i) The measure of ∠AOC = $130^0$.

(ii) The measure of ∠BOC = $50^0$

Step-by-step explanation:

Given: $AOB$ is a straight line and $2a-3b=10$, $a-b=20$.

To find: (i) ∠AOC

             (ii) ∠BOC

Step 1 of 3

Finding the values of $a$ and $b$.

Consider the given two conditions as follows:

$2a-3b=10$ . . . . . (1)

$a-b=20$ . . . . . (2)

From (2), we get

$a=20+b$ . . . . . (3)

Substitute the value $20+b$ for $a$ in the equation (1).

⇒ $2(20+b)-3b=10$

Now, simplify as follows:

⇒ $40+2b-3b=10$

⇒ $-b=10-40$

⇒ $-b=-30$

⇒ $b=30$

Now, substitute the value $b=30$ in equation (3), we get

$a=20+30$

$a=50$

Thus, $a=50$ and $b=30$.

Step 2 of 3

(I) Computing ∠AOC.

From the figure, ∠AOC = $2a+b$

⇒ ∠AOC = $2(50)+30$ (Since $a=50$ and $b=30$)

               = $100+30$

⇒ ∠AOC = $130^0$

Thus, the measure of ∠AOC = $130^0$.

Step 3 of 3

(II) Computing ∠BOC.

From the figure, AOB is a straight line.

⇒ ∠AOC + ∠BOC = 180° (Sum of linear pair is 180°)

⇒ 130° + ∠BOC = 180° (Since ∠AOC = 180°)

⇒ ∠BOC = 180° - 130°

               = 50°

Thus, the measure of ∠BOC = $50^0$.

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