Question
The angles of a quadrilateral are in A.P. whose common difference is 10º. Find the numbers.
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Answer
A quadrilateral has four angles. Given, four angles are in A.P. with common difference 10.
Let the four angles be, a - 3d, a - d, a + d, a + 3d with common difference = 2d.
2d = 10
$\text{d}=\frac{10}{2}=5$
In a quadrilateral, sum of all angles = 360º
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 360
4a = 360
$\text{a}=\frac{\text{360}}{4}=90^\circ$
$\therefore$ The angles are a - 3d, a - d, a + d, a + 3d with a = 90, d = 5
i.e. 90 - 3(5), 90 - 5, 90 + 3(5)
⇒ 75º, 85º, 95º, 105º.
Let the four angles be, a - 3d, a - d, a + d, a + 3d with common difference = 2d.
2d = 10
$\text{d}=\frac{10}{2}=5$
In a quadrilateral, sum of all angles = 360º
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 360
4a = 360
$\text{a}=\frac{\text{360}}{4}=90^\circ$
$\therefore$ The angles are a - 3d, a - d, a + d, a + 3d with a = 90, d = 5
i.e. 90 - 3(5), 90 - 5, 90 + 3(5)
⇒ 75º, 85º, 95º, 105º.
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