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Two angles of a quadrilateral are $3x\ -\ 4$ each and two are $3x\ +\ 10$ each. Find all four angles of the quadrilateral.
Given: Two angles of a quadrilateral are $3x\ -\ 4$ each and two are $3x\ +\ 10$ each.
To find: We have to find all four angles of the quadrilateral.
Solution:
$2(3x\ -\ 4)\ +\ 2(3x\ +\ 10)\ =\ 360°$
=> $6x\ -\ 8\ +\ 6x\ +\ 20\ =\ 360°$
=> $12x\ +\ 12\ =\ 360°$
=> $x\ +\ 1\ =\ \frac{360}{12}$
=> $x\ +\ 1\ =\ 30$
=> $x\ =\ 30\ -\ 1\ =\ 29°$
Now,
Two angles = $3x\ -\ 4$
Two angles = $3(29)\ -\ 4$
Two angles = $87\ -\ 4$
Two angles = $83°$
Remaining two angles = $3x\ +\ 10$
Remaining two angles = $3(29)\ +\ 10$
Remaining two angles = $87\ +\ 10$
Remaining two angles = $97°$
So, the four angles of the quadrilateral are 83°, 83°, 97°, and 97°.
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