In the given figure, $AB$ is a diameter of a circle with centre $O$ . If $ADE$ and $CBE$ are straight lines, meeting a $E$ such that $∠ BAD = 35^{o}$ and $∠ BED = 25^{o}$ . Then, which of the following is the measure of $∠ DCB$ , $∠ DBC$ , $∠ BDC$ respectively?

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55^{o}, 100^{o} 35^{o}

25^{o}, 120^{o} 35^{o}

35^{o}, 115^{o} 30^{o}

65^{o}, 135^{o} 255^{o}

answer is C.

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Detailed Solution

Consider the given question,
In the given figure, AB is a diameter of circle with centre O. If ADE and CBE are straight lines, meeting at E such that BAD = 35^o and BED = 25^o

In the given figure, AB is a diameter of circle with centre O. If ADE and CBE are straight lines, meeting at E such that BAD = 35^o and BED = 25^o

Here,
O is the centre of a circle with diameter AB.

AD BC

are chords which have been extended to intersect at E.

AB CD

are joined.

∠ BAD = 35^{o} ∠ BED = 25^{o} .

Now,

∠ BAD = ∠ DCB = 35^{o}

[since both of them have been subtended by the chord BD to the circumference at A and C]
Since,

∠ ACB

is an angle in a semicircle

∠ ACB = 90^{o}

∴ ∠ ACD = ∠ ACB - ∠ DCB

∠ ACD = 90^{o} - 35^{o}

∠ ACD = 55^{o}

Again,

∠ ABD = ∠ ACD = 55^{o}

[since both of them have been subtended by the chord AD to the circumference at B and C]
Now,
In triangle ACE,
By angle sum property,

∠ CAD + ∠ ACB + ∠ BEA = 180^{o}

∠ CAD = 180^{o} - (∠ ACB + ∠ BEA)

∠ CAD = 180^{o} - (90^{o} + 25^{o})

∠ CAD = 65^{o}

∴ ∠ BAC = ∠ CAD - ∠ BAD

∠ BAC = 65^{o} - 35^{o}

∠ BAC = 30^{o}

∠ BDC = ∠ BAC = 30^{o}

[since both of them have been subtended by the chord BC to the circumference at A and D]
Again,
ACBD is a cyclic quadrilateral,
We know that,
In a cyclic quadrilateral sum of opposite sides are supplementary.
Then,