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The angles of a quadrilateral are in AP whose common difference is $10^\circ$. Find the smallest angle of the quadrilateral.

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Answer:

$75^\circ$

Step by Step Explanation:

1. The angles of the quadrilateral are in AP with the common difference of $10^\circ$.
   
   Let the angles be $x, x + 10^\circ, x + 20^\circ,$ and $x + 30^\circ$.
2. We know that the sum of all angles of a quadrilateral is $360^\circ$. Thus $$\begin{aligned} & x + x + 10^\circ + x + 20^\circ + x + 30^\circ = 360^\circ \\ \implies & 4 x + 60^\circ = 360^\circ \\ \implies & 4 x = 360^\circ - 60^\circ \\ \implies & 4 x = 300^\circ \\ \implies & x = \dfrac { 300^\circ } { 4 } \\ \implies & x = 75^\circ \end{aligned}$$
3. Let us now substitute the value of $x$ to get the four angles. $$\begin{aligned} & x = 75^\circ \\ & x + 10^\circ = 75^\circ + 10^\circ = 85^\circ \\ & x + 20^\circ = 75^\circ + 20^\circ = 95^\circ \\ & x + 30^\circ = 75^\circ + 30^\circ = 105^\circ \end{aligned}$$
4. Hence, the smallest angle of the quadrilateral is $75^\circ$ .

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