Question

in fig ab//cd if angle 2=(2x+30) , angle 4=(x+2y) and angle 6=(3y+10) find angle 5​.​

Answer 1

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Answer 2

Step-by-step explanation:

Question:- In figure AB||CD if ∠2=(2x+30) ,∠4=(x+2y) and ∠6=(3y+10) find ∠5.

Solution:-

Given that AB∥CD

Now,

∠2 = ∠4 [Vert. opp. ∠s]

and ∠4 = ∠6 [Alternate int. ∠s]

=> ∠2 = ∠5 = ∠6

Now, ∠2 = ∠ 4

=> (2x + 30)° = (x + 2y)°

=> 2x-x-2y+30 = 0

=> x - 2y + 30 = 0 __________(1)

Also, ∠4 = ∠6

=> (x + 2y)° = (3y + 10)°

=> y - y - 10 = 0 _________(2)

Subtracting Eq.(1) from Eq.(2), we get

(x-2y+30)-(x-y-10) = 0

=> - y + 40 = 0 => y = 40

Substituting y = 40 in Eq.(2), we get

x - 40 - 10 = 0 => 50

∴ ∠4 = (x-2y)°= (50+2×40)°=130°

But, ∠4 + ∠5 = 180° [Co-interior ∠s]

=> 130° + ∠5 = 180°

=> ∠5 = 180° - 130°

= 50°

Answer:-

Hence, the measure of ∠5 is 50°.

:)