Math, asked by Shivaniverma2001, 7 months ago

in parallelogram ABCD, angleA=(4x-5) and(3x+10). find angleA and angleB​

Answers

Answered by manikandan52
31

Answer:

<A = 55°

<B = 125°

Step-by-step explanation:

<A = (4x-5) and (3x+10)

so,

<A = <C (in parallelogram opposite angles are equal)

(4x-5) = (3x+10)

4x-3x = 10+5

x = 15

<A = 4x-5

= 4(15)-5

= 60-5

= 55°

<B = <A+<B = 180° (in parallelogram sum of adjacent angles are 180°)

<B = 55° + <B = 180°

<B = 180° - 55° = 125°

<A = 55°

<B = 125°

Answered by Anonymous
22

Correction :-

  • In a parallelogram ABCD, ∠A = (4x - 5)° and ∠C = (3x + 10)°. Find ∠A and ∠B.

Answer :-

  • ∠A = 55°
  • ∠B = 125°

Solution :-

Here,

  • ∠A = (4x - 5)
  • ∠C = (3x + 10)

∠A = ∠C [ opposite angles of the parallelogram ]

→ (4x - 5) = (3x + 10)

→ 4x - 5 = 3x + 10

→ 4x - 3x = 10 + 5

→ x = 15

Now,

  • ∠A = 4(15) - 5 = 60 - 5 = 55°

We've to find ∠B

According to question :-

∠A + ∠B = 180° [ adjacent angles of a parallelogram ]

55° + ∠B = 180°

∠B = 180° - 55°

∠B = 125°

Hence, the values of A and B are 55° and 125° respectively.

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