In the fig abcd is a cyclic quadrilateral. if angle bcd=110degree and angleabd =60degree ,find the measure of angle dae and angle adb

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angle ADB =120°

angle DAE=70°

angle bac+angle bcd =180as sum of two opposite angles of a quadrilateral is 180 degrees

50+angle bcd =180

angle bcd =130

answer:

BCD= 70 because 50+60=110 180_110=70

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Step-by-step explanation:

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Step-by-step explanation:

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In ABCD is cyclic quadrilateral. So ∠ABC + z = 180°

At B, ∠ABC = 180 - 60° = 120°

=> z = 60°

As Arc ABC makes ∠AOC at O and ∠ z at D on the circle,

∠AOC = 2 z = 120°

In ΔBCQ, ∠BCQ = 180° - y - 60° = 120° - y

Hence, at C, ∠BCD = 180° - ∠BCQ = 60° + y

In ΔAPB, ∠PBA = 180° - ∠ABC = 60°

∠PAB = 180° - x - 60° = 120° - 2 y

Hence, ∠DAB = 180 - ∠PAB = 60° + 2y

In the cyclic quadrilateral ABCD, ∠DAB + ∠BCD = 120° + 3 y = 180°

So, y = 20° & x = 40°

Solution :

Given That ∠BAC = 50° and ∠DBC  = 60°

In the given fig. we see that ∠BAC and ∠BDC are in the same segment

Now, In triangle BDC ,

∠BDC + ∠BCD +∠DBC = 108°(By angle sum property )

∠BAC +∠BCD + ∠DBC = 180° (  ∵∠BDC = ∠BAC = 50°) ( angle in the same segment are equal)

⇔50° +60° +∠BCD = 180°

⇔∠BCD = 180° - 110° = 70°

Therefore ∠BCD = 70°

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