A circle with centre O and diameter PR is inscribed in quadrilateral A
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19 Jul 2023, 00:07
In triangle OQR, OQ = OR. (Both OQ and OR are radii of given circle)
So, angle opposite to these two sides would also be equal.
So, we can say, in triangle OQR.
Angle OQR = Angle ORQ
Hence angle OQR = 34°
Now consider quadrilateral ABCD, Side AD of quadrilateral is tangent to the circle. We
know, the angle between the radius of a circle and tangent is 90°
So, we can say angle OQD = 90°
From the given figure, we can say angle OQD = angle OQR + angle RQD
Hence 90°= 𝑥° + 34°
Hence 𝑥°= 56°
Ans. (56)