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In the given figure, O is the centre of the circle and ∠BCO = 30°. Find x and y. |
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Answer» From the figure we know that ∠AOD and ∠OEC form right angles So we get ∠AOD = ∠OEC = 90o We know that OD || BC and OC is a transversal From the figure we know that ∠AOD and ∠OEC are corresponding angles ∠AOD = ∠OEC We know that ∠DOC and ∠OCE are alternate angles ∠DOC = ∠OCE = 30o Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference. So we get ∠DOC = 2 ∠DBC It can be written as ∠DBC = ½ ∠DOC By substituting the values ∠DBC = 30/2 By division y = ∠DBC = 15o In the same way ∠ABD = ½ ∠AOD By substituting the values ∠ABD = 90/2 By division ∠ABD = 45o We know that ∠ABE = ∠ABC = ∠ABD + ∠DBC So we get ∠ABE = ∠ABC = 45o + 15o By addition ∠ABE = ∠ABC = 60o Consider △ ABE Using the angle sum property ∠BAE + ∠AEB + ∠ABE = 180o By substituting the values x + 90o + 60o = 180o On further calculation x = 180o – 90o – 60o By subtraction x = 180o – 150o So we get x = 30o Therefore, the value of x is 30o and y is 15o. |
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