Question

find the unknown angles (give reasons)​

Answer 1

Step-by-step explanation:

(i) In Fig (i),

x = y (Angles opposite to equal sides)

But x + y + 80° = 180° (Angles of a triangle)

⇒ x + x + 80° = 180°

⇒ 2x = 180° - 80° = 100°

⇒ x = 100°/2 = 50° ∴ y = x = 50°

Hence x = 50°, y = 50°

(ii) In Fig. (ii),

b = 40° (Angles opposite to equal sides)

But a + b + 40° = 180°

(Angles of a triangle)

⇒ a + 40° + 40° = 180°

⇒ a + 80° = 180°

⇒ a = 180° - 80° = 100°

Hence, a = 100° , b = 40°

(iii) In Fig. (iii)

x = y (Angles opposite to equal sides)

But x + y + 90° = 180°

(Angles of a triangle)

⇒ x + x + 90° = 180°

⇒ 2x + 90° = 180°

⇒ 2x = 180° - 90° = 90°

∴ x = 90°/2 = 45° ∴ y = x = 45°

Hence x = 45°, y = 45°

Answer 2

All the given questions are in linear

linear pairs will add up to 180°

1) ∠AOD + ∠DOC + ∠COB = 180°

⇒ (x+10)° + x° + (x+20)° = 180°

⇒ x + 10 + x + x + 20 = 180°

taking x's to one side and numericals on other side

⇒ x + x + x + 10 + 20 = 180°

⇒ 3x + 30 = 180°

⇒ 3x = 180 - 30

⇒ 3x = 150

⇒ x = $\frac{150}{3}$

x =  50°

∠AOD =  x + 10 = 50 + 10 ⇒ 60°

∠DOC = x = 50°

∠COB = x + 20 = 50 + 20 = 70°

----------------------------------------------------------------------------------------------

2)  Here we can solve this question by 3 methods

                                   1st Method

∠AOD + ∠DOB = 180°                                    ∵ Linear Pair

⇒ 150° + x = 180°

⇒ x = 180° - 150°

x = 30°

3x = 3 × 30° = 90°

3x = 3 × 30° = 90°

                                  2nd Method

∠AOC = ∠BOC = 3x

3x + 3x = 180°                                                               ∵ Linear Pair

⇒ 6x = 180

⇒ x = $\frac{180}{6}$

x = 30°

3x = 3 × 30° = 90°

3x = 3 × 30° = 90°

                                  3rd Method

By Seeing ∠BOC or ∠AOC, we can say that they are in 90° angle

∴ 3x = 90°

⇒ x =  $\frac{90}{3}$

x = 30°

----------------------------------------------------------------------------------------------

3) ∠AOC + ∠COB = 180°                                               ∵ Linear Pair

⇒ ( 6y + 30 ) + 4y = 180°

⇒ 6y + 30 + 4y = 180

⇒ 6y + 4y = 180 - 30

⇒ 10y = 150

⇒ y =  $\frac{150}{10}$

y = 15°

∠AOC = 6y + 30 = ( 6 × 15 ) + 30 = 90 + 30 ⇒ 120°

∠COB = 4y = 4 × 15 = 60°

----------------------------------------------------------------------------------------------

Hope It Helps You ! :D

By Abhiram5566

           

                                MARK ME AS BRAINLIEST ANSWER

----------------------------------------------------------------------------------------------