Question

THREE ANGLES OF A QUADRILATERAL ARE IN THE RATIO 3 : 5 : 7. THE DIFFERENCE OF THE LEAST AND THE GREATEST OF THESE ANGLES IS 76 DEGREE. FIND ALL FOUR ANGLES OF THE OUADRILATERAL​

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

$\textbf{\underline{Common\;Ratio\;be\;p}}$

Hence,

$\textbf{\underline{Ratio\;are\;3p,5p\;and\;7p}}$

Now,

$\textbf{\underline{Largest\;angle=7p}}$

Also

$\textbf{\underline{Smallest\;angle=5p}}$

It is given that difference of the least and greatest of these angles is 76

${\textbf{\boxed{\bigstar{{Hence\;Equation}}}}}$

7p - 3p = 76

4p = 76

$\tt{\rightarrow p=\dfrac{76}{4}}$

p = 19

$\textbf{\underline{Hence\;Three\;angles\;are:-}}$

3p = 3(19) = 57

5p = 5(19) = 95

7p = 7(19) = 133

${\boxed{\sf\:{Combined\;sum\;of\;angles\;of\;quadrilateral=360}}}$

${\boxed{\sf\:{So\;Fourth\;Angle}}}$

= 360 - ( 57 + 95 + 133 )

= 360 - ( 285 )

= 75

$\textbf{\underline{4\;angles\;are:-}}$

$\Large{\boxed{\sf\:{57, 75 ,95,133}}}$

Answer 2

Let take ratio as 3x 5x 7x
3x+5x+7x+75 = 360
15x = 360 - 75
X = 285/15
X = 19
3x= 57
5x= 95
7x= 113

The angles of a quadrilateral is 19 , 57 , 95 , 113