Question
The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.
✓
Answer
Let the angle be
$\text{a}-3\text{d},\ \text{a}-\text{d},\ \text{a}+3\text{d}$
Then,
Sum of all angles $=360^\circ$
$\text{a}-3\text{d}+\text{a}-\text{d}+\text{a}+\text{d}+\text{a}+3\text{d}=360^\circ$
$4\text{a}=360^\circ$
$\text{a}=90^\circ\ .....{ (1)}$
and
$(\text{a}-\text{d})-(\text{a}-3\text{d})=10$
$2\text{d}=10$
$\text{d}=5$
$\therefore$ The angle of the given quadrilateral are 75°, 85°, 95°, and 105°.
$\text{a}-3\text{d},\ \text{a}-\text{d},\ \text{a}+3\text{d}$
Then,
Sum of all angles $=360^\circ$
$\text{a}-3\text{d}+\text{a}-\text{d}+\text{a}+\text{d}+\text{a}+3\text{d}=360^\circ$
$4\text{a}=360^\circ$
$\text{a}=90^\circ\ .....{ (1)}$
and
$(\text{a}-\text{d})-(\text{a}-3\text{d})=10$
$2\text{d}=10$
$\text{d}=5$
$\therefore$ The angle of the given quadrilateral are 75°, 85°, 95°, and 105°.
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