The angles of a quadrilateral are in A.P. whose common difference… - Vidyadip

Question

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.

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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°.

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