a circle has radius root 2 cm . it is divided into 2 segments by a chord of 2cm. prove that the angle subtended by d chord at a point in major segment is 45.
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Let us solve this only using Simple principles of circle and triangles without going into trigonometry. As this qn is probably from those not knowing trigonometry.
see diagram.
Given OA = OB = √2 cm. Let C be midpoint of chord AB. AC = CB = 1 cm.
So OC is the perpendicular bisector of chord AB
So ΔOCA & OCB are right angle triangles. They are congruent as all the corresponding sides are equal.
OC = √(OA² - AC²) = √(2 - 1) = 1 cm
So ΔOCA & ΔOCB are isosceles triangles, as OC = AC = BC.
∠AOC = ∠BOC = (180-90)/2 = 45°
∠AOB = 90°
In a circle, a chord subtends an angle at the circle (at P) equal to half of that it subtends at the center O. Hence, it subtends 45° at the major segment.
see diagram.
Given OA = OB = √2 cm. Let C be midpoint of chord AB. AC = CB = 1 cm.
So OC is the perpendicular bisector of chord AB
So ΔOCA & OCB are right angle triangles. They are congruent as all the corresponding sides are equal.
OC = √(OA² - AC²) = √(2 - 1) = 1 cm
So ΔOCA & ΔOCB are isosceles triangles, as OC = AC = BC.
∠AOC = ∠BOC = (180-90)/2 = 45°
∠AOB = 90°
In a circle, a chord subtends an angle at the circle (at P) equal to half of that it subtends at the center O. Hence, it subtends 45° at the major segment.
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