Question

The exterior angles of a quadrilateral are y°, (2y+5)°,(y+15)° and
(3y-10)°, find y°

Answer 1

y+(2y+5)+(y+15)(3y-10)=360

7y +5+15-10=360

7y +10=360

7y=360-10

y=350/7

y=50°.

Answer 2

The correct answer is y= 50°.

Given:

The exterior angles of a quadrilateral are y°, (2y+5)°,(y+15)°, and

(3y-10)°.

To Find:

The value of y.

Solution:

A quadrilateral is a polygon comprising four sides, four vertices, and four angles.

An exterior angle of a quadrilateral is the angle between one side of the line extending from the adjacent side of the quadrilateral.

The sum of exterior angles of a quadrilateral is 360°.

We are given that the exterior angles of a quadrilateral are y°, (2y+5)°,(y+15)°, and (3y-10)°.

The sum of these (exterior) angles = 360°.

⇒ y + (2y+5)+(y+15)+(3y-10) = 360

Rearranging the terms, we have:

(y+2y+y+3y) + (5+15-10) = 360

⇒ 7y = 350

⇒ y = 50°.

Hence y = 50°.

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