Question

Heya !!

the question is here """"""""""

Answer 1

Hey !!!

Solution :-

1st of all let's consider point D orginated and cuts at point P and Q

given :- 4y = 5x

and , z = y - 30°

•°• 4y = 5x

4y/5 = x

but y = <APQ [ from corresponding angle ]

<APQ + <BPQ = 180 { linear pair]

y+ 4y/5 = 180°

9y / 5 = 180°

9y = 180° × 5

y = 900° / 9 = 100°

•°• Y = 100°

from given ,

z = y - 30°

z = 100° - 30°

z =70°

Again , i saw EF is linear pair

from linear pair ,

z + q = 180°

70°+q = 180°

<q = 110° Answer ✔

_______________________

Hope it helps you !!!

@Rajukumar111

Answer 2

$\mathfrak{ \large{HELLO \: \: DEAR,}}$

$\mathit{GIVEN \: \: THAT:- } \\ \\ \mathit{4Y = 5X , Z = Y - 30,} \\ \\ \mathit{X = 4Y/5 , Z = Y - 30} \\ \\ \\ \mathit{NOW,} \\ \\ \mathit{\angle Y = \angle Y_1 \: \: [ CORRESPONDING \: \: ANGELES]} \\ \\ \mathit{\angle Y_1 + \angle x = 180\degree \: \: [AB \: \: IS \: \: A \: \: STRAIGHT \: \: LINE]} \\ \\ \mathit{\angle Y + 4\angle y/5 = 180\degree} \\ \\ \mathit{\angle (5\angle Y + 4\angle Y)/5 = 180\degree} \\ \\ \mathit{9\angle y = 900\degree} \\ \\ \mathit{\angle Y = 900\degree/9\degree} \\ \\ \mathit{\angle Y = 100\degree}$



$\\ \\ \mathit{\angle Z + \angle q = 180\degree [ EF \: \: IS \: \: A \: \: STRAIGHT \: \: LINE]} \\ \\ \mathit{(\angle Y - 30\degree) + \angle q = 180\degree} \\ \mathit{\to\to\to\to\to\to\to \therefore \boxed{\angle Z = \angle Y - 30 \: \: AND, \: \: \angle Y = 100\degree}} \\ \\ \mathit{ 100\degree - 30\degree + \angle q = 180\degree} \\ \\ \mathit{\angle q = 180\degree - 100\degree + 30\degree} \\ \\ \mathit{\angle q = 80\degree + 30\degree} \\ \\ \mathit{\angle q = 110\degree}$


$\mathit{\underline{ I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}$