Question

In a pentagon ABCDE, BC is parallel to DE. IF angleBAE = 150, find angleABC and angleAED if they are in the
ratio 2:1
​

Answer 1

Answer:

Step-by-step explanation:

∠B+∠C=180°    (Since AB∥CD,sum of interior angle on same side of transversal is 180°)Let angle A , E and D be 3x , 4x and 5x respectively.We know , sum of all interior angles is equal to 540°.∠A+∠B+∠C+∠D+∠E=540°180°+3x+4x+5x=540°12x=360°x=30°∠E=4x=4(30°)=120°

Answer 2

Answer:

angle ABC =140°, angle ADE=70°

Step-by-step explanation:

Let ABCDE be a pentagon.

Given, BC || DE

Let ∠AED = x

Then ∠ABC = 2x [They are in the ratio 2:1]

Since BC || DE

⇒ ∠C + ∠D = 180°

We know that in a pentagon

sum of angles = (5 – 2) × 180°

∠A + ∠B + ∠C + ∠D + ∠E = 540°

150° + 2x + (∠C + ∠D) + x = 540°

150° + 2x + 180 + x = 540

3x = 540° – 330°

3x = 210°

⇒ x = 70°

and 2x = 2 × 70°

2x = 140°

∴ ∠ABC = 140°

∠AED = 70°