Question

in the given figure abcd is a cyclic quadrilateral with a center o and diameter ad. if bcd=130°.find giving reasons
∠dab, ∠adb, ∠aeb

Answer 1

$here \: is \: mate$

GIVEN:::::::

ABCD IS A CYCLIC QUADRILATERAL

ANGLE BCD = 130 DEGREE

TO FIND :::::::

THE ABOVE GIVEN ANGLES

FORMULA USED :::::

SUM OF OPPOSITE ANGLES OF A CYCLIC QUAD.

HOW TO FIND ::::"

GIVEN ANGLE BCD = 130

THEN,

ANGLE DAB = 180 - BCD = 180 - 130 = 50

( SUM OF OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL IS 180)

NOW,

ANGLE ABD = 90 DEGREE

(ANGLE SUBTENDED BY THE DIAMETER ON THE CIRCLE IS 90)

THEN,

ANGLE ADB = 180-(50 + 90)

ADB = 180 - 140 = 40 DEGREE

(ANGLE SUM PROPERTY)

AGAIN,

ANGLE BDA = ANGLE BEA

(ANGLES SUBTENDED BY THE SAME CHORD ON THE CIRCLE ARE EQUAL)

SO,

ANGLE AEB = 40 DEGREE

HOPE THIS HELPS YOU......❤...PLZ MARK AS THE BRAINLIEST........

Answer 2

Answer:

ABCD is a cyclic quadrilateral, thus:

/_C + /_A = 180° [Sum of opposite angles of a cyclic quadrilateral is 180°]

=> /_A = 180 - 130

=> /_A = 50°

/_BDA = /_BEA [Angles on the circle by the same chord]

In triangle BDA:

/_B + /_BDA + /_A = 180° [Angle Sum Property]

=> /_BDA = 180 - 90 - 50

=> /_BDA = 40°

Thus:

/_BDA = /_BEA = 40°

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