Question

find x+y for the figure given below *​

Answer 1

$\huge \colorbox{red}{given}$

==》AB || CD , <APQ = 50⁰ and <PRD = 127⁰

$\huge \colorbox{cyan}{to \: find = x + y}$

$\huge \fcolorbox{green}{re}{solution}$

==》We have AB || CD and PQ is the transversal.

==》Therefore , X = 50⁰ [Alternate interior angles are equal ]

In Triangle PQR

▪︎<PRD IS an exterior angle

▪︎<PRD= X + Y [ Exterior angles is equal to the sum of its interior opposite angles ]

$127 ^{0} = 50 ^{0} + y \\ \\ y = 127 - 50 \\ \\ y = 77 ^{0}$

▪︎x+y= 50+77= 127°

Hence,

▪︎ x+y= 127°

Note

==》An exterior angle of triangle is greater than either of its interior opposite angle .

.==》Theorem 1 says =》The sum of interior angles of a triangle is 180⁰ .

==》Theorem 2 says =》If a side of a triangle is produced ,then the exterior angles so formed is equal to the sum of the two interior opposite angles

Answer 2

$\huge\mathfrak{\red{Answer}}$

127°

Step-by-step explanation:

From the figure;

x= 50° [alternate interior angles]

As 127° forms the exterior angle for x and y

Therefore

x+y = 127 [by exterior angle property]

___________________________________

Let's check:-

x= 50

therefore

x+y = 127

50+y= 127

y= 127-50

=> y = 77°

___________________________

x+y= 50+77= 127°

Hence, x+y= 127°