Three angles of a quadrilateral are in the ratio $2 : 6 : 4$ . The difference of the least and the greatest of angles out of these angle is fourth angle. Find all the angles of the quadrilateral.

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Solution

$\begin{matrix} A s w e k n o w, \Rightarrow S u m o f a l l a n g l e s o f q u a d r i a l t e r a l i s 360^{\circ} \Rightarrow 2 x + 6 x + 4 x + P = 360^{\circ} \Rightarrow 12 x + P = 360^{\circ} \to (i) l e t l e a s t a n g l e b e 2 x a n d g r e a t e s t a n g l e b e 6 x, T h e i r d i f f e r e n c e (- 2 x + 6 x) = P, 4 x = P \Rightarrow f r o m e q u a t i o n, 12 x + 4 x = 360^{\circ}, 16 x = 360^{\circ}, x = {22.5}^{\circ} \Rightarrow∴ ∠ A = 2 x = 2 \times 22.55 = 45^{\circ}, ∠ B = 6 x = 135^{\circ}, ∠ C = 90^{\circ}, ∠ D = 90^{\circ} . \end{matrix}$

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