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What will be the value of ∠CAB, if in the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB?

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30 ∘

45 ∘

60 ∘

90 ∘

answer is A.

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Detailed Solution

It is given that O is the centre of the circle, BD = OD and CD ⊥ AB.
Question Image

We have to find the value of ∠CAB.

△ O D B

is an equilateral triangle.
In

△ O D B

, BD=OD (Given)….(1)
OD=OB [Radii of the same circle] …… (2)
So, OB=OD=BD [From (1) and (2)]
Hence,

△ O D B

is an equilateral triangle.

∴ ∠ B O D = ∠ O B D = ∠ O D B = 600

In ∆EBC and ∆EBD,
EB=EB [common side]

∠ C E B = ∠ D E B

[Each

90 ∘

]
CE=DE [In a circle, any perpendicular drawn on a chord bisects the chord]
Applying the test of congruency in triangle EBC and angle EBD because they are congruent to each other, i.e.,∆EBC≅∆EBD [By SAS congruence]

∴ ∠ E B C ≅ ∠ E B D

[By C.P.C.T]

⇒ ∠ E B C = ∠ O B D = 60 ∘ ∠ A C B = 90 ∘

[Since,

∠ O B D = 60 ∘

]
And,

∠ A C B = 90 ∘

[Angle in a semi-circle is a right angle]
Applying the sum of all angles of a triangle rule,
In

△ A C B

∠ C A B + ∠ C B A + ∠ A C B = 1800

[Sum of all angle of a triangle is

]

⇒ ∠ C A B + 600 + 900 = 1800 ⇒ ∠ C A B = 300

So, the value of

∠ C A B = 300

.
Therefore, option 1 is correct.

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